McGILL UNIVERSITY FACULTY OF SCIENCE DEPARTMENT OF MATHEMATICS AND STATISTICS MATH · PDF fileMcGILL UNIVERSITY FACULTY OF SCIENCE DEPARTMENT OF MATHEMATICS AND STATISTICS MATH 140

McGILL UNIVERSITY

FACULTY OF SCIENCE

DEPARTMENT OFMATHEMATICS AND STATISTICS

MATH 140 2006 09

CALCULUS 1

Information for Students(Fall Term, 2006/2007)

Pages 1 - 23 of these notes may be considered theCourse Outline for this course.

W. G. Brown

December 1, 2006

Information for Students in MATH 140 2006 09

Contents

1 General Information 11.1 Instructors and Times . . . . 11.2 Course Description . . . . . . 2

1.2.1 Calendar Description ofMATH 141 . . . . . . 2

1.2.2 Calendar Descriptionsof other beginning cal-culus courses . . . . . 2

1.2.3 How much of my pre-vious calculus course doI need to remember? . 3

1.2.4 Late transfer from MATH150 . . . . . . . . . . . 3

1.3 Tutorials; Tutors’ Coordinates 31.4 Evaluation of Your Progress . 5

1.4.1 Your final grade . . . 51.4.2 WeBWorK . . . . . 61.4.3 Written Assignments . 71.4.4 Quizzes at the Tutori-

als; Submission of Writ-ten Assignments at theQuizzes . . . . . . . . 7

1.4.5 Final Examination . . 91.4.6 Supplemental Assess-

ments . . . . . . . . . 91.4.7 Machine Scoring . . . 101.4.8 Plagiarism . . . . . . . 101.4.9 Keep all your graded

materials! . . . . . . . 111.5 Published Materials . . . . . 11

1.5.1 Required Text-Book . 111.5.2 Optional Reference Books 111.5.3 Recommended Video Ma-

terials . . . . . . . . . 131.5.4 Other Calculus Text-

books . . . . . . . . . 141.5.5 Website . . . . . . . . 15

1.6 Syllabus . . . . . . . . . . . . 161.7 Preparation and Workload . . 17

1.7.1 Prerequisites. . . . . . 17

1.7.2 Calculators . . . . . . 181.7.3 Self-Supervision . . . . 181.7.4 Escape Routes . . . . 191.7.5 Terminology . . . . . 19

1.8 High Technology and MATH140 . . . . . . . . . . . . . . . 201.8.1 Keep your e-mail ad-

dresses up to date . . 201.8.2 Use of Calculators and

Computer Algebra Sys-tems . . . . . . . . . . 21

1.8.3 Use of the Internet . . 211.9 Which problems should I work? 21

1.9.1 There are no public orprivate restricted listsof textbook problemsto which you can con-fine your preparationfor testing in MATH 140. 21

1.9.2 Are the type of prob-lems I find on WeB-WorK problems indica-tive of what I need toknow in this course? . 22

1.9.3 Problems discussed inthe lectures . . . . . . 22

1.9.4 And anyhow, the courseis not concerned onlywith problem solving. 22

1.9.5 Repairing your precal-culus foundations. . . 22

1.10 Last Weeks of Term (to beupdated) . . . . . . . . . . . . 22

2 About Quiz Q0 242.1 All students must write the

diagnostic quiz. . . . . . . . . 242.2 Late registration and medical

absences. . . . . . . . . . . . 242.3 Type of Quiz. . . . . . . . . . 242.4 Preparing for the quiz. . . . . 24


2.5 What to do if your grade isvery low. . . . . . . . . . . . . 24

3 W1, First Written Assignment 253.1 Certificate . . . . . . . . . . . 253.2 The assignment question . . . 25

4 W2, Second Written Assignment 264.1 Certificate . . . . . . . . . . . 264.2 The assignment questions . . 26

5 W3, Third Written Assignment 295.1 Certificate . . . . . . . . . . . 295.2 The assignment questions . . 29

6 Draft Solutions to W1, First Writ-ten Assignment 316.1 Certificate . . . . . . . . . . . 316.2 The assignment question . . . 316.3 Solutions . . . . . . . . . . . 32

7 Draft Solutions to Quiz Q1 387.1 Instructions to students . . . 387.2 Monday version . . . . . . . . 387.3 Wednesday version . . . . . . 407.4 Thursday version . . . . . . . 437.5 Friday version . . . . . . . . . 44

8 W4, Fourth Written Assignment 478.1 Certificate . . . . . . . . . . . 478.2 The assignment question . . . 47

9 Draft Solutions to W2, SecondWritten Assignment 499.1 Certificate . . . . . . . . . . . 499.2 The assignment questions . . 499.3 Solutions . . . . . . . . . . . 52

10 Draft Solutions to W3, Third Writ-ten Assignment 5710.1 Certificate . . . . . . . . . . . 5710.2 The assignment questions . . 5710.3 Solutions . . . . . . . . . . . 58

11 Draft Solutions to Quiz Q2 6211.1 Instructions to students . . . 62

11.1.1 Monday version . . . . 6211.1.2 Wednesday version . . 6411.1.3 Thursday version . . . 6611.1.4 Friday version . . . . . 67

12 Draft Solutions to Quiz Q3 7012.1 Instructions to students . . . 70

12.1.1 Monday version . . . . 7012.1.2 Wednesday version . . 7212.1.3 Thursday version . . . 7412.1.4 Friday version . . . . . 75

13 Draft Solutions to W4, Fourth Writ-ten Assignment 7813.1 Certificate . . . . . . . . . . . 7813.2 The assignment question . . . 78

14 Draft Solutions to Quiz Q4 8114.1 Instructions to students . . . 8114.2 Monday version . . . . . . . . 8114.3 Wednesday version . . . . . . 8414.4 Thursday version . . . . . . . 8614.5 Friday version . . . . . . . . . 88

A Information Specifically for Stu-dents in Lecture Section 001 1001A.1 Timetable for Lecture Section

001 of MATH 140 2006 09 . . 1001

B Supplementary Notes for Lecturesin Lecture Section 001 1003

C Information Specifically for Stu-dents in Lecture Section 002 2001C.1 Timetable for Lecture Section

002 of MATH 140 2006 09 . . 2001

D Notes for Lecture Section 002 2003D.1 Lecture style in Lecture Sec-

tion 002 . . . . . . . . . . . . 2003


D.2 Supplementary Notes for theLecture of September 6th, 2006 2004D.2.1 §1.1 Four Ways to Rep-

resent a Function. . . 2005D.3 Supplementary Notes for the

Lecture of September 11th, 2006 2009D.3.1 §1.1 Four Ways to Rep-

resent a Function (con-clusion). . . . . . . . . 2009

D.3.2 §1.2 Mathematical Mod-els: A Catalog of Es-sential Functions. . . . 2012

D.4 Supplementary Notes for theLecture of September 13th, 2006 2025D.4.1 §1.2 Mathematical Mod-

els: A Catalog of Es-sential Functions (con-clusion). . . . . . . . . 2025

D.4.2 Remedying deficienciesin your precalculus back-ground . . . . . . . . . 2030

D.5 Supplementary Notes for theLecture of September 18th, 2006 2031D.5.1 §1.3 New Functions from

Old Functions. . . . . 2031D.5.2 §1.4 Graphing Calcu-

lators and Computers 2032D.5.3 §1.5 Exponential Func-

tions . . . . . . . . . . 2033D.5.4 §1.6 Inverse Functions

and Logarithms . . . . 2034D.6 Supplementary Notes for the

Lecture of September 20th, 2006 2036D.6.1 §1.6 Inverse Functions

and Logarithms (con-tinued) . . . . . . . . 2036

D.6.2 §2.1 The Tangent andVelocity Problems. . . 2040

D.6.3 §2.2 The Limit of a Func-tion. . . . . . . . . . . 2041

D.6.4 §2.3 Calculating Lim-its Using the Limit Laws. 2043

D.7 Supplementary Notes for theLecture of September 25th, 2006 2046D.7.1 §1.6 Inverse Functions

and Logarithms (con-clusion) . . . . . . . . 2046

D.7.2 §2.5 Continuity. . . . . 2058D.8 Supplementary Notes for the

Lecture of September 27th, 2006 2059D.8.1 §2.4 The Precise Defi-

nition of a Limit (notexamination material in2006-07). . . . . . . . 2059

D.8.2 §2.5 Continuity (con-tinued) . . . . . . . . 2062

D.8.3 §2.6 Limits at Infinity;Horizontal Asymptotes. 2070

D.9 Supplementary Notes for theLecture of October 4th, 2006 2076D.9.1 §2.6 Limits at Infinity;

Horizontal Asymptotes(continued). . . . . . . 2076

D.9.2 §2.7 Tangents, Veloci-ties, and Other Ratesof Change. . . . . . . 2077

D.9.3 §2.8 Derivatives. . . . 2080D.9.4 §2.9 The Derivative as

a Function . . . . . . 2083D.9.5 2 Review . . . . . . . 2087D.9.6 Appendix A. Numbers,

Inequalities, and Ab-solute Values . . . . . 2090

D.9.7 Appendix B. Coordi-nate Geometry and Lines 2091

D.9.8 Appendix C. Graphs ofSecond-Degree Equations 2091

D.9.9 Appendix D. Trigonom-etry . . . . . . . . . . 2091

D.10 Supplementary Notes for theLecture of October 10th, 2006 2092D.10.1 §3.1 Derivatives of Poly-

nomials and Exponen-tial Functions . . . . . 2094


D.10.2 §3.2 The Product andQuotient Rules . . . . 2099

D.11 Supplementary Notes for theLecture of October 11th, 2006 2104D.11.1 §3.3 Rates of Change

in the Natural and So-cial Sciences. . . . . . 2104

D.11.2 §3.4 Derivatives of Trigono-metric Functions. . . . 2105

D.12 Supplementary Notes for theLecture of October 16th, 2006 2111D.12.1 §3.5 The Chain Rule. 2111D.12.2 §3.6 Implicit Differen-

tiation. . . . . . . . . 2117D.13 Supplementary Notes for the

Lecture of October 18th, 2006 2120D.13.1 §3.5 The Chain Rule

(conclusion) . . . . . . 2120D.13.2 §3.6 Implicit Differen-

tiation (conclusion) . . 2121D.14 Supplementary Notes for the

Lecture of October 23rd, 2006 2130D.14.1 §3.7 Higher Derivatives 2131D.14.2 §3.8 Derivatives of Log-

arithmic Functions . . 2138D.15 Supplementary Notes for the

Lecture of October 25th, 2006 2145D.15.1 §3.9 Hyperbolic Func-

tions . . . . . . . . . . 2145D.15.2 §3.10 Related Rates. . 2150

D.16 Supplementary Notes for theLecture of October 30th, 2006 2152D.16.1 §3.10 Related Rates (con-

clusion) . . . . . . . . 2152D.16.2 §3.11 Linear Approxi-

mations and Differentials 2157D.16.3 3 Review . . . . . . . 2161

D.17 Supplementary Notes for theLecture of November 1st, 2006 2166D.17.1 §4.1 Maximum and Min-

imum Values . . . . . 2166

D.18 Supplementary Notes for theLecture of November 6th, 2006 2174D.18.1 §4.2 The Mean Value

Theorem . . . . . . . 2174D.19 Supplementary Notes for the

Lecture of November 8th, 2006 2183D.19.1 §4.3 How Derivatives

Affect the Shape of aGraph . . . . . . . . . 2183

D.20 Supplementary Notes for theLecture of Monday, Novem-ber 13th, 2006 . . . . . . . . . 2190D.20.1 §4.4 Indeterminate Forms

and L’Hospital’s Rule 2190D.21 Supplementary Notes for the

Lecture of Wednesday, Novem-ber 15th, 2006 . . . . . . . . . 2200D.21.1 §4.5 Summary of Curve

Sketching . . . . . . . 2200D.21.2 §4.6 Graphing with Cal-

culus and Calculators 2212D.22 Supplementary Notes for the

Lecture of Monday, Novem-ber 20th, 2006 . . . . . . . . . 2214D.22.1 Sketch of Solutions to

Problems on the De-cember, 2003 Final Ex-amination . . . . . . . 2214

D.22.2 Sketch of Solutions toProblems on one of sev-eral versions of the De-cember, 2004 Final Ex-amination . . . . . . . 2224

D.22.3 §4.5 Summary of CurveSketching (conclusion) 2233

D.22.4 §4.7 Optimization Prob-lems . . . . . . . . . . 2236

D.22.5 §4.8 Applications to Busi-ness and Economics –OMIT . . . . . . . . . 2241

D.22.6 §4.9 Newton’s Method– OMIT . . . . . . . . 2241


D.23 Supplementary Notes for theLecture of November 22nd, 2006 2242D.23.1 §4.10 Antiderivatives . 2242

D.24 Supplementary Notes for theLecture of November 27th, 2006 2254D.24.1 Sketch of Solutions to

Problems on One Ver-sion of the December,2005 Final Examination 2254

D.25 Supplementary Notes for theLecture of November 29th, 2006 2262D.25.1 Sketch of Solutions to

Problems on One Ver-sion of the December,2005 Final Examina-tion (conclusion) . . . 2262

E Assignments from Previous Years 3001E.1 Fall 1998 Problem Assignments 3001E.2 Fall 1999 Problem Assignments 3001

E.2.1 First Fall 1999 Prob-lem Assignment, withSolutions . . . . . . . 3001

E.2.2 Second Fall 1999 Prob-lem Assignment . . . . 3011

E.2.3 Third Fall 1999 Prob-lem Assignment, withSolutions . . . . . . . 3012

E.2.4 Fourth Fall 1999 Prob-lem Assignment . . . . 3020

E.2.5 Fifth Fall 1999 Prob-lem Assignment . . . . 3021

E.3 2000/2001 Problem Assignments,with Solutions . . . . . . . . . 3023E.3.1 First 2000/2001 Prob-

lem Assignment, withSolutions . . . . . . . 3023

E.3.2 Second 2000/2001 Prob-lem Assignment, withSolutions . . . . . . . 3034

E.3.3 Third 2000/2001 Prob-lem Assignment, withSolutions . . . . . . . 3041

E.3.4 Fourth 2000/2001 Prob-lem Assignment, withSolutions . . . . . . . 3050

E.3.5 Fifth 2000/2001 Prob-lem Assignment, withSolutions . . . . . . . 3056

E.3.6 Sixth 2000/2001 Prob-lem Assignment, withSolutions . . . . . . . 3061

F Some Tests and Quizzes from Pre-vious Years 3065F.1 Fall 1998 Class Quiz, with So-

lutions . . . . . . . . . . . . . 3065F.2 Last Three Tutorial Quizzes

in 2000/2001 (many versions) 3068F.2.1 Fourth 2000/2001 Tu-

torial Quizzes . . . . . 3068F.2.2 Fifth 2000/2001 Tuto-

rial Quizzes . . . . . . 3071F.2.3 Sixth 2000/2001 Tuto-

rial Quizzes . . . . . . 3074F.3 First 2005/2006 Written As-

signment W1, with Sketch ofSolutions . . . . . . . . . . . 3079F.3.1 The assignment question 3079F.3.2 Solutions . . . . . . . 3079

F.4 Second 2006/2006 Written As-signment W2, with Sketch ofSolutions . . . . . . . . . . . 3082F.4.1 The assignment ques-

tions with solutions . 3082F.4.2 Solution to the “Fun

Problem” (Student werenot asked to submit asolution.): . . . . . . . 3085

F.5 Third 2005/2006 Written As-signment W3 . . . . . . . . . 3086F.5.1 Certificate . . . . . . . 3086


F.5.2 Instructions . . . . . . 3086F.5.3 The assignment ques-

tions . . . . . . . . . . 3087F.6 Fourth 2005/2006 Written As-

signment W4 . . . . . . . . . 3087F.6.1 Certificate . . . . . . . 3087F.6.2 Instructions . . . . . . 3088F.6.3 The assignment ques-

tions . . . . . . . . . . 3088F.7 Fifth 2005/2006 Written As-

signment W5 . . . . . . . . . 3088F.7.1 Certificate . . . . . . . 3088F.7.2 Instructions . . . . . . 3089F.7.3 The assignment question 3089

G Examinations from Previous Years 3090G.1 . . . . . . . . . . . . . . . . . 3090G.2 December 1996 Final Exami-

nation in 189-122A . . . . . . 3091G.3 December 1997 Final Exami-

nation in 189-140A . . . . . . 3092G.4 December 1998 Final Exami-

nation in 189-140A . . . . . . 3095G.5 May 1999 Supplemental Ex-

amination in 189-140A . . . . 3097G.6 December 1999 Final Exami-

nation in 189-140A . . . . . . 3098G.7 December 1999 Special Final

Examination in 189-140A . . 3100G.8 December 2000 Final exami-

nation in 189-140A . . . . . . 3101G.9 May 2001 Supplemental/Deferred

Examination in 189-140A . . 3102G.10 December 2001 Final Exami-

nation in 189-140A . . . . . . 3103G.11 May 2002 Supplemental/Deferred

Examination in 189-140A . . 3105G.12 December 2002 Final Exami-

nation in MATH 140 2002 09 3107G.13 May 2003 Supplemental/Deferred

Examination in MATH 140 200209 . . . . . . . . . . . . . . . 3109

G.14 December 2003 Final Exami-nation in MATH 140 2003 09 3111

G.15 May 2004 Supplemental/DeferredExamination in MATH 140 200309 . . . . . . . . . . . . . . . 3116

G.16 December 2004 Final Exami-nation in MATH 140 2004 09(One of several versions) . . . 3120

G.17 May 2005 Supplemental/DeferredExamination in MATH 140 200409 . . . . . . . . . . . . . . . 3125

G.18 December, 2005, Final Exam-ination in MATH 140 2005 09(one version) . . . . . . . . . 3129

G.19 May, 2006, Supplemental/DeferredExamination in MATH 140 200509 . . . . . . . . . . . . . . . 3133

H WeBWorK 4001H.1 Frequently Asked Questions (FAQ) 4001

H.1.1 Where is WeBWorK? 4001H.1.2 Do I need a password

to use WeBWorK? . 4001H.1.3 Do I have to pay an

additional fee to use WeB-WorK? . . . . . . . . 4002

H.1.4 When will assignmentsbe available on WeB-WorK? . . . . . . . . 4002

H.1.5 Do WeBWorK assign-ments cover the full rangeof problems that I shouldbe able to solve in thiscourse? . . . . . . . . 4002

H.1.6 May I assume that thedistribution of topicson quizzes and final ex-aminations will paral-lel the distribution oftopics in the WeBWorKassignments? . . . . . 4002


H.1.7 WeBWorK providesfor different kinds of“Display Mode”. Whichshould I use? . . . . . 4003

H.1.8 WeBWorK providesfor printing assignmentsin “Portable DocumentFormat” (.pdf), “PostScript”(.ps) forms. Which shouldI use? . . . . . . . . . 4003

H.1.9 What is the relation be-tween WeBWorK andWebCT? . . . . . . . . 4003

H.1.10 Which browser shouldI use for WeBWorK? 4004

H.1.11 What do I have to doon WeBWorK? . . . 4004

H.1.12 How can I learn howto use WeBWorK? . 4005

H.1.13 Where should I go ifI have difficulties withWeBWorK ? . . . . 4005

H.1.14 Can the WeBWorKsystem ever break downor degrade? . . . . . . 4005

H.1.15 How many attempts mayI make to solve a par-ticular problem on WeB-WorK? . . . . . . . . 4006

H.1.16 Will all WeBWorK as-signments have the samelength? the same value? 4006

H.1.17 Is WeBWorK a goodindicator of examina-tion performance? . . 4006

I Contents of the DVD disks forLarson/Hostetler/Edwards 5001

J References 6001J.1 Stewart Calculus Series . . . 6001J.2 Other Calculus Textbooks . . 6002

J.2.1 R. A. Adams . . . . . 6002

J.2.2 Larson, Hostetler, et al. 6002J.2.3 Edwards and Penney . 6003J.2.4 Others, not “Early Tran-

scendentals” . . . . . . 6003J.3 Other References . . . . . . . 6003

List of Tables

1 Instructors and Times . . . . 12 Schedule and Locations of Tu-

torials, as of December 1, 2006(subject to change) . . . . . . 4

3 Tutors’ Coordinates, as of De-cember 1, 2006 . . . . . . . . 5

4 Summary of Course Require-ments, as of December 1, 2006;(all dates are subject to change) 12

5 Summary of solutions to W1,MONDAY version . . . . . . 33

6 Summary of solutions to W1,WEDNESDAY version . . . . 34

7 Summary of solutions to W1,THURSDAY version . . . . . 36

8 Summary of solutions to W1,FRIDAY version . . . . . . . 37

9 Values of the TrigonometricFunctions for simple, non-negativesubmultiples of π. . . . . . . . 2017

10 Differentiation Rules from [1,§3.1] . . . . . . . . . . . . . . 2096

11 Differentiation Rules from [1,§3.2] . . . . . . . . . . . . . . 2100

12 Some Antiderivatives . . . . . 224313 1998 Problem Assignments . 3001

List of Figures

1 Showing a discontinuity in agraph . . . . . . . . . . . . . 27

2 Showing a discontinuity in agraph . . . . . . . . . . . . . 50


3 Graph of the Function(x − a)2 − b2

(x − a)2 + b2

and its horizontal asymptote,y = 1, when a = 2, b = 5 . . . 80

4 Graph of the Function sin x . 20195 Graph of the Function cos x . 20206 Graph of the Function tanx . 20217 Graph of the Function cot x . 20228 Graph of the Function sec x . 20239 Graph of the Function csc x . 202410 Invertible restriction of the Func-

tion sin x . . . . . . . . . . . 205111 Graph of the Inverse Sine Func-

tion . . . . . . . . . . . . . . 205212 Reflection (in red) of the re-

striction of sinx in the liney = x . . . . . . . . . . . . . 2053

13 Graph of the Function f(x) =√x2 + x + 1 −

√x2 − x . . . 2090

14 Graph of the Functionx2

x2 + 3and its horizontal asymptote,y = 1 . . . . . . . . . . . . . . 2186

15 Graph of the Function cos2 x−2 sin x . . . . . . . . . . . . . 2188

16 Graph of the Function x ln x . 2189

17 Graph of f(x) =

√

x

x − 5and

its horizontal asymptote . . . 220418 Graph of y = ex − 3e−x − 4x 2207

19 Graph of f(x) =sin x

xfor x >

0 . . . . . . . . . . . . . . . . 221020 Portion of the graph of y =

sin x− tan x, showing its ver-tical asymptotes . . . . . . . 2234

Information for Students in MATH 140 2006 09 1

1 General Information

Distribution Date: This preliminary version as of December 1, 2006(all information is subject to change)

Pages 1 - 23 of these notes may be considered the Course Outline for this course.

These notes may undergo minor corrections or updates during the term:the definitive version will be the version accessible at

http://www.math.mcgill.ca/brown/math140a.html

or on WebCT, at

http://www.mcgill.ca/webct or http://webct.mcgill.ca

1.1 Instructors and Times

INSTRUCTOR: Prof. N. Sancho Prof. W. G. Brown(Course Coordinator)

LECTURE SECTION: 1 2CRN: 415 418

OFFICE: BURN 1130 BURN 1224OFFICE HOURS: MW 11:00→12:00 M 15:00→16:00

(subject to change) F 10:00→11:00or by appointment

TELEPHONE: (514)-398-3823 (514)-398–3836E-MAIL:1 SANCHO@ BROWN@

MATH.MCGILL.CA MATH.MCGILL.CACLASSROOM: ADAMS AUD ADAMS AUD

CLASS HOURS: MWF 8:35–9:25 h. MW 16:35–17:55 h.

Table 1: Instructors and Times

1Please do not send e-mail messages to your instructors through the WebCT or WeBWorK2 systems;rather, use the addresses given in Table 1.1.

2E-mail messages generated by the Feedback command in WeBWorK should be used sparingly,and confined to specific inquiries about WeBWorK assignments.


1.2 Course Description

1.2.1 Calendar Description of MATH 141

MATH 1403 CALCULUS 1. (3 credits. 3 hours lecture; 1 hour tutorial. Prerequisite:High School Calculus. Not open to students who have taken MATH 120, MATH 122,MATH 139 or CEGEP objective 00UN or equivalent. Not open to students who havetaken or are taking MATH 130 or MATH 131, except by permission of the Departmentof Mathematics and Statistics. Each Tutorial section is enrolment limited.) Reviewof functions and graphs. Limits, continuity, derivative. Differentiation of elementaryfunctions. Antidifferentiation. Applications.

1.2.2 Calendar Descriptions of other beginning calculus courses

• MATH 139 Calculus (4 credits); Instructor = Professor J. J. Xu. Thiscourse (whose prerequisite is a course in functions) is intended for students whohave never had a course in calculus. Students may apply at the Department ofMathematics and Statistics before the end of the Course Change Period for au-thorization to register in this course; they must bring copies of their transcripts4.This course covers approximately the same material as MATH 140. MATH 139uses the same textbook as MATH 140.

• MATH 150 Calculus A (4 credits); Instructor = Dr. Igor Wigman. Thiscourse, together with either of its 4-credit sequels, MATH 151 Calculus B andMATH 152 Calculus E5, covers most of the material of courses MATH 140/MATH139, MATH 141 Calculus 2, and MATH 222 Calculus 3, in only two semesters6. Aprior or concurrent course in Vector Geometry (e.g. MATH 133) is recommended.7

3The previous designation for this course was 189-140, and the version given in the fall was labelled189-140A; an earlier number for a similar course was 189-122.

4Authorization for registration in MATH 139 2006 09 will be available beginning on Wednesday, 30August, 2006, and ending on Tuesday, 19 September, 2006. For details, see the following URL:

http://www.math.mcgill.ca/brown/incoming.htm

5Open only to students in the Faculty of Engineering6MATH 152 lacks one topic in MATH 222, as it is deferred to another Engineering Mathematics

course.7Note that MATH 150/151 uses a different textbook from MATH 140/141; the textbook is by R. A.

Adams [18], [16], [17], [19], [20], [21].


1.2.3 How much of my previous calculus course do I need to remember?

Since the gaps in students’ knowledge from their high school calculus courses are differentfor different students, we will be covering the entire syllabus of a first calculus course,i.e., the same material that is covered in MATH 139, described above. However, we doassume a working knowledge of the prerequisites for a first calculus course — includingfamiliarity with the content of a course on functions, including algebra and trigonometry.

1.2.4 Late transfer from MATH 150

Some students from MATH 150 may be permitted by their Faculty to transfer intoMATH 140 after the end of the Change of Course Period. If you are in this category,please send an e-mail message to Professor Brown as soon as your transfer has beenapproved.8

1.3 Tutorials; Tutors’ Coordinates

1. Every student must be registered in one lecture section and one tutorial section.Tutorials begin in the week of September 11th, 2006. The last tutorials in tutorialsections T009 – T020 will be during the week of November 27th, 2006. Table2 gives times, locations, and the tutor’s name for each of the tutorials; Table 3gives the tutors’ coordinates. The information in these tables is subject tochange. We try to publicize changes but sometimes we are not informedin advance.9

2. You are expected to write quizzes and submit written assignments onlyin the tutorial section in which you are registered.10 You do not have alicence to move from one tutorial section to another at will, even if youfind the time, location, or personnel of your tutorials either temporar-ily or permanently inconvenient; in the latter case the onus is on youto transfer formally to another tutorial section, to change your otherclasses, or to drop MATH 140 2006 09. Please remember that transfersmust be completed by the Course Change (drop/add) deadline (Septem-ber 19th, 2006), and are subject to the maximum capacities establishedfor each tutorial section11.

8This is to ensure that a WeBWorK account is opened for you in this course, and that your dateof entry to the course is recorded.

9The current room for your tutorial should always be available by clicking on “Class Schedule” onMINERVA FOR STUDENTS, http://www.mcgill.ca/minerva-students/.

10In some time slots there may be several tutorial sections, meeting in different rooms.11Your instructors do not have the ability to change the maximum capacities of tutorials.


All tutorial sections meet first during the week of September 11th, 2006

# CRN Day Begins Ends Room Tutor

T003 421 Mon 13:35 14:25 BURN 1214 K. DavisT004 422 Mon 14:35 15:25 BURN 1214 K. DavisT005 423 Mon 14:35 15:25 BURN 1B39 J. McKeownT006 424 Mon 15:35 16:25 BURN 1B23 J. McKeownT007 425 Mon 14:35 15:25 ARTS 145 N. TouikanT008 426 Mon 15:35 16:25 ARTS 145 N. TouikanT009 427 Wed 13:35 14:25 BURN 1214 C. BoudreaultT010 428 Wed 14:35 15:25 BURN 1214 C. BoudreaultT011 429 Wed 14:35 15:25 BURN 1B39 S. NashaatT012 430 Wed 15:35 16:25 BURN 1214 A. BakerT013 431 Wed 14:35 15:25 BURN 1B36 A. BakerT014 432 Wed 13:35 14:25 BURN 1B24 S. NashaatT015 433 Thurs 16:05 16:55 ENGMC 12 B. CordyT016 434 Thurs 17:05 17:55 ENGMC 12 B. CordyT017 435 Wed 15:35 16:25 BURN 1B24 Y. ZhaoT018 436 Wed 16:35 17:25 BURN 1214 Y. ZhaoT019 437 Fri 14:35 15:25 BURN 1B39 S. MiscioneT020 438 Fri 15:35 16:25 BURN 1B23 S. Miscione

Table 2: Schedule and Locations of Tutorials, as of December 1, 2006 (subject to change)

3. The tutorials in MATH 140 are short. About half of the total tutorial time isdevoted to testing, through quizzes every two or three weeks, and subsequent dis-cussion of the solutions to the quiz and written assignment problems.

The remaining time will be devoted to brief discussion of the solution of specifickinds of problems. Students must not assume that they will be exposed in lecturesand tutorials to detailed model solutions for every type of Calculus 1 problem. It isessential that you supplement these classes with serious work on your own, carefullyreading the textbook and solving problems therein. If you encounter difficulties,take them to the tutors during one of their many office hours: you may attend theoffice hours of any tutor in the course, and are not restricted to those of the tutorof the tutorial in which you are registered.


Tutor E-mail address Office Office HoursBURN Day Begins Ends Day Begins Ends

Baker, A. [email protected] 1035 Th 14:00 17:00Boudreault, C. [email protected] 1115 M 14:00 17:00

Cordy, B. [email protected] 1032 T 11:30 14:30Davis, K. [email protected] 1134 Th 09:30 10:30 Th 13:30 15:30

McKeown, J. [email protected] 1218 W 12:30 15:30Miscione, S. [email protected] 1036 T 10:30 11:30 Th 10:30 12:30Nashaat, S. [email protected] 1132 M 11:00 12:30 F 11:00 12:30Touikan, N. [email protected] 1008 T 15:00 18:00Zhao, Y. [email protected] 1035 F 14:15 17:15

During her/his office hours, a tutor is available to all students in the course,not only to the students of her/his tutorial section.

Table 3: Tutors’ Coordinates, as of December 1, 2006

1.4 Evaluation of Your Progress

1.4.1 Your final grade

(See Table 4, p. 12) Your grade in this course will be a letter grade, based on a percentagegrade computed from the following components:

1. Assignments submitted over the Web: WeBWorK homework assignments (cf.§1.4.2) — counting together for a total of 10% of the final grade. All WeBWorKassignments must be completed by their posted expiration dates and times. (cf.§1.4.1). We plan to have 7 Assignments, denoted by A1, . . .A7.

2. Materials graded by your Tutor:

• Four Written Assignments — counting together for 4%. The Written Assign-ments will be denoted by W1, . . . , W4.

• Four Quizzes given at the tutorials — counting together for 16% of the finalgrade. The Quizzes will be denoted by Q1, . . . , Q4. There will also be acompulsory diagnostic quiz, Q0. Quizzes Q0 and Q4 will be 45 minutes long;Q1, Q2, Q3 will each be 20 minutes long. Quizzes Q1 — Q4 will each countfor 4% of the 16% allocated to the quizzes; Q4 counts equally with the earlierquizzes, even though it is longer; the grade on Q0 does not count — but,


if you do not write Q0, your other quiz grades will not count.12 QuizQ0 will be written at your first tutorial. Be there!

3. The final examination — counting for 70%.

Where a student’s performance on the final examination is superior to her performanceon the tutorial quizzes, 16% of the final examination percentage will replace the quizgrades in the calculations. It is not planned to permit the examination grade to replacethe grades on WeBWorK assignments or on written assignments.

1.4.2 WeBWorK

The WeBWorK system, developed at the University of Rochester — is designed toexpose you to a large number of drill problems, and where plagiarism is discouraged.WeBWorK is accessible only over the Web. Details on how to sign on to WeBWorKare contained in Appendix H to these notes, page 4001.

WeBWorK assignments carry a due date and time; only answers submitted by thedue date and time will count. The WeBWorK assignments will be labelled A1, . . ., A7.Assignments A1 and A2 together review some precalculus concepts.

Numbers of permitted attempts at WeBWorK questions: The assignmentshave limits to the numbers of times a student may attempt a problem. However, for eachassignment An there will be a companion “Practice” Assignment Pn (n = 1, 2, . . . , 7)with an unlimited number of attempts at similar problems, but in which the specificdata will be different. You may prepare yourself on the Practice assignment beforeattempting the actual assignment. The practice assignments DO NOT COUNT in yourterm mark, even though a grade is recorded.. Practice assignment Pn is usually due 1 weekbefore assignment An; but there are exceptions, so please consult the timetable. Anotherassignment which will not count will be Practice Assignment P0, which is intended tointroduce you to the WeBWorK system.

Due dates and times for WeBWorK assignments The due dates for WeBWorKassignments will be on specified Sundays, about 23 : 30h (with the exception of As-signment A3, which will be due late in the evening on the Monday which is Canada’sThanksgiving Day); last minute changes in the due dates may be announced either onWeBWorK, on WebCT, or by an e-mail message13 As mentioned in the WeBWorK

12 It is essential that you write Q0, since it can indicate areas where you need to do remedial work atthe beginning of the term. (An announcement was posted on WebCT in September that the compulsoryrequirement for Q0 was being relaxed.)

13Be sure that your e-mail addresses are correctly recorded. See 1.8.1, p. 20 of these notes.


FAQ (cf. Appendix §H) if you leave your WeBWorK assignment until the hours closeto the due time on the due date, you should not be surprised if the system is slow to re-spond. This is not a malfunction, but is simply a reflection of the fact that other studentshave also been procrastinating! To benefit from the speed that the system can deliverunder normal conditions, do not delay your WeBWorK until the last possible day! Ifa systems failure interferes with the due date of an assignment, arrangements may bemade to change that date, and an e-mail message may be broadcast to all users (to thee-mail addresses on record), or a note posted in the course announcements on WebCT;but slowness in the system just before the due time will not normally be considered asystems failure.14

Precalculus WeBWorK assignments: Assignments A1 and A2 are based on precal-culus material. These assignments have been scheduled to be due before the last day tochange courses; they can help you decide whether you are ready for MATH 140.

1.4.3 Written Assignments

Written assignments W1, W2, W3, W4 will be posted on WebCT, one or two weeksbefore they are due; they may also be included in updated versions of these notes.These assignments will contain one or more problems for which you will be expected towrite full solutions, modelled on similar types of solutions in worked examples in thetextbook, or solved problems in the Student Solutions Manual. You will be assigned anindividualized version of the problems. You should bring your solutions to the TutorialQuiz the following week, and submit them with your quiz paper; they will be graded byyour tutor and returned to you with that graded quiz. Please do not attempt any othermethods for submitting your written assignments, to minimize the risk of loss.

1.4.4 Quizzes at the Tutorials; Submission of Written Assignments at theQuizzes

1. There will be 5 quizzes, numbered Q0, Q1, Q2, Q3, Q4, administered at the tutorials.These quizzes will be graded, and returned (together with the written assignmentthat will be handed in with quiz papers Q1 — Q4). The primary purpose of a quizis to diagnose possible gaps in your understanding, not to drill on examinationskills.

No provision is being made for students who miss one of Q1 — Q4. The gradingformula permits the quiz component of the final grade to be replaced by the final

14Should you find that the system is responding slowly, do not submit your solutions more than once;you may deplete the number of attempts that have been allowed to you for a problem: this will not beconsidered a systems failure.


examination grade, if this is to a student’s advantage.

2. The grade you obtain on quiz Q0, given at your first tutorial, during theweek of September 11th, 2006, DOES NOT COUNT IN YOUR TERMMARK. But your other quiz grades will not count unless you write Q0.Be there!15

3. The remaining quizzes will be based on current topics in the syllabus of the course,most16 of which topics will have been discussed in the lectures before the quiz;the quizzes are not based directly on WeBWorK assignments. To prepare fora quiz you should be working exercises in the textbook based on the materialcurrently under discussion at the lectures, and you should have attempted any openWeBWorK assignments. But, unlike the WeBWorK assignments — where theemphasis is on correct answers alone — students may be expected to provide fullsolutions to some or all problems on quizzes.17

4. Submission of Written Assignments at Quizzes (excluding quiz Q0) In-dividualized written assignments W1, W2, W3, W4 will be published about a weekbefore they are to be handed in at a tutorial quiz; you should download or copyeach written assignment from the web site, write out your solutions at home, bringthe completed assignment with you to the tutorial, and hand it in with your quizpaper — no other submission method is acceptable. The assignment must be readyfor submission when you arrive, as you will not be provided time in the tutorialroom to complete it. You will enclose the written assignment in your folded answerpaper.

5. Your tutors will normally bring graded quizzes and graded assignments submittedwith them to the tutorial to be returned to you. University regulations do notpermit us to leave unclaimed materials bearing names and student numbers inunsupervised locations; you may be able to recover an unclaimed quiz from thetutor who graded it, during her/his regular office hours.

6. “Raw” versus “Scaled” Quiz Grades. After he receives all quiz data fromT.A.’s, Professor Brown may make small upward adjustments in the grades in

15See footnote 12.16but possibly not all17In Math 140 and Math 141 the general rule is that full solutions are expected to all problems, unless

you receive explicit instructions to the contrary: ALWAYS SHOW YOUR WORK! The solutions in theStudent Solution Manual [3] to the textbook can serve as a guide to what should be included in a “full”solution.

Please do not ask your tutors to provide information about the content of coming quizzes: they willnot have that information in advance.


certain tutorial sections, in order to reduce variations between sections, based onanalysis of the average or median performance.

7. Posting of grades. Grades on Quizzes and Written Assignments will eventuallybe uploaded to WebCT, where you will be able to check that your grades havebeen properly recorded. Report any error by e-mail to Professor Brown and toyour T.A., within one month of the date it has been posted.

8. Plagiarism. (cf. §1.4.8) When a T.A. discovers evidence of cheating, she/he isobliged to bring it to the attention of the instructors, who may have to forward thematerials to the Associate Dean of the Student’s Faculty for possible disciplinaryaction. Such actions have been taken in the past in this course.

9. Students with Disabilities who wish to write any of their quizzes at the OSDoffice need to make prior arrangements with that office approximately 8 days beforeany quiz.

1.4.5 Final Examination

A 3-hour-long final examination will be scheduled during the regular examination periodfor the fall term (December 7th, 2006 through December 22nd, 2006). You are advisednot to make any travel arrangements that would prevent you from being present oncampus at any time during this period.

1.4.6 Supplemental Assessments

Supplemental Examination. There will be a supplemental examination in this course.For information about Supplemental Examinations, see

http://www.mcgill.ca/artscisao/departmental/examination/supplemental/.

Note, in particular, that a Supplemental Examination may be written only by a studentwho has obtained a grade of F or D as a grade in the course, and that the grade on theSupplemental Examination counts in your average as though you have taken the courseagain — without a term work component! )

There is No Additional Work Option. “Will students with marks of D, F, or Jhave the option of doing additional work to upgrade their mark?” No. (“AdditionalWork” refers to an option available in certain Arts and Science courses, but not availablein MATH 140 2006 09.)


1.4.7 Machine Scoring

“Will the final examination be machine scored?” The final examination will not bemachine scored.

Answer-only Problems It is possible that some of the problems on your final exam-ination will request that the answer only be given, and will not carry part marks whichcould be based on the work leading up to the answer.

Multiple Choice Questions Do not confuse machine scoring with multiple choicequestions. We may use multiple choice questions in situations where it is expedient; forexample, Diagnostic Quiz Q0 will have multiple choice questions. The grade on this quizdoes not count in your final grade, but you are required to write the quiz. Thesemultiple choice questions will be graded manually by your TA, not by machine.

1.4.8 Plagiarism

While students are not discouraged from discussing methods for solving WeBWorKassignment problems with their colleagues, all the work that you submit — whetherthrough WeBWorK or written assignments, or on tutorial quizzes or the final exami-nation must be your own. The Senate of the University requires the following messagein all course outlines:

“McGill University values academic integrity. Therefore all students must under-

stand the meaning and consequences of cheating, plagiarism and other academic

offences under the Code of Student Conduct and Disciplinary Procedures. (See

http://www.mcgill.ca/integrity for more information).

“L’universite McGill attache une haute importance a l’honnetete academique. Il

incombe par consequent a tous les etudiants de comprendre ce que l’on entend

par tricherie, plagiat et autres infractions academiques, ainsi que les consequences

que peuvent avoir de telles actions, selon le Code de conduite de l’etudiant et des

procedures disciplinaires. (Pour de plus amples renseignements, veuillez consulter

le site http://www.mcgill.ca/integrity).”

It is a violation of University regulations to permit others to solve yourWeBWorK problems, or to extend such assistance to others; you could beasked to sign a statement attesting to the originality of your work. TheHandbook on Student Rights and Responsibilities18 states in ¶A.I.15(a) that

18http://upload.mcgill.ca/secretariat/greenbookenglish.pdf


“No student shall, with intent to deceive, represent the work of another personas his or her own in any academic writing, essay, thesis, research report,project or assignment submitted in a course or program of study or representas his or her own an entire essay or work of another, whether the material sorepresented constitutes a part or the entirety of the work submitted.”

You are also referred to the following URL:

http://www.mcgill.ca/integrity/studentguide/

1.4.9 Keep all your graded materials!

Grades will be posted on WebCT during the term; check periodically to see that yourgrades are recorded correctly, and advise Professor Brown and/or your TA if you find agrade missing or incorrectly recorded. You may need to produce the graded assignmentor quiz, so these should be retained until the end of the term.19

1.5 Published Materials

1.5.1 Required Text-Book

The textbook for the course is J. Stewart, SINGLE VARIABLE CALCULUS: Early

Transcendentals, Fifth Edition, Brooks/Cole (2003), ISBN 0-534-39330-6, [1].This book is the first half of J. Stewart, CALCULUS: Early Transcendentals,

Fifth Edition, Brooks/Cole (2003), ISBN 0-534-39321-7, [2]; this edition covers thematerial for Calculus 3 (MATH 222) as well, but is not the text-book for that courseat the present time. The textbook will be sold in the McGill Bookstore bundled withits Student Solutions Manual (see below). The ISBN number for the entire bundle is0-17-617367620.

1.5.2 Optional Reference Books

Students are urged to make use of the Student Solution Manual:

• D. Anderson, J. A. Cole, D. Drucker, STUDENT SOLUTIONS MANUAL

FOR STEWART’S SINGLE VARIABLE CALCULUS: Early Transcen-

dentals, Fifth Edition, Brooks/Cole (2003), ISBN 0-534-39333-0, [3]. Thisbook is also sold “bundled” with either version of the text book; we expect theBookstore to stock the bundle numbered ISBN 0-17-6173676 [4].

19The justification for quizzes and assignments is mainly as learning experiences: for that reason aloneyou should be retaining all materials that could help you prepare for your final examination.

20The bundle also includes ISBN 0-17-6173676 ACP Calculus Formula Card.


Item # Due Date Details

P0 DOES NOT COUNT: introduces WeBWorKP1 17 Sept 06 DOES NOT COUNT; practice for A1

P2 17 Sept 06 DOES NOT COUNT; practice for A2

WeBWorK A1 26 Sept 06Assignments A2 26 Sept 06(cf. §1.4.2) P3 01 Oct 06 DOES NOT COUNT; practice for A3

10% A3 09 Oct 06P4 15 Oct 06 DOES NOT COUNT; practice for A4

A4 22 Oct 06P5 29 Oct 06 DOES NOT COUNT; practice for A5

A5 05 Nov 06P6 12 Nov 06 DOES NOT COUNT; practice for A6

A6 19 Nov 06P7 26 Nov 06 DOES NOT COUNT; practice for A7

A7 03 Dec 06 A1–A7 count equally, but may have differentnumbers of problems.

Written W1 with Q1 Download; complete at home; hand in with Q1

Assignments W2 with Q2 Download; complete at home; hand in with Q2

(cf. §1.4.3) W3 with Q3 Download; complete at home; hand in with Q3

4% W4 with Q4 Download; complete at home; hand in with Q4

Quizzes Q0 11–15 Sept 06 The grade on Quiz Q0 does not count, but you(cf. §1.4.4) Q1 25–29 Sept 06 must write the quiz (for diagnostic purposes)!

16% or 0% Q2 16–20 Oct 06 Quizzes Q1 — Q4 count equally.Q3 30 Oct-3 NovQ4 20–24 Nov 06

Final Exam 07–22 Dec 06 Date of exam to be announced by Faculty

70% or 86%

SupplementalExam

01–02 May 07 Only for students who do not obtain standing atthe final. Supplemental exams count in your av-erage like taking the course again; exam countsfor 100%.

Table 4: Summary of Course Requirements, as of December 1, 2006; (all dates are subjectto change)

The publishers of the textbook and Student Solutions Manual also produce

• a “Study Guide”, designed to provide additional help for students who believethey require it: R. St. Andre, STUDY GUIDE FOR STEWART’S SIN-


GLE VARIABLE CALCULUS: Early Transcendentals, Fifth Edition,Brooks/Cole (2003), ISBN 0-534-39331-4, [7]. (The “Study Guide” resembles theStudent Solution Manual in appearance: be sure you know what you are buying.)

• a “Companion” which integrates a review of pre-calculus concepts with the contentsof Math 140, including exercises with solutions: D. Ebersole, D. Schattschneider,A. Sevilla, K. Somers, A COMPANION TO CALCULUS. Brooks/Cole (1995),ISBN 0-534-26592-8 [32].

1.5.3 Recommended Video Materials

Use of the following materials is recommended, but is not mandatory21.

Larson/Hostetler/Edwards DVD Disks A set of video DVD disks produced foranother calculus book, [22] Calculus Instructional DVD Program, for use with (interalia) Larson/Hostetler/Edwards, Calculus of a Single Variable: Early TranscendentalFunctions, Third Edition [23] is produced by the Houghton Mifflin Company. Copieshave been requested to be placed on reserve in the Schulich Library. In Appendix §Iof these notes there are charts that indicate the contents of these disks that pertain toMATH 140.

Text-specific videos for Stewart’s Calculus, early transcendentals, 5th editionThe publisher of Stewart’s Calculus has produced a series of videotapes, [8] Video Outlinefor Stewart’s Calculus (Early Transcendentals), Fifth Edition. These will initially beavailable for reserve loan at the Schulich Library. There may not be VCR viewingequipment freely available in the library; the intention is that interested students borrowa tape for viewing on their own equipment at home.

Tools for Enriching Calculus This is a CD-ROM included with new copies of Stew-art’s Calculus. From [10, Introduction]: “Tools for Enriching Calculus (TEC) enhances atopic that is covered in the textbook by providing both broader and deeper coverage of thoseaspects for which technology is particularly useful. The basic format of most modules is apoint-and-click laboratory environment in which you can easily visualize functions and theirderivatives, experiment with suggested examples and exercises, explore your own choices ofexamples, and perhaps even test some of your own conjectures. You need to be a well-preparedand active player to reap the benefits from these approaches.

21No one will check whether you have used any of these aids; a student can obtain a perfect gradein the course without ever consulting any of them. No audio-visual or calculator aid can replace thesystematic use of paper and pencil as you work your way through problems. But the intelligent use ofsome of these aids may assist your in understanding the subject matter.


“First, you need to read the textbook materials carefully to gain an understanding of theessential ideas. Next, you need to read the introductory material for each TEC module, whichexplains the basic mathematical approaches and describes how to use the module. Each modulehas several examples which will familiarize you with its basic features. When you have finishedreviewing this material, and have some paper and pencil in hand, you are ready to get the mostbenefit from using the module. You can improve your understanding of the topic by exploringmathematical questions that you find puzzling, and checking your ideas for solutions using themodule. Try to work through some of the exercises in the module to gauge your understandingof the topic. Be willing to use pencil and paper to first guess what the answer might be beforeseeing an electronic graph...

“Another important TEC feature is the homework22 hints. Hints have been created for

several selected exercises in each section of your textbook to help you understand some key

points in finding solutions for these exercises. Similar to a good instructor or teaching assistant,

these hints ask you questions that will allow you to make progress toward a solution without

giving you the actual answer. You need to actively pursue each hint with pencil and paper and

fill in many of the computations and details. If you can complete the solution after reading

only one or two hints, you can feel proud of your achievements. If you still have questions after

completing all of the hints for a problem, your work should help you to better understand the

solution presented in the Student Solutions Manual.”

Interactive Video Skillbuilder CD for Stewart’s Calculus: Early Transcen-dentals, 5th Edition [9] This CD-ROM is included with new copies of the textbook.It contains, after an enlightening “pep-talk” by the author, a discussion of some of theworked examples in the text-book, followed by a quiz for each section in the book. Somestudents may find the animations of the examples helpful, although the examples are allworked in the book. You might wish to try some of the quiz questions using paper andpencil, and then check your answers with those given on the CD. It is not recommendedthat you attempt to enter your answers digitally, as this is a time-consuming process,and uses a different input method from your WeBWorK assignments, which serve thesame purpose.

1.5.4 Other Calculus Textbooks

While students may wish to consult other textbooks, or other editions of the recom-mended textbooks, instructors and teaching assistants in Math 140 will normally referonly to the prescribed edition of the prescribed textbook for the course. Other bookscan be very useful, but the onus is on you to ensure that your book covers the syllabus to

22Note that homework refers to odd-numbered problems in the exercises in [1], not to problems onWeBWorK .


at least the required depth; where there are differences of terminology, you are expectedto be familiar with the terminology of the textbook.23

In your previous calculus course(s) you may have learned methods of solving problemsthat appear to differ from those you find in the current textbook. Your instructors willbe pleased to discuss any such methods with you personally, to ascertain whether theyare appropriate to the present course. In particular, any methods that depend uponthe use of a calculator, or the plotting of multiple points, or the tabulation of functionvalues, or the inference of a trend from a graph should be regarded with scepticism.

1.5.5 Website

These notes, and other materials distributed to students in this course, will be accessibleat the following URL:


The notes will be in “pdf” (.pdf) form, and can be read using the Adobe Acrobat reader,which many users have on their computers. This free software may be downloaded fromthe following URL:

http://www.adobe.com/prodindex/acrobat/readstep.html 24

The questions on some old examinations will also be available as an appendix to thesenotes on the Web.25 It is expected that most computers in campus labs should have thenecessary software to read the posted materials.

Where revisions are made to distributed printed materials — for example these in-formation sheets — we expect that the last version will be posted on the Web.

The notes and WeBWorK will also be available via a link from the WebCT URL:

http://webct.mcgill.ca

but other features of WebCT26 have not yet been implemented.

23There should be multiple copies of the textbook on reserve in the Schulich library.24At the time of this writing the current version appears to be 7.0.8.25There is no reason to expect the distribution of problems on quizzes or in assignments and exami-

nations from previous years be related to the frequencies of any types of problems on the examinationthat you will be writing at the end of the term.

26cf. Appendix H to these notes, p. 4001


1.6 Syllabus

In the following list section numbers refer to the text-book [1]. The syllabus will includeall of Chapters 1, 2, 3, 4, with omissions, as listed below.27

Chapter 0: A Preview of Calculus. This is motivational material, and may not allbe discussed in the lectures. Read it.

Chapter 1: Functions and Models. In §1.1, two of the four ways to define a functionare not useful in general: a table of values can be used to define a function only ifits domain is finite, and the value of the function is prescribed for every point inthe domain; it is normally not acceptable to define a function by a graph, unlessthe nature of the graph can be described without any ambiguity. Some parts of§1.2 may not be discussed in the lectures, but you should read the whole section— in particular the definitions of various kinds of functions — as this terminologymay be used from time to time. Omit §1.4.

Mathematical Induction (Because time has been needed for review of othertopics, this topic will not be examination material in 2006-07.28) The Principle ofMathematical Induction [1, p. 81] will be required in proving theorems and solvingproblems that involve the positive integers (=“the natural numbers”).

Chapter 2: Limits and Derivatives.

Chapter 3: Differentiation Rules. Omit §3.3; but you are encouraged to read theparts of §3.3 that pertain to your own fields of interest.

Chapter 4: Applications of Differentiation. In §4.5 you may skip the discussion of“slant asymptotes” [1, bottom p. 322 – p. 323]. Omit §4.6, §4.8, §4.9.

Exercises that require technology Students are not expected to be able to solveexercises that require the use of calculators or computers. You may wish to trysuch problems, as a challenge, as some of them can be solved with clever use ofpaper and pencil.

27If a textbook section is listed below, you should assume that all material in that section is examina-tion material even if the instructor has not discussed every topic in his lectures ; however, the instructorsmay give you information during the term concerning topics that may be considered subsidiary.

Do not assume that a topic is omitted from the syllabus if it has not been tested in aWeBWorK assignment or a quiz, or if it has not appeared on any of the old examinationsin the course! Some topics to not lend themselves to this type of testing; others may have beenomitted simply because of lack of space, or oversight. By the same token, you need not expect everytopic in the course to be examined on the final examination.

28Added 26 September 06


Problems Plus The exercises and other material that appear in [1, Principles of Prob-lem Solving, pp. 80-85], and in the “Problems Plus” subsections following the laterchapters are to be omitted.

The following appendices in the textbook contain some prerequisite material for thiscourse:

Appendix A: Intervals, Inequalities, and Absolute Values. (see information forChapters 1 and 3 above)

Appendix B: Coordinate Geometry and Lines.

Appendix C: Graphs of Second-Degree Equations. You are expected to be famil-iar with the material concerning the circle, [1, pp. A16-A17]. The remainder of thematerial should be familiar to most students, but will not be assumed.

Appendix D: Trigonometry. You are assumed to be familiar with the material in [1,pp. A24–A31].

Please do not ask the tutors to provide information as to which topics should beemphasized. Unless you are informed otherwise by the instructors in the lecture sectionsor published notes — printed, or mounted on the Web — you should assume that allmaterials listed are included in the syllabus. You are not expected to be able to reproduceproofs of the theorems in the textbook.

1.7 Preparation and Workload

1.7.1 Prerequisites.

All students must have completed a course in functions — in particular, familiaritywith trigonometric functions is assumed; it is your responsibility as a student to verifythat you have this necessary background to benefit from this course. While a course inhigh school calculus is the stated “prerequisite” for this course, stronger students havingonly the prerequisite for MATH 139 (viz., “a course in functions”) have normally beenpermitted to register at their own risk in MATH 140. Some of the prerequisites arereviewed in Appendices A, B, C, D of your textbook [1]; solutions to odd-numberedproblems can be found in the Student Solution Manual [3].

Because weakness in pre-calculus topics can contribute to failure of a student inMATH 140 or MATH 141, students will be tested on prerequisites in both the WeB-WorK assignments A1, A2 and quiz Q0. Students who do not believe they can remedydeficiencies on their own might wish to consider MATH 11229. MATH 112 2006 09 isscheduled at the same time as Section 1 of MATH 140 2006 09.

29MATH 112 Fundamentals of Mathematics (3) (Fall. Not open to students who have taken CEGEP


1.7.2 Calculators

The use of calculators is not permitted in either quizzes or the examination in this course.Students whose previous mathematics courses have been calculator-oriented would beadvised to make particular efforts to avoid the use of a calculator in solving problemsin this course, in order to develop a minimal facility in manual calculation. This meansthat you are urged to do all arithmetic by hand. If, for a problem presented to youin this course, you find that you cannot solve it without using a calculator,then you are almost certainly solving the problem in an unacceptable way.

Do not assume that, in excluding the use of calculators, your instructors are “Lud-dites”30! We believe that you will master the materials in Calculus 1 and 2 best by relyingon manual calculations. In some more advanced mathematics and statistics courses youwill be encouraged to use calculators and computers.

1.7.3 Self-Supervision

MATH 140 is not a high-school course, and McGill is not a high school. Themonitoring of your progress before the final examination is largely your own responsibil-ity. While the tutors and instructors are available to help you, they cannot do so unlessand until you identify the need for help. WeBWorK and quizzes are designed to assistyou in doing this.

Time Demands of your Other Courses. Be sure to budget enough time to attendlectures and tutorials, for private study, and for the solution of many problems. Don’t betempted to divert calculus study time to courses which offer instant gratification. Whilethe significance of the tutorial quizzes in the computation of your grade is minimal, theseare important learning experiences, and can assist you in gauging your progress in thecourse. This is not a course that can be crammed for: you must work steadily throughthe term if you wish to develop the facilities needed for a strong performance on the finalexamination.

Working Problems on Your Own. An effective way to master the calculus isthrough working large numbers of problems from the textbook. Your textbook wasselected partly because of the availability of an excellent Student Solutions Manual [3];this manual has brief but complete solutions to most of the odd-numbered exercises in

course 201-101. Open only to those students who are deficient in a pre-calculus background.) Equationsand inequalities, graphs, relations and functions, exponential and logarithmic functions, trigonometricfunctions and their use, mathematical induction, binomial theorem, complex numbers.

30opposed to technological innovation


the textbook. The skills you acquire in solving textbook problems could have much moreinfluence on your final grade than either WeBWorK or the quizzes.

The real uses of WeBWorK and the quizzes. Students often misunderstand thetrue significance of WeBWorK assignments and the quizzes. While both contributeto your grade, they help you estimate the quality of your progress in the course. Takeproper remedial action if you are obtaining low grades on quizzes31, or if you requiremany attempts before being able to solve a problem on WeBWorK. However, whileboth WeBWorK and the quizzes have a role to play in learning the calculus, neitheris as important as reading your textbook, working problems yourself, and attending andlistening at lectures and tutorials.

Does a high grade on WeBWorK indicate the likelihood of a high grade on thefinal examination? NO! The primary purpose of the WeBWorK assignments is asan aid to learning; but, as your work is not being done under examination conditions, youshould not use the WeBWorK grades as indicators of your likely examination grade.The grades on the quizzes are somewhat more useful for that purpose.

1.7.4 Escape Routes

At any time, even after the last date for dropping the course, students who are experi-encing medical or personal difficulties should not hesitate to consult their advisors or theStudent Affairs office of their faculty. Don’t allow yourself to be overwhelmed by suchproblems; the University has resource persons who may be able to help you.

1.7.5 Terminology

Do not be surprised if your instructors and tutors use different terminology from whatyou have heard in your previous calculus course, particularly if that course was at a highschool. Sometimes the differences are purely due to different traditions in the professions.

“Negative x”. Your instructors and tutors will often read a formula −x as minus x,not as negative x. To a mathematician the term negative refers to real numbers whichare not squares, i.e. which are less than 0, and −x can be positive if x itself is negative.

However, mathematicians will sometimes refer to the operation of changing a signas the replacement of x by “its negative”; this is not entirely consistent with the usualpractice, but is an “abuse of language” that has crept into the professional jargon.

31An unwise remedy would be is to miss the quizzes, and thereby avoid an unwelcome message.


Inverse trigonometric functions; fn. A formula like sin−1 x will be read as theinverse sine of x — never sine to the minus 1 or sine to the negative 1 . However, if wewrite sinn x, where n is a positive integer, it will always mean (sin x)n. These conventionsapply to any of the functions sin, cos, tan, cot, sec, csc; they also apply to the hyperbolicfunctions, which we will meet in [1, §3.9]: sinh, cosh, tanh, coth.... We will usually notwrite exponents on general functions, so a formula like f 2(x) does not have an obviousmeaning, and we will avoid writing it when f is other than a trigonometric or hyperbolicfunction.

Logarithms. If you were taught to interpret log x as being the logarithm to base 10,you should now forget that convention, at least in your mathematics courses; nowadaysmathematicians rarely use logarithms to base 10.32 Most often, if your instructor speaksof a logarithm, and writes log x, he will be referring to the base e, i.e. to loge; that is,he is referring to the function that calculus books call ln. When a logarithm to someother base is intended, it will either be denoted by an explicit subscript, as log2, or somecomment will be made at the beginning of the discussion, such as “all logarithms in thisdiscussion are to the base 2”.

Your instructors try to think like mathematicians even when lecturing to their classes,and to use the language and terminology we use when talking to each other.

1.8 High Technology and MATH 140

1.8.1 Keep your e-mail addresses up to date

• From http://www.mcgill.ca/email-policy/: “The UEA (normally a variation [email protected]) is an ‘alias’ address that points to an actualexisting mailbox — all e-mail sent to the UEA is simply delivered to this exist-ing mailbox. By default, all e-mail addressed to students’ UEA is delivered totheir McGill e-mail box. Students can arrange to have their UEA deliver theire-mail to a mailbox other than their McGill e-mail box (there are online self-serve tools to facilitate this) — note that it is the student’s responsibility to en-sure that this alternate mailbox is viable. It is recommended that students usetheir McGill e-mail box.” You can enter or change a forwarding e-mail addressby going to http://webmail.mcgill.ca, and logging in to your student mailbox atpo-box.mcgill.ca.

32Your calculator keyboard may be using this convention, but you won’t be using the calculator inthis course.


• The WeBWorK system permits you to designate any e-mail address; that is alsothe address that will be used if you send a FEEDBACK message.

1.8.2 Use of Calculators and Computer Algebra Systems

You are urged to do all calculations manually, and to avoid the use of calculators andcomputer algebra systems until you have completed MATH 141. You should not use acalculator or computer in the solution of WeBWorK problems, as it prevents you fromdeveloping skill for detecting errors in manual calculations — a skill that you will needfor the quizzes and final examination. It is also foolish to use a calculator to producedata to enter into WeBWorK, since WeBWorK will accept the data that you wouldtype into your calculator, and will do the calculations internally for you.

1.8.3 Use of the Internet

Students are expected to be able to access materials through the Internet, wheneverrequired. Here are some of the uses that are expected:

• To access and submit WeBWorK (§1.4.2) assignments

• To access the individualized problems for a student’s Written Assignments

• To access WebCT, where course grades and announcements will be mounted

• To access the web page for the course — also available through WebCT — wherethese and other notes will be available in .pdf form; the site also contains notesand examinations from previous years.

• In Lecture Section 002 the instructor will post notes regularly. Use of these notesis optional, but they will probably be available only as .pdf files.

1.9 Which problems should I work?

1.9.1 There are no public or private restricted lists of textbook problems towhich you can confine your preparation for testing in MATH 140.

The syllabus is defined in terms of sections of the textbook, and the intention is that youshould be able to solve most of the problems in the designated sections, as well as othertypes of questions that could be placed before you. Of course, there are certain questionsin the book that are not appropriate to this course, e.g., a small number of problemsnear the ends of the exercise sets, which can be difficult and require some assistance; orproblems requiring the use of computers; or problems that involve concepts that havebeen excluded from the course.


1.9.2 Are the type of problems I find on WeBWorK problems indicative ofwhat I need to know in this course?

Only partially! The WeBWorK problems are simply a sampling, and some importanttopics don’t appear at all in the WeBWorK assignments. The course is not defined bythe WeBWorK problems.

1.9.3 Problems discussed in the lectures

The problems cited by your instructors could be appropriate for their use at that time,or could be a useful introduction to a topic, or could be otherwise worthy of attention.If your instructor does refer to a problem, it would be wise to consider it, as his pastor subsequent lectures could refer to it. But you should not assume that every suchproblem will ultimately end up on a test or examination. We are trying to introduce youto a beautiful area of mathematics, and we sincerely wish that you will excel on testsand examinations; but we are not teaching a high school course, and do not encouragememorization or rote-learning.

1.9.4 And anyhow, the course is not concerned only with problem solving.

We are also trying to convey to you an understanding of basic theorems of the subject.We want you to learn precisely what the theorems are saying, and how these resultscan be applied. At the ends of the textbook chapters there are “Concept Checks” andTrue-False Quizzes that can help you evaluate your understanding. These questions areusually not in a form that lends itself to examinations, but they are excellent questionsto help you evaluate your progress. Some of the questions could be reformulated asacceptable test questions.

1.9.5 Repairing your precalculus foundations.

Some students may find, through WeBWorK assignments A1 and A2 or Quiz Q0, thatthey need remedial work to rebuild the foundations required for a first calculus course.If you are in this category, you are urged to expedite that study, so that it has minimaleffect on your progress through MATH 140. One of our goals in MATH 140 is to prepareyou for MATH 141, so that you will not have to face similar deficiencies when you enterthat course.

1.10 Last Weeks of Term (to be updated)

Tutor Office Office Hours 28 Nov - 06 DecBURN Weekday Month Date Begins Ends


(updated to December 1, 2006)

During her/his office hours, a tutor is available to all students in the course,not only to the students of her/his tutorial section.


2 About Quiz Q0

Distribution Date: Wednesday, September 6th, 2006

2.1 All students must write the diagnostic quiz.

It is essential that all students in MATH 140 2006 09 write the diagnostic quiz, Q0. Whilethe grade you obtain on the quiz does not count in your term mark, you are required towrite the quiz, so that you (and your instructors) can assess your level of preparation forthe course, and determine which areas of precalculus you need to strengthen early in theterm. Students who do not write Q0 will receive a grade of 0 on their otherquizzes.

2.2 Late registration and medical absences.

Students who registered in MATH 140 2006 09 after September 10th, 2006, and studentswith a valid medical certificate of absence should contact Professor Brown without delay.

2.3 Type of Quiz.

This quiz will be multiple choice. Since the purpose of the quiz is to diagnose weaknesses,it was decided to make the quiz somewhat challenging: you will have to work quickly, asyou will have only 45 minutes. But, again, the grade you obtain does not count inyour term mark; the grade will, however, be recorded, and will eventually be visibleto you on WebCT. It does not become part of the official McGill record for the course.

2.4 Preparing for the quiz.

Most of the material is discussed briefly in [1, Appendices A, B, D]. However, it is noteasy to cram for a test of this type.

2.5 What to do if your grade is very low.

If you aren’t able to analyze your performance yourself, your TA or your instructorcan help you. If your grade is essentially 0, you should consider whether you require aprecalculus course to prepare you. MATH 112 2006 09 is given only in the fall term,MWF 08:35–09:25. Rather than seeking to blame your previous educational system, thisis a time to see positive remedies, so that your time at McGill is productive. If youare planning to remedy deficiencies yourself, you could use, for example, the book [14],which we have asked to be kept on reserve in the Schulich Library.


3 W1, First Written Assignment

Distribution Date: Mounted on WebCT on Monday, September 4th, 2006Your completed solution to this assignment should be submitted, with a copy of this

question sheet at your Tutorial, during the week September 26-30, 2006. Please slip theassignment into your folded quiz answer paper to Quiz Q1. All materials must bearyour name and/or student number. No other method of submission is acceptable.

3.1 Certificate

Your assignment will not be graded unless you attach or include the following completedcertificate of originality, signed in ink:

I have read the information on the web page

http://www.mcgill.ca/integrity/studentguide/,

and assert that my work submitted for W1, A1 and A2

does not violate McGill’s regulations concerning plagiarism.

Signature(required) Date(required)

3.2 The assignment question

You are given formulæ for functions f and g.

1. Determine a formula for each of the functions f ◦ f , f ◦ g, g ◦ f , g ◦ g, simplified asmuch as you can.

2. Determine for each of the functions f , g, f ◦f , f ◦g, g ◦f , g ◦g, its domain. Wherethe domain is not all of R, you must explain your work — it’s not enough just towrite down the answers.

TUTORIAL DAY f(x) g(x)

MONDAY, 25 Sept., 2006 x2 − 4√

x − 1

WEDNESDAY, 27 Sept., 2006 x − 14x

1x

THURSDAY, 28 Sept., 2006 e−x 1ln x

FRIDAY, 29 Sept., 2006 sin 1x

arcsin x2

While you are expected to submit solutions for only the problem assigned to your tutorialday, you should also try the other 3 (but not hand in solutions).


4 W2, Second Written Assignment

Release Date: Mounted on WebCT on Sunday, September 24th, 2006Your completed solution to this assignment should be submitted, with a copy of thisquestion sheet at your Tutorial, during the week October 16-20, 2006. Please slip theassignment into your folded quiz answer paper to Quiz Q2. All materials must bearyour name and/or student number. No other method of submission is acceptable.

4.1 Certificate




and assert that my work submitted for W2, A3, A4 does not violateMcGill’s regulations concerning plagiarism.


4.2 The assignment questions

1. This first problem concerns a pair of functions that may not be discussed at the lec-tures, except incidentally. The purpose of the problem is to allow you to investigatethe functions yourself. You are not asked for rigorous proofs in this problem.

This problem involves, in part, sketching some graphs. The functions in questionhave discontinuities. Please use the following convention. If f has a discontinuityat x = a, please enlarge the point (a, f(a)); this convention is particularly usefulwhere a function is, for example, continuous from the right but not from the left —in that way the reader can see immediately what type of discontinuity is present.Another way of showing such discontinuities is through judicious use of square andround brackets. For example, the graph of

f(x) =

{

−1 if −4 ≤ x < 01 if 0 ≤ x ≤ 4

could be sketched as in Figure 1 on page 27. You are expected to solve this problem


-6sor

-6[

)

Figure 1: Showing a discontinuity in a graph

by using information from your graphs — you are not expected to provide rigorousjustifications for all statements. You do not need to use graph paper — just tryto have your graphs approximately to scale, so that the properties of the functionscan be clearly visible.

Definition 4.1 The floor function ⌊x⌋ is defined as follows: ⌊x⌋ is the largestinteger n such that n ≤ x. (Your textbook [1, Example 10, p. 110] calls thisfunction the Greatest Integer Function, and denotes it by [[x]].)

The ceiling function ⌈x⌉ is defined to be the smallest integer n such that n ≥ x.(This function used to be called the Least Integer Function.)

(a) Sketch a graph of each of the functions ⌊x⌋, ⌈x⌉ for −2 < x < 2. (You don’tneed to use graph paper for these problems, but you may if you wish.)

(b) Sketch graphs of the functions f(x) = ⌊2x⌋ + ⌈2x⌉ and g(x) = ⌊2x⌋ − ⌈2x⌉for −1 ≤ x ≤ 1.

(c) Describe the following sets:

i.

{

a

∣

∣

∣

∣

limx→a−

⌊x⌋ does not exist

}

=

ii.

{

a

∣

∣

∣

∣

limx→a−

⌈x⌉ does not exist

}

=

iii.

{

a

∣

∣

∣

∣

limx→a+


}

=

iv.

{

a

∣

∣

∣

∣

limx→a+


}

=

v.

{

a

∣

∣

∣

∣

limx→a−

⌊2x⌋ = 2a

}

=


vi.

{

a

∣

∣

∣

∣

limx→a−

⌊x⌋ = limx→a−

⌈x⌉}

=

vii.

{

a

∣

∣

∣

∣

limx→a−

⌊x⌋ = limx→a+

⌊x⌋}

=

(The “set-builder notation” used here is described in [1, Appendix A, p. A3].)

2. Prove the statement assigned for your tutorial day by using the Principle of Math-ematical Induction. For this problem you are expected to provide a very carefulproof. (One version of this problem was on the final examination in MATH 1402005 09.)

Students whose tutorial is on Monday: Let f(x) = x2ex. Prove carefully bymathematical induction that

dnf

dxn(x) =

(

x2 + 2nx + (n − 1)n)

· ex

for all positive integers n.

Students whose tutorial is on Wednesday: Let f(x) = x2e−x. Prove care-fully by mathematical induction that

dnf

dxn(x) = (−1)n

(

x2 − 2nx + (n − 1)n)

· e−x


Students whose tutorial is on Thursday: Let f(x) = 2xe2x. Prove carefullyby mathematical induction that

dnf

dxn(x) = 2n (2x + n) · e2x


Students whose tutorial is on Friday: Let f(x) = 2xe−2x. Prove carefully bymathematical induction that

dnf

dxn(x) = (−2)n (2x − n) · e−2x



5 W3, Third Written Assignment

Release Date: Mounted on WebCT on Tuesday, October 17th, 2006Your completed solution to this assignment should be submitted, with a copy of thisquestion sheet at your Tutorial, during the week October 30 - November 03, 2006.

Please slip the assignment into your folded quiz answer paper to Quiz Q3. All materialsmust bear your name and/or student number. No other method of submission is

acceptable.

5.1 Certificate




and assert that my work submitted for W3, A4 does not violate McGill’sregulations concerning plagiarism.



1. A function y = y(x) is defined implicitly by the following equation. You are,showing all your work, to determine the values of y′(x) and y′′(x) at the givenpoint.

Tutorial Day Defining Equation Evaluate at the Point

Monday x3 + y4 = 2xy (x, y) = (1, 1)

Wednesday x4 + y3 = 2xy (x, y) = (1, 1)

Thursday x3 − y4 = 2xy (x, y) = (−1, 1)

Friday −x4 + y3 = 2xy (x, y) = (1,−1)

2. The six hyperbolic functions are defined in terms of sinh and cosh, which, in turn,are defined in terms of exponentials. These six functions satisfy identities thatresemble properties of the trigonometric functions. Beginning with each function


expressed in terms of exponentials, and showing all your work, prove the identityassigned below to your Tutorial day:

Tutorial Day Identity to be Proved

Monday cosh(x + y) = cosh x · cosh y + sinh x · sinh y

Wednesday tanh(x − y) =tanh x − tanh y

1 − tanh x · tanh y

Thursday ddx

tanh x = sech2x

Friday ddx

csch x = −csch x · coth x


6 Draft Solutions to W1, First Written Assignment

Distribution Date: Released to students on October 21st, 2006Draft version, subject to correction

Your completed solution to this assignment was to be submitted, with a copy of thisquestion sheet at your Tutorial, during the week September 26-30, 2006. You were

asked to “Please slip the assignment into your folded quiz answer paper to Quiz Q1.All materials must bear your name and/or student number. No other method of

submission is acceptable.”

6.1 Certificate












MONDAY, 25 Sept., 2006 x2 − 4√

x − 1

WEDNESDAY, 27 Sept., 2006 x − 14x

1x

THURSDAY, 28 Sept., 2006 e−x 1ln x

FRIDAY, 29 Sept., 2006 sin 1x

arcsin x2


While you are expected to submit solutions for only the problem assigned to your tutorialday, you should also try the other 3 (but not hand in solutions).

6.3 Solutions

Marking Scheme: TOTAL = 20 MARKS

MONDAY: 1. Since f is a polynomial, there are no real numbers where it is unde-fined. Domain(f) = R

2. Since g(x) =√

x − 1, it is defined only where the argument of the square rootis non-negative, i.e., where x ≥ 1: this is the Domain(g).

3. f has been shown to be defined for all x. We compose it with itself, findingthat f(f(x)) = (x2−4)2 −4 = x4 −8x2 +12. This is also a polynomial, whichis defined for all x, so Domain(f) = R.

4. We have already shown that the domain of g is the interval [1,∞). In thatinterval

f(g(x)) = |x − 1| − 4 = x − 5 .

This last is the formula for f ◦g, but it is valid only where x ≥ 1. Even thoughthe function x − 5 has all of R as its domain, Domain(f ◦ g) = [1,∞).

5. The domain of f is unrestricted. Hence f(x) − 1 is defined for all x, and isalways equal to x2 − 5. But we cannot evaluate

√x2 − 5 unless x2 − 5 ≥ 0,

i.e., unless |x| ≥√

5.

Domain(g ◦ f) = (−∞,−√

5] ∪ [√

5,∞)

6. When we compose g with itself, the first application of g will be defined onlyon the domain of g, i.e., for x ≥ 1. But the second application of g will give√√

x − 1 − 1, and will not be defined unless the argument of the last squareroot applied is non-negative. Thus we require that

√x − 1 ≥ 1, which implies

that(√

x − 1)2 ≥ 12, i.e., that |x − 1| ≥ 1, which is equivalent to x ≥ 2 or

x ≤ 0. We have already required that x ≥ 1. This condition is inconsistentwith x ≤ 0, and is implied by x ≥ 2; hence

Domain(g ◦ g) = {x|x ≥ 2} = [2,∞) .

WEDNESDAY: 1. In the development of the real number system we do not assigna meaning to 1

0; we can’t! No matter what real number r we would want to

choose as the value of 10, we find that one of our other basic properties of


function formula domain

f x2 − 4 R

g√

x − 1 [1,∞)f ◦ f x4 − 8x2 + 12 R

f ◦ g x − 5 [1,∞)

g ◦ f√

x2 − 5 R − (−√

5, +√

5)

g ◦ g√√

x − 1 − 1 [2,∞)

Table 5: Summary of solutions to W1, MONDAY version

the real number system would be violated if we used r as the value of thisquotient. We resolve this issue by restricting fractions a

bto cases where b 6= 0.

Thus Domain(g) = R − {0}.2. The same reasoning tells us that 1

4xis defined only for x 6= 0; when we

subtract this function from the function x, which is defined everywhere, weobtain another function defined away from 0; thus Domain(f) = R − {0}.

3. We wish to compose f with itself. For the first application we require thatx 6= 0. The value of that first application, i.e.,

x − 1

4x=

4x2 − 1

x=

(2x − 1)(2x + 1)

x=

1

4·(

x − 12

) (

x + 12

)

x

must not be equal to 0 if it is to be the point where we apply the function fagain. Thus we have to insist that x be different from each of ±1

2, and also

different from 0:

Domain(f ◦ f) = R −{

−1

2, 0, +

1

2

}

.

In that domain the function has the following formula:

(f ◦ f)(x) =

(

x − 1

4x

)

− 1

4(

x − 14x

) =16x4 − 12x2 + 1

4x(4x2 − 1).

4. To follow g by f we first require that x 6= 0, in order to apply g. The valueof g(x) is never 0, so there is no restriction for the application of f . Thus thedomain of f ◦ g is R − {0}, and the formula is

(f ◦ g)(x) =1

x− 1

4 · 1x

=4 − x2

4x.


5. To follow f by g, we need to arrange first that f be defined: that needs x 6= 0.Then we need to arrange that the value of f be non-zero, in order that it liein the domain of g. The value of f(x) in general is 4x2−1

x= (2x−1)(2x+1)

x, so it

will be non-zero if and only if x is different from ±12. Thus

Domain(g ◦ f) = R −{

−1

2, 0, +

1

2

}

.

A simplified formula for the function is given by

(g ◦ f)(x) =1

x − 14x

=4x

4x2 − 1.

Note that, even though this rational function is defined when x = 0, theformula does not apply to g ◦ f at that point, since 1

4xis not defined there.

6. To apply g twice in succession we obtain

(g ◦ g)(x) = g(g(x)) =11x

= x .

However, the equation does not apply when x = 0, since 1x

is not definedthere. The domain of the composition g ◦ g is R − {0}.


f x − 14x

R − {0}g 1

xR − {0}

f ◦ f 16x4−12x2+14x(4x2−1)

R −{

−12, 0, +1

2

}

f ◦ g 4−x2

4xR − {0}

g ◦ f 4x4x2−1

R −{

−12, 0, +1

2

}

g ◦ g x R − {0}

Table 6: Summary of solutions to W1, WEDNESDAY version

THURSDAY: 1. The exponential is defined for all exponents. Whatever value wechoose for x, −x is in the domain of the exponential, so the domain of f is R.

2. The logarithm is defined only for positive numbers. So we cannot ventureoutside of (0,∞). For x in the interval (0, 1) the logarithm is negative; atx = 1 it is equal to 0; and for x > 1 it is positive. Since we will be taking thereciprocal, 1

ln x, we have to exclude the point 1. Thus

Domain(f) = (0, 1) ∪ (1,∞) .


3. Since the domain of f is the whole line, we can apply the function a secondtime indiscriminately, and we have

Domain(f ◦ f) = R .

4. To follow g by f we first must arrange that g be applicable. That requiresthat x be in the union (0, 1) ∪ (1,∞). It matters not what the value of g isat x, since f is defined everywhere. Hence

Domain(f ◦ f) = (0, 1) ∪ (1,∞) .

The formula does not admit simplification,

(f ◦ g)(x) = e−1

ln x .

5. f and g are mutual inverses: If we follow one by the other we obtain theidentity function, i.e., x. The logarithm of f(x) is, therefore −x. Thus farthere is no restriction on x. However, in calculating g, we wish to take areciprocal: for this we require that the logarithm be different from 0: −x 6= 0if and only if x 6= 0. Thus

Domain(g ◦ f) = R − {0} .

The formula for the composition function is simply 1−x

or − 1x.

6. We wish to compose g with itself. In the first application we have to requirethat x be contained in the union (0, 1) ∪ (1,∞). The value of this functionmust be restricted: it must be in the domain of the natural logarithm function,and it must be such that the natural logarithm there is non-zero. To be in thedomain of the function ln we require that 1

ln xbe positive, equivalently that

ln x be positive, equivalently that x > 1. And, we have to exclude that pointx = e, since g(e) = 1, and a second application of g would not be possible,since ln 1 = 0 and 1

0is undefined. Thus

Domain(g ◦ g) = (1, e) ∪ (e,∞) .

The formula for the function does not admit much simplification:

(g ◦ g)(x) =1

ln(

1ln x

) = − 1

ln ln x.

FRIDAY: 1. 1x

is defined for all x 6= 0. Then the sine is defined everywhere, so thedomain of f is R − {0}.



f e−x Rg 1

ln x(0, 1) ∪ (1,∞)

f ◦ f e−e−x

R

f ◦ g e−1

ln x (0, 1) ∪ (1,∞)g ◦ f − 1

xR − {0}

g ◦ g − 1ln ln x

(1, e) ∪ (e,∞)

Table 7: Summary of solutions to W1, THURSDAY version

2. The arcsine function is defined only on the interval [−1, 1]. That is the domainof g.

3. To compose f with itself the first application requires only that x 6= 0. Fora second application we require only that sin 1

x6= 0, which is equivalent to

1x6= nπ where n is any integer. Hence we have

Domain(f ◦ f) = R −{

0,±1

π,± 1

2π,± 1

3π, . . .

}

.

4. To follow g by f we first must restrict x to be in the domain of g, i.e.,[−1 ≤ x ≤ 1]. Then we must also restrict x so that the value g(x) 6= 0, i.e.,that arcsin x 6= 2(0), which is equivalent to excluding the point x = 0. Hence

Domain(f ◦ g) = [−1, 0) ∪ (0, 1] .

The formula is simply

(f ◦ g)(x) = sin

(

2

arcsin x

)

.

and does not admit significant simplifications.

5. The values of the sine function are always in the interval [−1, 1], and thearcsine is defined at all such points. To follow f by g we must, however, firstrestrict x to be non-zero, in order that f be applicable. Hence

Domain(g ◦ f) = R − {0} .

The formula (g ◦ f)(x) =arcsin(sin 1

x)2

does admit simplifications. For example,when |x| > 2

π, (g ◦ f)(x) = 1

2x.



f sin 1x

R − {0}g arcsin x

2

f ◦ f sin(

sin 1x

)

R − {0} −{

± 1nπ

|n = 1, 2, 3, 4, . . .}

f ◦ g sin(

2arcsinx

)

[−1, 0) ∪ (0, 1]

g ◦ farcsin(sin 1

x)2

R − {0}g ◦ g

arcsin( arcsin x2 )

2[−1, 1]

Table 8: Summary of solutions to W1, FRIDAY version

6. The arcsine function takes its values in the interval[

−π2, π

2

]

; hence g takesits values in the interval

[

−π4, π

4

]

, which is entirely contained in the interval[−1, 1]. This means that a second application of the arcsine function is alwayspossible. Hence the domain of g ◦ g is still [−1, 1].


7 Draft Solutions to Quiz Q1

Distribution Date: Released to students 21 October, 2006Draft, subject to correction.

The following sketches of solutions will be given for one specific set of data for each ofthe versions of the quiz.

7.1 Instructions to students

1. Show all your work. Marks are not given for answers alone.

2. You must enclose this question sheet in your folded answer sheet.

3. Time = 20 minutes

4. No calculators are permitted.

7.2 Monday version

1. By using appropriate algebraic operations, compute the following limit or showthat it does not exist:

limx→4

x − 4

5 −√

x + 21

Don’t use L’Hospital’s rule or epsilon-delta methods!

Solution:

(a) As x → 4 the numerator has limit 0, and the limit of the denominator,5 −

√x + 21, is also 0; so the Quotient Law may not be used.

(b) Rationalize the denominator and simplify the ratio:

x − 4

5 −√

x + 21=

(x − 4)(5 +√

x + 21)

(5 −√

x + 21)(5 +√

x + 21)

=(x − 4)(5 +

√x + 21)

25 − (x + 21)

=(x − 4)(5 +

√x + 21)

4 − x

= −(5 +√

x + 21)


(c) Now the Limit Laws may be used:

limx→4

(−5 −√

x + 21) = limx→4

(−5) −√

limx→4

(x + 21)

= −5 −√

25 = −10

If a student used L’Hospital’s Rule, she should have received a grade of 0, as it wasexplicitly forbidden to use that method, and we often make a similar restriction onexamination questions.

2. Find a formula for the inverse of the function

f(x) =1 + 6ex

2 + 7ex

What is the range (image) of the inverse function?

Solution:

(a) Name the “dependent” variable y:

y =1 + 6ex

2 + 7ex

(b) Solve for ex in terms of y:

ex =1 − 2y

7y − 6

(c) Using the logarithm, solve further for x in terms of y:

x = ln1 − 2y

7y − 6

(d) Describe the inverse function:

f−1(y) = ln1 − 2y

7y − 6

The usual convention would be to rename the independent variable of theinverse function as x, but this step is not needed if the preceding step wascarried out using another name (y) for the independent variable:

f−1(x) = ln1 − 2x

7x − 6


(e) The student is to give the image of the inverse function. The instruction (1)is clear: Marks are not given for answers alone.

This problem could be solved directly or indirectly. To solve it indirectly,recall that the image of f−1 is the domain of f . The value of f(x) is definedas a fraction, defined everywhere except where the denominator is 0, i.e.,everywhere except where 2 + 7ex = 0; equivalently, except where ex = −7

2.

But an exponential cannot be negative, so the denominator is non-zero for allvalues of x, and the domain of f is R.

To find the image directly is harder; I would not expect students in MATH140 to be able to articulate a convincing argument of this type. The ratio1−2x7x−6

is undefined when x = 67; it is positive only where 1

2< x < 6

7; as we

wish to take a logarithm, the function is defined only for x in this interval.As x →

(

12

)+, the fraction approaches 0 from the right, and f−1(x) → −∞.

As x →(

67

)−, the fraction approaches ∞ and f−1(x) → +∞. The function is

continuous, and thus (by the Intermediate Value Theorem) takes on all realvalues; that is, the image of f−1 is R.

7.3 Wednesday version


limt→0

(

1

5t− 1

7t2 + 5t

)


Solution:

(a) Algebraic manipulations before evaluating the limit:

1

5t− 1

7t2 + 5t=

1

5t− 1

(7t + 5)t

=

(

1

5− 1

7t + 5

)

· 1

t

=7t

5(7t + 5)· 1

t

=7t

5(7t + 5)t


(b) Divide numerator and denominator by factor t, which is not zero, as we arecomputing the limit as t → 0:

7t

5(7t + 5)t=

7

5(7t + 5)

for t near, but not equal to 0.

(c) Apply the limit laws — I don’t expect students to give their reasons as pre-cisely as the following:

limt→0

(

7

5(7t + 5)

)

=limt→0

7

limt→0

5(7t + 5)

=7

25

TA’s were instructed not to give any marks if either

• the student uses L’Hospital’s Rule; or

• the student computes a difference like ∞−∞.


f(x) =7 − 2 ln x

6 + 1 ln x.

What is the range (image) of the inverse function?

Solution:

(a) Name the dependent variable y:

y =7 − 2 ln x

6 + 1 ln x

(b) Solve for ln x in terms of y:

ln x =7 − 6y

y + 2.

(c) Using exponential, solve for x in terms of y:

x = e

7 − 6y

y + 2 .


(d) Describe the inverse function precisely:

f−1(y) = e

7 − 6y

y + 2 .

The usual convention would be to rename the independent variable of theinverse function as x, but this step is not needed if the preceding step wascarried out using another name (y) for the independent variable:

f−1(x) = e

7 − 6x

x + 2 .

(e) The student is to give the image of the inverse function. The instruction (1)is clear: Marks are not given for answers alone!

This problem could be solved directly or indirectly. To solve it indirectly, recallthat the image of f−1 is the domain of f . The value of f(x) is a fraction,defined everywhere except where the denominator is 0, i.e., everywhere exceptwhere 6 + 1 lnx = 0; equivalently, where ln x = −6

1, i.e., where x = e−6. The

point e−6 must be deleted from the maximum domain of the function ln, i.e.,from the interval (0, +∞). Thus the domain of f is

(

0, e−6)

∪(

e−6, +∞)

,

and this must be the image, or range of the inverse function.

To find the image directly is harder. If a student attempts this approach, usegood judgement in assigning marks. One approach would observe that

limx→−2−

e7−6xx+2 = 0

limx→−∞

e7−6xx+2 = e−6

limx→−2+

e7−6xx+2 = +∞

limx→+∞

e7−6xx+2 = e−6.

The Intermediate Value Theorem, applied to the interval (−∞,−2) yieldsfunction values in the interval (−∞, e−6); applied to the interval (−2, +∞)it yields function values in the interval (e−6, +∞). It’s unrealistic to expectCalculus 1 students to argue in this way.


7.4 Thursday version


limx→4

1x− 1

4

x2 + 3x − 28


Solution:


1x− 1

4

x2 + 3x − 28=

4−x4x

x2 + 3x − 28

=4−x

x

4(x + 7)(x − 4)

=(4 − x)

4x(x + 7)(x − 4)

(b) Divide numerator and denominator by the factor x− 4, which cannot be 0 aswe are computing the limit x → 4:

(4 − x)

4x(x + 7)(x − 4)=

−1

4x(x + 7)

for x near, but not equal to 4.

(c) Now apply the Limit Laws:

limx→4

1x− 1

4

x2 + 3x − 28= lim

x→4

( −1

4x(x + 7)

)

=limx→4

(−1)

limx→4

(4x(x + 7))

=−1

4 · 4(4 + 7)= − 1

176


f(x) = x2 + 8x + 25

defined for all x ≥ 0. What is the domain of the inverse function?

Solution:


(a) Name the dependent variable:

y = x2 + 8x + 25 .

(b) Solve for x in terms of y, keeping in mind that x has been restricted to lie inthe interval x ≥ 0. Students may use the “quadratic formula”, although Iprefer to “complete the square”:

y = x2 + 8x + 25 = (x + 4)2 + 9

⇔ x + 4 = ±√

y − 9

⇔ x = −4 ±√

y − 9.

But the lower choice of sign would make x < 0, which has been excluded.Thus x = −4 +

√y − 9.

(c) Even with the resolution of the double sign there is still restriction on y thatis required to ensure that x ≥ 0. While the expression

√y − 9 is meaningful

for y ≥ 9, we need to ensure that it is no less than 4; this requires thaty ≥ 25. (This observation could have been made earlier, by examining thesum x2 + 8x + 25 when x ≥ 0.)

(d) Give a clear formula for the inverse function:

f−1(y) = −4 +√

y − 9 (y ≥ 25)

orf−1(x) = −4 +

√x − 9 (x ≥ 25) .

(e) The domain of the inverse function is not the interval [9, +∞), even thoughthat is where

√x − 9 is meaningful! The function is defined only on the image

of f , which excluded the points in [9, 25). Thus the domain of f−1 is [25,∞).

7.5 Friday version


limx→6

x2 − 36√6 −√

x


Solution:



x2 − 36√6 −√

x=

(x − 6)(x + 6)√6 −√

x

=(√

x +√

6)(√

x −√

6)(x + 6)√6 −√

x

We could also have proceeded by factorizing the numerator as a difference of4th powers of

√x and

√6.

(b) Division of numerator and denominator by the factor√

x−√

6, which is non-zero, since we are evaluating the limit as x → 6, but need not permit x = 6:

(√

x +√

6)(√

x −√

6)(x + 6)√6 −√

x=

(√

x +√

6)(x + 6)

−1= −(

√x +

√6)(x + 6) ,

for x near but not equal to 6.

(c) Now apply the Limit Laws:

limx→6

(

−(√

x +√

6)(x + 6))

= − limx→6

(√

x +√

6) · limx→6

(x + 6)

= −2√

6 · (6 + 6) = −24√

6 .


f(x) = x2 − 10x + 21

defined for all x ≤ 0. What is the domain of the inverse function?

Solution:

(a) Name the dependent variable:

y = x2 − 10x + 21 .

(b) Solve for x in terms of y, keeping in mind that x has been restricted to lie inthe interval x ≤ 0. Students may use the “quadratic formula”, although Iprefer to “complete the square”:

y = x2 − 10x + 21 = (x − 5)2 − 4

⇔ x − 5 = ±√

y + 4

⇔ x = 5 ±√

y + 4.

But the upper choice of sign would make x > 0, which has been excluded.Thus x = 5 −√

y + 4.


(c) Even with the resolution of the double sign, there is still further restrictionon y that is required to ensure that x ≤ 0. While the expression

√y + 4 is

meaningful for y ≥ −4, we need to ensure that it is no less than 5; this requiresthat y ≥ 21. (This observation could have been made earlier, by examiningthe sum x2 − 10x + 21 when x ≤ 0.)

(d) Give a clear formula for the inverse function:

f−1(y) = 5 −√

y + 4 (y ≥ 21)

orf−1(x) = 5 −

√x + 4 (x ≥ 21) .

(e) The domain of the inverse function is not the interval [−4, +∞), even thoughthat is where

√x + 4 is meaningful! The function is defined only on the image

of f , which excluded the points in [−4, 21). Thus the domain of f−1 is [21,∞).


8 W4, Fourth Written Assignment

Distribution Date: Mounted on WebCT on Monday, October 30th, 2006Your completed solution to this assignment should be submitted, with a copy of this

question sheet at your Tutorial, during the week November 20-24, 2006. Please slip theassignment into your folded quiz answer paper to Quiz Q4. All materials must bearyour name and/or student number. No other method of submission is acceptable.

8.1 Certificate








Consider the function f(x) =(x − a)2 − b2

(x − a)2 + b2on the interval a − 2b ≤ x ≤ a + b, where a

and b are constants determined from your student number, as follows:

Starting from the left, list the non-zero digits in your student num-ber. Ignore the first of these. The second is a; the third is b.

(It could happen that a = b.) Showing all your work, modelled on the solution in thetextbook of Example 8, page 284,

1. Find the global (absolute) maximum and minimum values of the function

f(x) =(x − a)2 − b2

(x − a)2 + b2

using the “Closed Interval Method” on the interval a − 2b ≤ x ≤ a + b.

2. Apply the First Derivative Test to any critical points in the interval.


3. Apply the Second Derivative Test to any critical points in the interval.

4. Explain how you know that f does not have a global maximum on the interval−∞ < x < +∞.

5. Sketch the graph of f , showing the extrema you have found.


9 Draft Solutions to W2, Second Written Assign-

ment

Release Date: Published on WebCT on November 11th, 2006.Students’ completed solutions to this assignment were to be submitted, with a copy ofthis question sheet at Tutorials during the week October 16-20, 2006. Students wereasked to “Please slip the assignment into your folded quiz answer paper to Quiz Q2.

All materials must bear your name and/or student number. No other method ofsubmission is acceptable.”

9.1 Certificate




and assert that my work submitted for W2, A3, A4 does not violateMcGill’s regulations concerning plagiarism.



1. This first problem concerns a pair of functions that may not be discussed at the lec-tures, except incidentally. The purpose of the problem is to allow you to investigatethe functions yourself. You are not asked for rigorous proofs in this problem.

This problem involves, in part, sketching some graphs. The functions in questionhave discontinuities. Please use the following convention. If f has a discontinuityat x = a, please enlarge the point (a, f(a)); this convention is particularly usefulwhere a function is, for example, continuous from the right but not from the left —in that way the reader can see immediately what type of discontinuity is present.Another way of showing such discontinuities is through judicious use of square andround brackets. For example, the graph of

f(x) =

{

−1 if −4 ≤ x < 01 if 0 ≤ x ≤ 4


could be sketched as in Figure 2 on page 50. You are expected to solve this problem

-6sor

-6[

)

Figure 2: Showing a discontinuity in a graph

by using information from your graphs — you are not expected to provide rigorousjustifications for all statements. You do not need to use graph paper — just tryto have your graphs approximately to scale, so that the properties of the functionscan be clearly visible.

Definition 9.1 The floor function ⌊x⌋ is defined as follows: ⌊x⌋ is the largestinteger n such that n ≤ x. (Your textbook [1, Example 10, p. 110] calls thisfunction the Greatest Integer Function, and denotes it by [[x]].)

The ceiling function ⌈x⌉ is defined to be the smallest integer n such that n ≥ x.(This function used to be called the Least Integer Function.)

(a) Sketch a graph of each of the functions ⌊x⌋, ⌈x⌉ for −2 < x < 2. (You don’tneed to use graph paper for these problems, but you may if you wish.)

(b) Sketch graphs of the functions f(x) = ⌊2x⌋ + ⌈2x⌉ and g(x) = ⌊2x⌋ − ⌈2x⌉for −1 ≤ x ≤ 1.

(c) Describe the following sets:

i.

{

a

∣

∣

∣

∣

limx→a−


}

=

ii.

{

a

∣

∣

∣

∣

limx→a−


}

=

iii.

{

a

∣

∣

∣

∣

limx→a+


}

=

iv.

{

a

∣

∣

∣

∣

limx→a+


}

=


v.

{

a

∣

∣

∣

∣

limx→a−

⌊2x⌋ = 2a

}

=

vi.

{

a

∣

∣

∣

∣

limx→a−


⌈x⌉}

=

vii.

{

a

∣

∣

∣

∣

limx→a−

⌊x⌋ = limx→a+

⌊x⌋}

=

(The “set-builder notation” used here is described in [1, Appendix A, p. A3].)

2. Prove the statement assigned for your tutorial day by using the Principle of Math-ematical Induction. For this problem you are expected to provide a very carefulproof. (One version of this problem was on the final examination in MATH 1402005 09.)

Students whose tutorial is on Monday: Let f(x) = x2ex. Prove carefully bymathematical induction that

dnf

dxn(x) =

(

x2 + 2nx + (n − 1)n)

· ex


Students whose tutorial is on Wednesday: Let f(x) = x2e−x. Prove care-fully by mathematical induction that

dnf

dxn(x) = (−1)n

(

x2 − 2nx + (n − 1)n)

· e−x


Students whose tutorial is on Thursday: Let f(x) = 2xe2x. Prove carefullyby mathematical induction that

dnf

dxn(x) = 2n (2x + n) · e2x


Students whose tutorial is on Friday: Let f(x) = 2xe−2x. Prove carefully bymathematical induction that

dnf

dxn(x) = (−2)n (2x − n) · e−2x



9.3 Solutions

1. (a) The graph of ⌊x⌋ consists of horizontal line segments of length 1, each includ-ing its left end-point but not its right; on the x-axis the graph includes theinterval [0, 1), and the next portion of the graph to the right is 1 unit higher,etc.

The graph of ⌈x⌉ also consists of horizontal line segments of length 1, thistime each including its right end-point but not its right; on the x-axis thegraph includes the interval (−1, 0], and the next portion of the graph to theright is 1 unit higher, etc.

(b) The graph of ⌊x⌋+⌈x⌉ consists of open line segments of length 1 lacking bothend points, together with points at even heights — the value of the functionat an integer n is 2n; the pattern is centred around the origin, which is a pointof the graph. The given function was, however ⌊2x⌋ + ⌈2x⌉: the effect is tocompress the previous graph horizontally — the line segments have length 1

2.

The function ⌊2x⌋ − ⌈2x⌉ is mostly constant: its value is −1 at every pointwhich is not half an integer; at the half-integers the function value is 0.

(c) To solve this problem, observe from the graphs or otherwise that

limx→a+

⌊x⌋ = ⌊a⌋ (1)

limx→a−

⌊x⌋ =

{

⌊a⌋ if a is not an integer⌊a⌋ − 1 if a is an integer

(2)

limx→a−

⌈x⌉ = ⌈a⌉ (3)

limx→a+

⌈x⌉ =

{

⌈a⌉ if a is not an integer⌈a⌉ + 1 if a is an integer

(4)

i.

{

a

∣

∣

∣

∣

limx→a−


}

is empty. The floor function is constant

“near” any point x; for points which are not integers, the constant valueis the same on both sides, and the limit of the function is the functionvalue. At integers the constant value on the left side is 1 unit less thanthe constant value on the right. This means that the limit from eitherside exists, since, in a one sided “neighbourhood” of the point on one sidethe function behaves “locally” like a constant. Thus the answers to thisquestion and, by analogous reasoning, the next 3 are all the same: theset of points where the one-sided limits do not exist is empty!

ii.

{

a

∣

∣

∣

∣

limx→a−


}

is empty.


iii.

{

a

∣

∣

∣

∣

limx→a+


}

is empty.

iv.

{

a

∣

∣

∣

∣

limx→a+


}

is empty.

v.

{

a

∣

∣

∣

∣

limx→a−

⌊2x⌋ = 2a

}

= By (2) the defining condition for this set is

⌊2a⌋ = 2a when 2a is not an integer; and⌊2a⌋ − 1 = 2a when 2a is an integer.

When 2a is not an integer, ⌊2a⌋ < 2a; when 2a is an integer, ⌊2a⌋ = 2a.Thus equality cannot hold at any time!

vi. The defining condition for

{

a

∣

∣

∣

∣

limx→a−


⌈x⌉}

is

⌊a⌋ = ⌈a⌉ if a is not an integer; and⌊a⌋ − 1 = ⌈a⌉ if a is an integer,

conditions that can never hold, since, in the first case

⌊a⌋ < a < ⌈a⌉

and, in the second case,

⌊a⌋ − 1 = a − 1 < a = ⌈a⌉ .

vii. The condition limx→a−

⌊x⌋ = limx→a+

⌊x⌋ is satisfied precisely when a is not an

integer.

2. The second problem involved the Principle of Mathematical Induction. While thistopic was discussed in detail in MATH 140 2005 09, we did not have time fora thorough treatment in the lectures this year. Nevertheless, we believe it is anappropriate topic for a written assignment: the students had to read up aboutthe principle from their textbook, look up worked examples, and then answer thequestion as posted. Because the topic has not seen significant discussion in thelectures, I would like the grading to be generous; students have been assured thatthis topic will not be on the final examination this year.

Students whose tutorial is on Monday: f(x) = x2ex. Let Sn denote the state-ment

dnf

dxn(x) =

(

x2 + 2nx + (n − 1)n)

· ex , (5)

for positive integers n.


“Anchor” of Induction, or “Base Step”: Applying the Product Rule ofDifferentiation, we obtain

df

dx(x) = (2x)ex + (x2)ex

=(

x2 + 2(1)x + (0)(1))

ex

which proves S1.

Induction Hypothesis: Assume that Sn is true.

Induction Step: Then

dn+1

dxn+1f(x) =

d

dx

(

dnf

dxn(x)

)

=d

dx

{(

x2 + 2nx + (n − 1)n)

ex}

by the Induction Hypothesis

= (2x + 2n) ex +(

x2 + 2nx + (n − 1)n)

ex

by the Product Rule

=(

x2 + (2n + 2)x + ((n − 1)n + 2n))

ex

=(

x2 + 2(n + 1)x + n(n + 1))

ex

proving Sn+1.

Conclusion: Hence, by Mathematical Induction, Sn is true for all positiveintegers n.

Students whose tutorial is on Wednesday: f(x) = x2e−x. Let Sn denote thestatement

dnf

dxn(x) =

(

x2 − 2nx + (n − 1)n)

· e−x , (6)



df

dx(x) = (2x)e−x − (x2)e−x

=(

−x2 + 2(1)x + (0)(1))

e−x

= (−1)1(

x2 − 2(1)x + (0)(1))

e−x

which proves S1.




dn+1

dxn+1f(x) =

d

dx

(

dnf

dxn(x)

)

=d

dx

{

(−1)n(

x2 − 2nx + (n − 1)n)

e−x}


= (−1)n (2x − 2n) e−x − (−1)n(

x2 − 2nx + (n − 1)n)

e−x

by the Product Rule

= (−1)n+1(

x2 − (2n + 2)x + ((n − 1)n + 2n))

e−x

= (−1)n+1(

x2 − 2(n + 1)x + n(n + 1))

e−x

proving Sn+1.


Students whose tutorial is on Thursday: f(x) = 2xe2x. Let Sn denote thestatement

dnf

dxn(x) = 2n(2x + n) · e2x , (7)



df

dx(x) = 2e2x + 2(2x)e2x

= 21(2x + 1)e2x

which proves S1.



dn+1

dxn+1f(x) =

d

dx

(

dnf

dxn(x)

)

=d

dx

{

2n(2x + n)e2x}


= 2n (2) e2x + 2n (2x + n) · 2 · e2x

by the Product Rule

= 2n (4x + (2n + 2)) e2x

= 2n+1 (2x + (n + 1)) e2x


proving Sn+1.


Students whose tutorial is on Friday: f(x) = 2xe−2x. Let Sn denote thestatement

dnf

dxn(x) = 2n(2x − n) · e−2x , (8)



df

dx(x) = 2e−2x − 2(2x)e−2x

= 21(−2x + 1)e−2x = (−2)1(2x − 1)e−2x

which proves S1.



dn+1

dxn+1f(x) =

d

dx

(

dnf

dxn(x)

)

=d

dx

{

(−2)n(2x − n)e−2x}


= (−2)n (2) e−2x + (−2)n (2x − n) · (−2) · e−2x

by the Product Rule

= (−2)n (−4x + (2n + 2)) e−2x

= (−2)n+1 (2x − (n + 1)) e−2x

proving Sn+1.



10 Draft Solutions to W3, Third Written Assign-

ment

Release Date: November 11th, 2006; subject to correction.Your completed solution to this assignment was to be submitted, with a copy of this

question sheet at your Tutorial, during the week October 30 - November 03, 2006. Youwere asked to “Please slip the assignment into your folded quiz answer paper to QuizQ3. All materials must bear your name and/or student number. No other method of


10.1 Certificate




and assert that my work submitted for W3, A4 does not violate McGill’sregulations concerning plagiarism.



1. A function y = y(x) is defined implicitly by the following equation. You are,showing all your work, to determine the values of y′(x) and y′′(x) at the givenpoint.

Tutorial Day Defining Equation Evaluate at the Point

Monday x3 + y4 = 2xy (x, y) = (1, 1)

Wednesday x4 + y3 = 2xy (x, y) = (1, 1)

Thursday x3 − y4 = 2xy (x, y) = (−1, 1)

Friday −x4 + y3 = 2xy (x, y) = (1,−1)

2. The six hyperbolic functions are defined in terms of sinh and cosh, which, in turn,are defined in terms of exponentials. These six functions satisfy identities that


resemble properties of the trigonometric functions. Beginning with each functionexpressed in terms of exponentials, and showing all your work, prove the identityassigned below to your Tutorial day:

Tutorial Day Identity to be Proved

Monday cosh(x + y) = cosh x · cosh y + sinh x · sinh y

Wednesday tanh(x − y) =tanh x − tanh y


Thursday ddx

tanh x = sech2x

Friday ddx

csch x = −csch x · coth x

10.3 Solutions

1. Students whose tutorial is on Monday: Differentiating the defining equation“implicitly” with respect to x, we obtain

3x2 + 4y3 · y′ = 2(y + x · y′) , (9)

which we evaluate at (x, y) = (1, 1), to obtain y′(1) = −12. We differentiate

equation (9) implicitly with respect to x to obtain

6x + 12y2 · (y′)2 + 4y3 · y′′ = 2(y′ + y′ + xy′′) (10)

and substitute the values (x, y, y′) =(

1, 1,−12

)

, obtaining that y′′(1) = −4.(We could have solved equation (9) for

y′ =3x2 − 2y

2x − 4y3,

and then differentiated with respect to x to obtain

y′′(x) =(6x2 − 24xy3 + 4y) + (−16y3 + 36x2y2 − 4x)y′

(2x − 4y3)2 .

Then we could again have substituted (x, y, y′) =(

1, 1,−12

)

. That, however,would not have made the calculations any easier.)

Students whose tutorial is on Wednesday: Differentiating the defining equa-tion “implicitly” with respect to x, we obtain

4x3 + 3y2 · y′ = 2(y + x · y′) , (11)


which we evaluate at (x, y) = (1, 1), to obtain y′(1) = −2. We differentiateequation (11) implicitly with respect to x to obtain

12x2 + 6y2 · (y′)2 + 3y2 · y′′ = 2(y′ + y′ + xy′′) (12)

and substitute the values (x, y, y′) = (1, 1,−2), obtaining that y′′(1) = −44.(We could have solved equation (11) for y′ and then differentiated with respectto x; that, however, would not have made the calculations any easier.)

Students whose tutorial is on Thursday: Differentiating the defining equation“implicitly” with respect to x, we obtain

3x2 − 4y3 · y′ = 2(y + x · y′) , (13)

which we evaluate at (x, y) = (−1, 1), to obtain y′(1) = 12. We differentiate

equation (13) implicitly with respect to x to obtain

6x − 12y2 · (y′)2 − 4y3 · y′′ = 2(y′ + y′ + xy′′) (14)

and substitute the values (x, y, y′) =(

1, 1, 12

)

, obtaining that y′′(1) = −112.

(We could have solved equation (13) for y′ and then differentiated with respectto x; that, however, would not have made the calculations any easier.

Students whose tutorial is on Friday: Differentiating the defining equation “im-plicitly” with respect to x, we obtain

−4x3 + 3y2 · y′ = 2(y + x · y′) , (15)

which we evaluate at (x, y) = (1,−1), to obtain y′(1) = 2. We differentiateequation (15) implicitly with respect to x to obtain

−12x2 + 6y · (y′)2 + 3y2 · y′′ = 2(y′ + y′ + xy′′) (16)

and substitute the values (x, y, y′) = (1, 1, 2), obtaining that y′′(1) = 44. (Wecould have solved equation (15) for y′ and then differentiated with respect tox; that, however, would not have made the calculations any easier.

2. Students whose tutorial is on Monday: I believe it will be easier to start thisproof on the right side of the identity, which is the more complicated.

cosh x · cosh y + sinh x · sinh y

=ex + e−x

2· ey + e−y

2+

ex − e−x

2· ey − e−y

2

=(ex + e−x) (ey + e−y) + (ex − e−x) (ey − e−y)

4

=(ex+y + ex−y + e−x+y + e−x−y) + (ex+y − ex−y − e−x+y + e−x−y)

4

=2ex+y + 2e−(x+y)

4= cosh(x + y) . �


Students whose tutorial is on Wednesday: In this example as well, I believeit will be easier to start this proof on the right side of the identity, which isthe more complicated.

tanh x − tanh y


=

sinh x

cosh x− sinh y

cosh y

1 − sinh x

cosh x· sinh y

cosh y

=sinh x · cosh y − cosh x · sinh y

cosh x · cosh y − sinh x · sinh y

=

ex − e−x

2· ey + e−y

2− ex + e−x

2· ey − e−y

2ex + e−x

2· ey + e−y

2− ex − e−x

2· ey − e−y

2

=(ex+y + ex−y − e−x+y − e−x−y) − (ex+y − ex−y + e−x+y − e−x−y)

(ex+y + ex−y + e−x+y + e−x−y) − (ex+y − ex−y − e−x+y + e−x−y)

=2ex−y − 2e−x+y

2ex−y + 2e−x+y

=ex−y − e−(x−y)

ex−y + e−(x−y)

=

ex−y − e−(x−y)

2ex−y + e−(x−y)

2

=sinh(x − y)

cosh(x − y)= tanh(x − y)

Students whose tutorial is on Thursday:

d

dxtanh x =

d

dx

(

sinh x

cosh x

)

=d

dx

ex − e−x

2ex + e−x

2

=

d

dx

(

ex − e−x

ex + e−x

)

=(ex + e−x)(ex + e−x) − (ex − e−x)(ex − e−x)

(ex + e−x)2

=(e2x + 2 + e−2x) − (e2x − 2 + e−2x)

(ex + e−x)2=

4

(ex + e−x)2

=

(

2

ex + e−x

)2

=

(

1

cosh x

)2

= (sech x)2 = sech2x


Students whose tutorial is on Friday:

d

dxcsch x =

d

dx

(

1

sinh x

)

=d

dx

(

2

ex − e−x

)

= −2ex + e−x

(ex − e−x)2

= −ex + e−x

2(

ex − e−x

2

)2

= −(

1

sinh x

)

·(

cosh x

sinh x

)

= −csch x · coth x �



Distribution Date: November 11th, 2006(subject to correction)

The following draft sketches of solutions will be given for one specific set of data for eachof the versions of the quiz.






11.1.1 Monday version

1. By using appropriate algebraic manipulations, compute the following limit or showthat it does not exist:

limx→−∞

(√x2 + 9x −

√x2 + 4x

)


Solution:

(a) If a student attempts to evaluate the limit of the difference by using theDifference Law, you should give ZERO for the entire problem. The differencemust be transformed BEFORE applying any of the Limit Laws!

(b) Multiply and divide by a function which will rationalize the given function(at the expense of introducing a complicated denominator):

(√x2 + 9x −

√x2 + 4x

)

·(√

x2 + 9x +√

x2 + 4x)

(√x2 + 9x +

√x2 + 4x

)

(c) Simplify the numerator:

(√x2 + 9x −

√x2 + 4x

)

·(√

x2 + 9x +√

x2 + 4x)

= (x2+9x)−(x2+4x) = 5x


(d) The function has now been transformed from a difference of the form ∞−∞,which we can’t evaluate, to a ratio of the form ∞

∞ , which we also can’t evaluate.But we can divide numerator and denominator by a factor which will thenpermit us to use the Quotient Law. To do this we need a factor like x in thedenominator. This is obtained using the property of exponents that

√AB =

√A ·

√B

where A and B are non-negative. In this case we have

√x2 + 9x =

√

x2

(

1 +9

x

)

√x2 + 4x =

√

x2

(

1 +4

x

)

(e) Using the fact that√

x2 = |x| we have

(√x2 + 9x −

√x2 + 4x

)

=x

|x| ·5

√

1 + 9x

+√

1 + 4x

(f) Now, as x → −∞, the first factor → −1, and the second → 52; so the limit is

−52.

You should not give full marks if the student’s solution doesn’t make it clear howshe is using the fact that the limit is taken as x → −∞; that is, if the argumentwritten down could apply equally well to the limit at +∞, then something ismissing in the solution, and full marks have definitely not been earned!

2. Use the limit definition of the derivative to compute the derivative of

f(x) =1

(x + 8)2

at x = −7. Don’t use any of the differentiation rules of Chapter 3 of Stewart!

Solution:

(a) Give the definition of the derivative as a limit (several alternatives are possi-ble):

f ′(−7) = limh→0

f(−7 + h) − f(−7)

h


(b) Express that general limit in the particular case of the given function

f ′(−7) = limh→0

1(−7+h)2

− 1(−7)2

h

(c) Simplify the fraction algebraically:

limh→0

1(−7+h)2

− 1(−7)2

h= lim

h→0

(−7)2 − (−7 + h)2

h(−7)2(−7 + h)2

(d) Simplify the numerator by expanding the squares:

limh→0

(−7)2 − (−7 + h)2

h(−7)2(−7 + h)2= lim

h→0

14h − h2

h(−7)2(−7 + h)2

(e) Divide numerator and denominator by h:

limh→0

14 − h

(−7)2(−7 + h)2

(f) Now the Quotient Law may be used:

limh→0

14 − h

(−7)2(−7 + h)2=

14 − 0

(−7)2(−7 + 0)2=

14

492=

2

343.

Don’t take off marks if students can’t compute the numerical powers withouta calculator.

11.1.2 Wednesday version

1. Show that the equation3x = 2x + 3

has at least one solution between x = −2 and x = 0. Justify all your steps!

Solution:

(a) The first step in a problem like this is to identify a function to which to applythe Intermediate Value Theorem. One such function — not the only one —is f(x) = 3x − (2x − 3).

(b) A next step is to identify the closed interval on which to apply the theorem.Here the data have been given, the student must simply show that she hasextracted this information from the problem: the interval is [−2, 0].


(c) The student must at least OBSERVE that the function is continuous on thegiven closed interval. That is, a word like continuous or continuity shouldappear explicitly in the solution.

(d) The function must be evaluated at the end-points of the given interval, andit must be shown that the two values are on opposite sides of 0:

f(−2) =1

9− (−4 + 3) =

10

9> 0

f(0) =1

1− (0 + 3) = −2 < 0

(e) Conclude by applying the Intermediate Value Theorem that the function as-sumes the value 0 somewhere in the interval.

2. Let

f(x) =2x2 + 2x + 1

4ex.

Compute f ′(x) and find the equation of the tangent line to the graph of f(x) atx = 0. You may use all the differentiation rules you know, but you are expectedto show all steps of your solution.

Solution:

(a) There could be different ways in which the differentiation rules can be imple-mented; only one is given here. Steps must be shown in the solution — donot give full marks if only the answer is given!

f ′(x) =(4x + 2) · (4ex) − (2x2 + 2x + 1) · (4ex)

(4ex)2

=−2x2 + 2x + 1

4ex.

(b) Hence f ′(0) = 14.

(c) When x = 0, f(x) = 14.

(d) An equation for the tangent line at the point (x, y) =(

0, 14

)

on the graph off is

y − 1

4=

1

4(x − 0)

or x − 4y = −1.


11.1.3 Thursday version

1. Let

f(x) =

9

xif x < −3

ab if x = −3

−6 + b + ln(x + 4) if x > −3

.

Showing all your work, determine what values must be chosen for a and b in orderto make this function continuous at x = −3?

Solution:

(a) Find the limit from the left at x = −3. This entails identifying which line ofthe definition is to be used. Here lim

x→−3−f(x) = 9

−3= −3.

(b) Find the limit from the right at x = −3. This entails identifying which lineof the definition is to be used. Here lim

x→−3+f(x) = −6 + b + ln(−3 + 4) =

−6 + b + 0 = b − 6.

(c) Continuity requires that the limit exists; that is equivalent to the existence ofthe 2 one-sided limits, and the equality of these 2 one-sided limits :

−3 = b − 6

so b = 3.

(d) Continuity requires that the limit be equal to the function value. Use thepreceding facts to observe the value of the 2-sided limit to be −3.

(e) Finally, the limit must be equal to the function value:

−3 = ab ⇒ −3 = 6a ⇒ a = −1

2.

2. Let

f(x) =−4 + 9ex

−3 + 7ex.

Compute f ′(x) and determine the equation of the tangent line to the graph of fat x = 0.

Solution:


(a) Applying the Differentiation Rules, we obtain

f ′(x) =(9ex) · (−3 + 7ex) − (−4 + 9ex) (7ex)

(−3 + 7ex)2

=ex

(−3 + 7ex)2 .

(b) When x = 0, f ′(0) = 142 .

(c) When x = 0, f(x) =−4 + 9

−3 + 7=

5

2.

(d) Hence an equation for the line through the point x = 0 and having the correctslope is

y − 5

2=

1

16(x − 0) .

Or, equivalently, x − 16y = −40.

11.1.4 Friday version

1. Use the limit definition of the derivative to compute the derivative of

f(x) =√

3x + 4 .

Don’t use any of the differentiation rules of Chapter 3 of Stewart!

Solution:

(a) Give the definition of the derivative as a limit (several alternatives are possi-ble):

f ′(x) = limh→0

f(x + h) − f(x)

h

(b) Express that general limit in the particular case of the given function

f ′(x) = limh→0

√

3(x + h) + 4 −√

3x + 4

h

(c) Simplify the fraction algebraically:

limh→0

√

3(x + h) + 4 −√

3x + 4

h= lim

h→0

√

3x + (3h + 4) −√

3x + 4

h


(d) Rationalize the numerator by multiplying the fraction by

√

3x + (3h + 4) +√

3x + 4√

3x + (3h + 4) +√

3x + 4

(e) Simplify the numerator

limh→0

(

√

3x + (3h + 4) −√

3x + 4

h

)(

√

3x + (3h + 4) +√

3x + 4√

3x + (3h + 4) +√

3x + 4

)

= limh→0

(3x + 3h + 4) − (3x + 4)

h(√

3x + (3h + 4) +√

3x + 4)

= limh→0

3h

h(√

3x + (3h + 4) +√

3x + 4)

(f) Divide numerator and denominator by h:

limh→0

3√

3x + (3h + 4) +√

3x + 4

(g) Now the Quotient Law may be used (or continuity may be invoked):

limh→0

3√

3x + (3h + 4) +√

3x + 4=

3√

3x + 4) +√

3x + 4=

3

2√

3x + 4

2. Let f be a differentiable function defined for all real numbers. Assume it is knownthat f(0) = 9. Compute the derivative of the function

g(x) =xf(x) − 8

ex

at x = 0.

Solution:

(a)

g′(x) =(5xf(x) − 8)′ · ex − (5xf(x) − 8) · (ex)

(ex)2

by the Quotient Rule

=(5xf(x))′ · ex − (5xf(x) − 8) · (ex)

(ex)2


by the Sum Rule

=(5f(x) + 5xf ′(x)) · ex − (5xf(x) − 8) · (ex)

(ex)2

by the Product Rule

=5(f(x) + xf ′(x) − xf(x)) + 8

ex

(b)

g′(0) =5f(0) + 8

1= 5 · 9 + 8 = 53.



Distribution Date: November 11th, 2006; (subject to correction)

The following draft sketches of solutions will be given for one specific set of data foreach of the versions of the quiz; in some problems the variations of the versions involvedifferent functions, so please be careful.






12.1.1 Monday version

1. In answering each of the following problems, show all your work, and simplify youranswer as much as possible.

(a) Find f ′(x), when f(x) = arcsin(4x − 1).

(b) Evaluate limx→0

sin 5x

sin 2xwithout using L’Hospital’s Rule.

Solution:

(a) i.

f ′(x) =1

√

1 − (4x − 1)2· d

dx(4x − 1)

ii. Evaluating the 2nd factor:

f ′(x) =1

√

1 − (4x − 1)2· 4

iii. Simplifying the first factor:

f ′(x) =1√

8x − 16x2· 4 =

2

2x − 4x2.

(b) Remember, if the student’s proof is essentially using L’Hospital’s Rule, thenshe should not receive any marks.


i. Transform so that numerator and denominator involve limits of sin x/x:

sin 5x

sin 2x=

sin 5x

5xsin 2x

2x

· 5x

2x

ii.

limx→0

sin 5x

sin 2x= lim

x→0

sin 5x

5xsin 2x

2x

· 5x

2x

=limx→0

sin 5x

5x

limx→0

sin 2x

2x

· limx→0

5x

2x

=lim5x→0

sin 5x

5x

lim2x→0

sin 2x

2x

· limx→0

5

2

iii.

limx→0

sin 5x

sin 2x=

1

1· 5

2=

5

2

2. Suppose that4x2 + 9y2 = 36 . (17)

(a) Find y′ only by implicit differentiation. Your solution will involve both x andy.

(b) Assuming y > 0, solve the given equation explicitly for y, and differentiatethe solution with respect to x to obtain an explicit formula for y′.

(c) Substitute the solution you found for y in part (b) into your derivative y′

found in part (a), and show that the result is the same as the value you foundfor y′ in (b).

Solution:

(a) Differentiate equation (17) implicitly with respect to x:

4 · 2x · dx

dx+ 9 · 2y dy

dx= 0


which can be solved to yield

dy

dx= −4x

9y.

(b) If y > 0,

y =2

3

√9 − x2 . (18)

Differentiating the explicit formula for f in (18) yields

dy

dx=

2

3· 1

2· 1√

9 − x2· (−2x) = −2x

3· 1√

9 − x2.

(c)

−4x

9y= − 4x

9(

23

√9 − x2

)

= −2

3· 1√

9 − x2

which is the value that we determined in part (b) by explicit differentiation.

12.1.2 Wednesday version


(a) Find f ′(x), when f(x) = arctan(√

x2 + 1 − x)

.


sin 7x2

sin 3x2without using L’Hospital’s Rule.

Solution:

(a) i.

f ′(x) =1

1 +(√

x2 + 1 − x)2 · d

dx

(√x2 + 1 − x

)

ii.

f ′(x) =1

1 +(√

x2 + 1 − x)2 ·

(

1

2· 2x√

x2 + 1− 1

)

iii. Final simplification:

f ′(x) = −1

2· 1

x2 + 1



i. Transform so that numerator and denominator involve limits of functionsof the form sin x/x:

sin 7x2

sin 3x2=

sin 7x2

7x2

sin 3x2

3x2

· 7x2

3x2

ii.

limx→0

sin 7x2

sin 3x2= lim

x→0

sin 7x2

7x2

sin 3x2

3x2

· 7x2

3x2

=limx→0

sin 7x2

7x2

limx→0

sin 3x2

3x2

· limx→0

7x2

3x2

=lim

7x2→0

sin 7x2

7x2

lim3x2→0

sin 3x2

3x2

· limx→0

7

3

iii.

limx→0

sin 7x2

sin 3x2=

1

1· 7

3=

7

3

2. Suppose that

x2 + 64y2 =(

x2 + 2y2 − 24x)2

.

Determine an equation — simplified as much as possible — for the tangent to thiscurve at the point with coordinates (0, 4).

Solution:

(a) Differentiate the given equation implicitly with respect to x:

2x + 64 · 2y · y′ = 2(

x2 + 2y2 − 24x)

· (2x + 4yy′ − 24)

(b) Substitute (x, y) = (0, 4) in the equation for y′ to determine the value of thatderivative when x = 4:

0 + 512y′ = 2(32)(16y′ − 24) ⇒ y′ = 3 .


Alternatively, one could determine a general formula for y′ for any point onthe curve, and then make the substitution.

(c) Give the equation with of the line through (4, 3) with the given slope:

y − 4 = 3(x − 0) or y = 3x + 4 .

12.1.3 Thursday version


(a) Find f ′(x), when f(x) = x arcsin x +√−x2 + 1.

Solution:

i.

f ′(x) = 1 · arcsin x + x · 1√−x2 + 1

+d

dx

√−x2 + 1

ii.

f ′(x) = 1 · arcsin x + x · 1√−x2 + 1

+1

2· 1√

−x2 + 1· (−2x)

iii. Final simplification:

f ′(x) = 1 · arcsin x + x · 1√−x2 + 1

− x · 1√−x2 + 1

= arcsin x .


sin(cos 2x)

sec 2xwithout using L’Hospital’s Rule.

Solution: For this problem l’Hospital’s Rule could not apply, since the limit ofthe denominator exists, and is not zero. You certainly should not give morethan 0 if a student appeals to l’Hospital’s Rule here!

i. The limit of the denominator is sec 0 = 1, by the continuity of the secantfunction.

ii. The limit of the numerator is

limx→0

sin(cos 2x) = sin(limx→0

cos 2x)

since the sine function is continuous at limx→0

cos 2x

= sin 1.

iii. By the Quotient Law, limx→0

sin(cos 2x)

sec 2x=

limx→0

sin(cos 2x)

limx→0

sec 2x=

sin 1

1= sin 1.


2. Suppose that3(x2 + y2)2 = 25

(

x2 − y2)

.


Solution:

(a) Differentiate the given equation implicitly with respect to x:

3 · 2 · (x2 + y2) · (2x + 2yy′) = 25 (2x − 2yy′)

(b) Solve for

y′ =x

y· 25 − 6x(x2 + y2)

6(x2 + y2) + 25.

and evaluate at (x, y) = (2, 1):

y′ = − 2

11.

Alternatively, substitute (x, y) = (2, 1) before solving for the value of thederivative at the point.

(c) Give the equation with of the line through (2, 1) with the given slope:

y − 1 =

(

− 2

11

)

(x − 2) or 2x + 11y = 15 .

12.1.4 Friday version

1. In each of the following problems, show all your work, and simplify your answer asmuch as possible.

(a) Find f ′(x), when f(x) =(√

−x2 + 1)

arccos x.

(b) Evaluate limt→0

t3

tan3(8t)without using L’Hospital’s Rule.

Solution:

(a) i.

f ′(x) =1

2

1√−x2 + 1

· (−2x) · arccos x +√−x2 + 1 · d

dxarccos x

ii.

f ′(x) =1

2

1√−x2 + 1

· (−2x) · arccos x +√−x2 + 1 · −1√

1 − x2


iii. Simplification:

f ′(x) =1

2

1√−x2 + 1

· (−2x)) · arccos x +√−x2 + 1 · −1√

1 − x2

= −x arccos x√1 − x2

− 1


i. Transform so that numerator and denominator involve limits of functionsof the form sin t

tor tan t

t:

t3

tan3 8t=

1(

tan 8t

8t

)3 · 1

83=

cos3 8t(

sin 8t

8t

)3 · 1

83

ii. Now prepare to use the Quotient Law for Limits:

limt→0

t3

tan3 8t= lim

t→0

cos3 8t(

sin 8t

8t

)3 · 1

83

=limt→0

cos3 8t

limt→0

(

sin 8t

8t

)3 · 1

83

=limt→0

cos3 8t(

limt→0

sin 8t

8t

)3 · 1

83

=

(

lim8t→0

cos 8t)3

(

lim8t→0

sin 8t

8t

)3 · 1

83

iii. Complete the computations and simplify:

=13

13· 1

83=

1

512

2. Suppose thaty2(y2 − 16) = x2(x2 − 5) .



Solution:

(a) Prior to implicit differentiation, prepare by replacing the products of polyno-mials by sums of constant multiples of powers:

y4 − 16y2 = x4 − 5x2 .

(b) Differentiate implicitly with respect to x:

4y3y′ − 16yy′ = 4x3 − 10x .

(c) Determine a general formula for the derivative:

y′ =x

y· 2x2 − 5

2(y2 − 2)

(d) Evaluate y′ at the given point:

y′(0) = 0 .

(e) Give the equation of the line with that slope through the given point: y = 4.


13 Draft Solutions to W4, Fourth Written Assign-

ment

Distribution Date: Monday, November 27th, 2006Completed solutions to this assignment were due to be submitted, with a copy of this

question sheet at Tutorials during the week November 20-24, 2006. Students wereinstructed to “Please slip the assignment into your folded quiz answer paper to QuizQ4. All materials must bear your name and/or student number. No other method of


13.1 Certificate








Consider the function f(x) =(x − a)2 − b2

(x − a)2 + b2on the interval a − 2b ≤ x ≤ a + b, where a

and b are constants determined from your student number, as follows:

Starting from the left, list the non-zero digits in your student num-ber. Ignore the first of these. The second is a; the third is b.

(It could happen that a = b.) Showing all your work, modelled on the solution in thetextbook of Example 8, page 284,

1. Find the global (absolute) maximum and minimum values of the function

f(x) =(x − a)2 − b2

(x − a)2 + b2

using the “Closed Interval Method” on the interval a − 2b ≤ x ≤ a + b.


2. Apply the First Derivative Test to any critical points in the interval.

3. Apply the Second Derivative Test to any critical points in the interval.

4. Explain how you know that f does not have a global maximum on the interval−∞ < x < +∞.

5. Sketch the graph of f , showing the extrema you have found.

Solution:

1.

f(x) =(x − a)2 − b2

(x − a)2 + b2= 1 − 2b2

(x − a)2 + b2

⇒ f ′(x) =4(x − a)b2

((x − a)2 + b2)2

⇒ f ′′(x) =4b2(

b −√

3(x − a)) (

b +√

3(x − a))

((x − a)2 + b2)3

Since f ′ is defined for all x, we can set f ′(x) equal to 0 to find all critical points:the only solution is x = a; f(a) = −1.

The values of f at the end-points of the given interval are 35

and 0. The globalmaximum is attained at x = a − 2b, with the value 3

5; the global minimum is at

x = a, with the value −1.

2. For x < a, f ′(x) < 0, while, for x > a, f ′(x) > 0. (Usually we should onlybe looking “within a small neighbourhood” of x = a, but, for this function, thederivative has the same sign throughout the ray to the left of x = a, and the samesign throughout the ray to the right of x = a.) Thus f ′ changes from negative topositive at the critical point, and the point must be a local (relative) minimum.

3. f ′′(a) =−4b2(−b)(b)

(b2)3=

4

b2> 0, implying that the critical point is a local (relative)

minimum.

4. As x → ±∞, f(x) → 1. But, for all x, 1 − f(x) = 2b2

(x−a)2+b2> 0. What could the

global maximum be? If you claimed the maximum occurred at some point x = x0;then, by taking x > x0, I would have a point where f(x) > f(x0), and x0 couldn’tbe a maximum point. And you can’t claim the maximum value is 1, since the value1 is never attained. There is no maximum!

5. see Figure 3 on page 80


1

0

0.5

-0.5

x

20100-10-20

-1

Figure 3: Graph of the Function(x − a)2 − b2

(x − a)2 + b2and its horizontal asymptote, y = 1, when

a = 2, b = 5



Distribution Date: Mounted on WebCT on December 01, 2006Draft, subject to correction.

The following sketches of solutions will be given for one specific set of data for each ofthe versions of the quiz.




3. Time = 45 minutes.


14.2 Monday version

1. [10 MARKS] Find a function

f(x) = ax2 + bx + c (19)

where a, b and c are constants to be determined, if f has all of the followingproperties:

(a) f(1) = 0

(b) f ′(3) = −13

(c) f ′′(−7) = −4

You must show a FULL solution—a formula for the function alone could be worthzero marks.

Solution:

(a) Imposition of the first boundary condition yields

0 = f(1) = a + b + c . (20)

(b) Differentiation of equation (19) yields

f ′(x) = 2ax + b . (21)

Imposition of the second boundary condition yields

−13 = f ′(3) = 2a(3) + b . (22)


(c) Differentiation of equation (21) yields

f ′′(x) = 2a . (23)

Imposition of the third boundary condition yields

−4 = f ′′(−7) = 2a , (24)

implying that a = −2.

(d) Back substitution of this value in equation (22) yields

−13 = 2(−2)(3) + b ,

implying that b = −1. Back substitution of the values of a and b in equation(20) yields

0 = (−2) + (−1) + c ,

which we may solve to obtain c = 3. The function we seek is now completelydetermined:

f(x) = −2x2 − 1x + 3 .

2. [10 MARKS] Use logarithmic differentiation to find the derivative of the functiony = x4x.

Solution:

(a) First apply logarithms to both sides of the given defining equation:

ln y = 4x ln x .

(b) Next differentiate implicitly with respect to x:

y′

y= 4 lnx + 4(x)

1

x= 4 lnx + 4 ,

implying thaty′ = 4y · (lnx + 1) .

(c) Now restore the known explicit formula for y(x):

y′ = 4x4x · (ln x + 1) .

3. [10 MARKS] If sinh x = −1, find the value of ex and then the value of cosh x.

Solution:


(a) In terms of exponentials, the given equation yields ex−e−x

2= −1 which implies

that e2x − 1 = −2ex, which may be viewed as a quadratic equation,

(ex)2 + 2ex − 1 = 0 ,

implying that ex = −1 ±√

2. But, of the two values given, one must bediscarded, as an exponential cannot be negative. Thus we have shown that

ex =√

2 − 1 .

(b) We may now determine the value of the hyperbolic cosine:

cosh x =ex +

1

ex

2

=

√2 − 1 + 1√

2−1

2

=

√2 − 1 + 1√

2−1·√

2+1√2+1

2

=

√2 − 1 +

√2+11

2

=√

2 .

(The instructions clearly stated that it was necessary to find ex first. Otherwise onecould have applied the identity cosh2 x−sinh2 x = 1 to determine that cosh x = ±2;then one could have observed that the hyperbolic cosine, being a positive multipleof the sum of exponentials, must be positive, etc.)

4. [10 MARKS] Find the global (absolute) extrema of the function

y =ln(4x)

x

on the closed interval [1, 8]. You must show all of your work—correct numericalanswers alone may not earn any marks.

Solution:

(a) To find the critical points, differentiate

y′(x) =1 − ln(4x)

x2.


(b) Observe that there are no points in the given interval where the function failsto be differentiable.

(c) Determine the critical points by setting the derivative equal to 0: ln 4x = 1 ⇒x = e

4< 1, which is not in the interval. Thus there are NO critical points!

(d) Compare the values of the function at the end points:

y(1) = ln 4 = 2 ln 2

y(8) =ln 32

8=

5

8ln 2

so the global maximum is at 1, and the maximum value is ln 4; the globalminimum is at 8, and the minimum value is 5

8ln 2.

14.3 Wednesday version

1. [10 MARKS] Find a function

f(x) = a cos x + b sin x + c (25)

where a, b and c are constants to be determined, if f has all of the followingproperties:

(a) f(0) = −1

(b) f ′(−π4) = −2

√2

(c) f ′′(π) = 1

You must show a FULL solution—a formula for the function alone could be worthzero marks.

Solution:

(a) Imposition of the first boundary condition yields

−1 = f(0) = a + b(0) + c = a + c . (26)

(b) Differentiation of equation (25) yields

f ′(x) = −a sin x + b cos x . (27)

Imposition of the second boundary condition yields

−2√

2 = f ′(

−π

4

)

= −a · −1√2

+ b · 1√2

, (28)

implying thata + b = −4 . (29)


(c) Differentiation of equation (27) yields

f ′′(x) = −a cos x − b sin x . (30)

Imposition of the third boundary condition yields

1 = f ′′(π) = −a(−1) + b(0) , (31)

implying that a = 1.

(d) Back substitution of this value in equation (29) yields

1 + b = −4 ,

implying that b = −5. Back substitution of the values of a and b in equation(26) yields

−1 = 1 + c ,

which we may solve to obtain c = 4. The function we seek is now completelydetermined:

f(x) = 1 cosx − 5 sin x + 4 .

2. [10 MARKS] Use logarithmic differentiation to find the derivative of the functiony = x2/x.

Solution:


ln y =2

x· ln x .


y′

y= 2 ·

1x· x − (ln x) · 1

x2,

implying that

y′ = 2y · 1 − ln x

x2.


y′ = 2x2x · 1 − ln x

x2.


3. [10 MARKS] If tanh x = 35, find the value of ex and then the value of cosh x.

Solution:

(a) Write the given equation in terms of exponentials:

ex − e−x

ex + e−x=

3

5.

(b) Solve for ex: 2e2x = 8 ⇒ ex = ±2 ⇒ ex = +2 since exponentials cannot benegative.

(c) Now express cosh x in terms of exponentials, and, through that, determinethe value of cosh x.

cosh x =ex + e−x

2=

2 + 12

2=

5

4.

4. [10 MARKS] Use a linear approximation to estimate the number√

99.9. Youmust show all your work, giving the precise values assigned to any variables orincrements.

Solution:

(a) Select function, e.g., f(x) =√

x. Also select the reference point, and, therebythe increment: a = 100, ∆x = −0.1. Thus f(a) = 10.

(b) Determine the derivative of f , and evaluate at a: f ′(x) = 12√

x. At a = 100,

f ′(a) = 12√

100= 1

20.

(c) Complete the approximation:

f(a + ∆x) ≈ f(x) + f ′(a) · ∆x

= 10 +1

20· (−0.1)

= 10 − 0.025 = 9.975 .

14.4 Thursday version

1. [10 MARKS] The volume of a right circular cone is V = 13πr2h, where r is the

radius of the base and h is the height. Assume that the height of such a coneremains constant at 6 centimeters, while its volume is expanding at a constantrate of 4 cubic centimeters per second. Find the rate of change of the radius whenthe radius is 2 centimeters. Simplify your answer as much as is possible without acalculator.

Solution: V = π3hr2 ⇒ dV

dt= π

3h · 2r · dr

dt⇒ dr

dt=

dVdt

23πrh

= 423π(2)(6)

= 13π

.


2. [10 MARKS] Use logarithmic differentiation to find the derivative of the functiony = x9 cos x.

Solution:


ln y = 9 cos x · ln x .


y′

y= 9(− sin x) · ln x + 9 cosx · 1x ,

implying thaty′ = 9y · ((− sin x) · ln x + 9 cosx · 1x) .


y′ = 9x9 cos x · ((− sin x) · ln x + 9 cosxx) .

3. [10 MARKS] Find the global (absolute) extrema of the function

y = x√

2 − x2

on the closed interval [−1, 1]. You must show all of your work—correct numericalanswers alone may not earn any marks.

Solution:

(a) Find the derivative of the function:

y′ =√

2 − x2 + x · 1

2

1√2 − x2

=2(1 − x)(1 + x)√

2 − x2

(b) Determine the critical points of the function on the given interval: f ′(x) =0 ⇒ x = ±1; there are no critical points of the type where the derivative doesnot exist. And the only critical points are at the end-points of the interval.Our definition of critical point does not include end-points; but it doesn’tmatter, because end-points must be checked separately in any case.

(c) Under the Closed Interval Method we need check only the end-points. Sincey(−1) = −1 and y(+1) = +1, the global maximum — of value +1 — occursat x = +1; while the global minimum — of value −1 — occurs at x = −1.


4. [10 MARKS] Use a linear approximation to estimate the number ln(1.05). Youmust show all your work, giving the precise values assigned to any variables orincrements.

Solution:

(a) Select function, e.g., f(x) = ln x. Also select the reference point, and, therebythe increment: a = 1, ∆x = +0.05. Thus f(a) = 0.

(b) Determine the derivative of f , and evaluate at a: f ′(x) = 1x. At a = 1,

f ′(a) = 11

= 1.


f(a + ∆x) ≈ f(x) + f ′(a) · ∆x

= 0 +1

1· (0.05)

= 0.05 .

14.5 Friday version

1. [10 MARKS] The volume of a right circular cone is V = 13πr2h, where r is the

radius of the base and h is the height. Assume that the volume of such a coneremains constant at 4 cubic centimeters, while its radius is increasing at a constantrate of 2 centimeters per second. Find the rate of change of the height when theradius is 3 centimeters. Simplify your answer as much as is possible without acalculator.

Solution: V = π3· r2h ⇒ h = 3V

π· 1r2 = 12

π· 1

r2 . Hence dhdt

= −24π· 1r3 · dr

dt= − 8

9π.

2. [10 MARKS] Use logarithmic differentiation to find the derivative of the functiony = (ln x)6/x.

Solution:


ln y = 6ln ln x

x.


y′

y= 6

( 1x ln x

· x − ln ln x · 1x2

)

,


implying that

y′ = 6y

( 1ln x

− ln ln x

x2

)

.


y′ = 6(lnx)6x

( 1ln x

− ln ln x

x2

)

.

3. [10 MARKS] Use a linear approximation to estimate the number (0.99)6. Youmust show all your work, giving the precise values assigned to any variables orincrements.

Solution:

(a) Select function, e.g., f(x) = x6. Also select the reference point, and, therebythe increment: a = 1, ∆x = −0.01. Thus f(a) = 1.

(b) Determine the derivative of f , and evaluate at a: f ′(x) = 6x5. At a = 1,f ′(a) = 1

1= 1.


f(a + ∆x) ≈ f(x) + f ′(a) · ∆x

= 0 +1

1· (0.05)

= 0.05 .

4. [10 MARKS] Show that the polynomial 5x4 − 9x − 4 has exactly 2 real roots.Carefully justify all your steps and name the theorems you’re using!

Solution:

Information for Students in Lecture Section 1 of MATH 140 2006 09 1001

A Information Specifically for Students in Lecture

Section 001

A.1 Timetable for Lecture Section 001 of MATH 140 2006 09

Distribution Date: Wednesday, September 21st, 2006This revision as of December 1, 2006. (All information is subject to change.)

This WeekMONDAY WEDNESDAY FRIDAY (Until Sun. 23:59)

SEPTEMBER

04 LABOUR DAY 06 §1.1, App. D 08 §1.1, App. D

All Tutorial Sections meet first during the week of September 11, 2006.Quiz Q0, at your first tutorial, diagnoses some precalculus deficiencies, and is compulsory!

11 §1.2 (parts) 13 §1.3 15 §1.5 Q0, P1, P2

Course changes must be completed by Tuesday, September 19, 2006

18 §1.6 (briefly),Math. Induction,§2.1

20 §2.2, §2.3 22 §2.4, §2.5 A1, A2

Deadline for withdrawal from course with fee refund = September 24, 2006

25 §2.6, §2.7 27 §2.8, §2.9 29 §3.1 Q1, W1, P3

OCTOBER

02 §3.2 04 §3.3 06 §3.4 (A3 due 09 Oct.23:30h)

Next week only Monday Lectures and tutorials T003–T008 move to Tuesday, October 10th, 2006.

10 (Tuesday) X 11 §3.5 13 §3.5, §3.6 P4, A3 (Mon.23:30h)

16 §3.6 18 §3.7 20 §3.8 Q2, W2, A4

Deadline for withdrawal (with W) from course = Oct. 22, 2006

23 §3.9 25 §3.10 27 §3.11 P5

30 §4.1 Q3, W3, A5

Notation:An = WeBWorK Assignment An due at 23:30h on Sunday this weekPn = WeBWorK Practice Assignment Pn due at 23:30h on Sunday this week

(The Practice Assignments do not count in your grade.)R© = Read Only

Qn, Wn = Hand in Wn at Quiz Qn in the tutorials this week; (there is no W0).X = reserved for eXpansion or review

Section numbers refer to the text-book.

UPDATED TO December 1, 2006


This WeekMONDAY WEDNESDAY FRIDAY (Until Sun. 23:59)

NOVEMBER

01 §4.1 03 §4.2 Q3, W3, A5

06 §4.3 08 §4.4 10 §4.4, §4.5 P6

13 §4.5 — omit“slant asymp-totes”

15 §4.5 — omit“slant asymp-totes”

17 §4.7 A6

20 §4.7 22 §4.10 24 §4.10 Q4, W4, P7

Next week’s tutorial is the last.

27 X 29 X A7

DECEMBER

Last lectures in both lecture sections are on Monday, December 4th, 2006.

01 X

4 X





Supplementary Notes for Students in Lecture Section 1 of MATH 140 2006 09 1003

B Supplementary Notes for Lectures in Lecture Sec-

tion 001

Prof. Sancho may, should the need arise, place supplemental materials on the web page

http://www.math.mcgill.ca/∼sancho/math140a.html


C Information Specifically for Students in Lecture

Section 002

C.1 Timetable for Lecture Section 002 of MATH 140 2006 09

Distribution Date: Wednesday, September 21st, 2006This revision as of December 1, 2006. (All information is subject to change.)

This WeekMONDAY WEDNESDAY (Until Sun. 24:00)

SEPTEMBER

04 LABOUR DAY 06 §1.1, App. D

All Tutorial Sections meet first during the week of September 11, 2006.Quiz Q0, at your first tutorial, diagnoses some precalculus deficiencies, and is compulsory!

11 §1.1, App. D, §1.2 (parts) 13 §1.2, §1.3 Q0, P1, P2

Course changes must be completed by Tuesday, September 19, 2006

18 §1.5, §1.6 (briefly) 20 §2.2, §2.3 A1, A2

Deadline for withdrawal from course with fee refund = September 24, 2006

25 §2.5 27 §2.4, §2.6 Q1, W1, P3

OCTOBER

02 NO LECTURE IN LECTURESECTION 002

04 §2.7, §2.8, §2.9 (A3 delayed until09 Oct., 23:30h)

Next week only Monday lectures and tutorials T003–T008 move to Tuesday, October 10th, 2006.

10 (Tuesday) , §3.1, §3.2, §3.3 11 §3.4 P4, A3 (Mon.23:30h)

16 §3.5, §3.6 18 §3.5, §3.6 Q2, W2, A4

Deadline for withdrawal (with W) from course = Oct. 22, 2006

23 §3.7, §3.8 25 §3.9, §3.10 P5

30 §3.10, §3.11, X Q3, W3, A5






Notes for Lecture Section 002, MATH 140 2006 09 2002

This WeekMONDAY WEDNESDAY (Until Sun. 24:00)

NOVEMBER

01 §4.1 Q3, W3, A5

06 §4.2 08 §4.3 P6

13 §4.4 15 §4.5 — omit “slant asymp-totes”

A6

20 §4.7 22 §4.10 Q4, W4, P7

Next week’s tutorial is the last.

27 X 29 X A7

DECEMBER

Last lectures in both lecture sections are on Monday, December 4th, 2006.

4 X






D Notes for Lecture Section 002

Release Date: Wednesday, September 6th, 2006

D.1 Lecture style in Lecture Section 002

Lecture content The timetable on pages 2001-2002 will show you approximately whatI plan to discuss at each lecture. I suggest that you look through the material in advance.If you have time to try some of the exercises, and find some that cause you difficulty,you are welcome to bring them to my attention; perhaps I may be able to work some ofthese examples into the lecture.

What goes on the chalkboard? — Should I take Notes? Your instructor believesstrongly that students should not spend the lecture hour feverishly copying notes for fearof missing some essential topic; in this course most of what you need to know is containedin the textbook. You should take notes, but you should be trying to think at the sametime. The chalkboard will be used for

• statement/illustration of specific definitions and theorems

• sketching solutions to problems, or classes of problems

• a scratchpad

Some of this material will be useful to you in learning the material in the course. Evenwhen the material on the board is equivalent to something in your textbook, the act ofwriting may help you remember it. But much of the material will be restatements ofyour textbook, so you should normally not panic if you miss something.

Graphs Several topics in the syllabus, culminating in [1, §4.5], are concerned withsketching of curves. Our emphasis is on qualitative properties of the graphs of functions,but not on the production of extremely precise graphs. You can expect to see me draw onthe chalkboard sketches that are extremely crude approximations of functions, sometimeseven caricatures of the true graph. Mathematicians do not base proofs on sketches ofgraphs — the role of a sketch is usually only to assist the reader to visualize the verbal orsymbolic reasoning which accompanies it. Sometimes a graph is used help one discover aphenomenon, but the result would not be acceptable to a mathematician unless it couldbe proved in a non-graphical way.33

33This is why I usually avoid in the textbook problems that appear to be making inferences fromgraphs.


These supplementary notes The following notes were prepared by me for my lec-tures. The section and paragraph headings follow the order of topics in the textbook.While some of the comments or explanations may be helpful in understanding the book,the notes are not required reading for examination purposes. Sometimes it may happenthat the discussion of a topic or an exercise evolves during the lecture into one whichrequires more detail than is practical to write on the chalkboard. In such cases you maybe referred at the following lecture to supplementary material that will be contained inthe notes placed on the Web. Such evolutions are spontaneous and not planned, andcannot be announced in advance.

Timing and corrections The notes will usually not be posted until after the lecture.While I do try to check the notes before posting them, there will inevitably be errors:if you see something that doesn’t look right, please ask. The notes will be progressivelycorrected as misprints and other errors are discovered.

D.2 Supplementary Notes for the Lecture of September 6th,

2006

Release Date: Wednesday, September 6th, 2006subject to correction

Tutorials in MATH 140 2006 09 begin next week. A 45-minutelong diagnostic test, Q0, will be written at your first tutorial.While the grade on this test does not count, writing the quiz iscompulsory34 for all students in the course. You must writeit in your assigned tutorial room. The quiz will be graded andreturned to you, so that you may identify areas of prerequisiteswhere you may need to do remedial work.

The supplementary notes for the first lectures are unusually long, as the text coversmuch material in an introductory way, or as review of material that you are expected toknow well already. I will certainly not had time to discuss all of these issues in class. Thebeginning of the lecture will be spent in discussing details of the course, as containedin pages 1–23 of these notes, and the WeBWorK system, as described in pages 4001 –4006 of these notes.

Textbook Chapter 0. A PREVIEW OF CALCULUS.

This chapter is motivational. Most of the issues raised will be considered in depth laterin MATH 140 or MATH 141.


Textbook Chapter 1. FUNCTIONS AND MODELS.

Many of the concepts in this chapter should be review for most students, since they shouldhave appeared in a good pre-calculus course; some of the terminology and notation maybe new. Some of the material in [1, §1.6] may be new.

In discussing [1, §1.2] I will also review material from [1, Appendix D]; if any of thismaterial is new to you, you should take steps to reinforce your background in trigonom-etry before a deficiency impedes your progress. A very small number of students mayfind themselves hopelessly deficient in both high school trigonometry and algebra, andmight seriously consider transferring from MATH 140 2006 09 to MATH 112 2006 09,a course given only in the fall term; or delaying MATH 140 until the second term, andtaking other steps to rebuild precalculus foundations.

D.2.1 §1.1 Four Ways to Represent a Function.

Before we discuss the subject matter, I wish to make it clear that no one is going toexpect you to answer silly questions like “How many ways are there to represent afunction?” Aside from that, some of the ways described in this section are problematic.The purpose of the section is to prepare you for subsequent discussions in which functionswill be specified in a variety of ways.

Functions To specify a function we need two sets. The first, called the domain ofthe function, is the set on which the function “acts”. The second is the set where thefunction’s values are located. Some authors call this “target” the codomain.

The textbook defines [1, p. 12] the range of a function f as “the set of all possiblevalues of f(x) as x varies throughout the domain.” Many mathematicians now avoid theword range35 and prefer to call the set of values the image of f .

When we write f(x), we normally mean the point to which the point x in the domainis mapped by the function f ; sometimes we may, carelessly, talk about “the functionf(x)”, but the intention is to reserve this notation for the general point of the “range”(image). The “variable” x is sometimes called the independent variable; if we writey = f(x), older terminology would call y the dependent variable.

One part of the definition of a function is non-negotiable: we always require that,for every point x of the domain, there be exactly one point to which it is mapped: thismeans never more than one image point; and never fewer than one, i.e., every point inthe domain must be mapped somewhere: there must never be any doubt about the imageof every single point of the domain. We paraphrase the first of these observations in the“Vertical Line Test” (see below); from the second of these observations we see that the

35One reason for avoiding the use of this word is that some authors have used the same term for thetarget, which could include points which are not values of the function.


domain of a function is precisely the projection — the shadow — cast by the graph onthe x-axis (if one imagines the sun at the “point” (0,∞) on the y-axis.

When a function f has domain A, modern mathematical notation will speak of thefunction36 f : A → B, where B is the “target” or codomain. The choice of a codomainis sometimes unclear in a particular context, since there is often no harm in enlargingthe set; when we say that B is the codomain we do not insist that every point in Bbe realized as the image under the function of a point in the domain A; all we ask isthat the image or range should be contained in the codomain. There are good algebraicreasons for being fussy about the codomain, but students are not likely to see them inMATH 140 or MATH 141, except possibly in connection with inverse functions (to beintroduced in [1, §1.6]).

Suppose that f : A → B is a given function. If we have a formula for computingf(x) when x is given, a useful notation is to denote the function by x 7→ f(x); thus,for example, the square root function could be denoted by x 7→ √

x. (Note the peculiarsymbol — a horizontal arrow with a vertical foot.)

Graph of a function The graph of f : A → B is the set of points in37R

2 whose firstcoordinate x ranges over all points in the domain, where the second coordinate is thecorresponding value f(x). This set may be interpreted “graphically”, as points in theplane, and that representation may be “sketched”. The “sketch” obtained may be usefulin solving problems, but mathematicians never base a serious proof on a sketch; if wemake a sketch, it is only as a visual aid. You are expected to follow this practice in yourwritten solutions.

Theorem D.1 (Vertical Line Test [1, p. 17]) A set S of points in R2 is the graph

of a function if and only if no vertical line passes through two distinct points in S.

Proof: The condition is necessary (the “only if” part of the theorem) because the presencein S of two points with the same x-coordinate (the abscissa) would entail that that valueof x was mapped on to38 two distinct points. Otherwise we can interpret S as the graphof a function whose domain is the set of all numbers that appear as an x-coordinate forsome point in S: every point in this domain is mapped on to exactly one point, so thefunction is “well defined”. For each point x of this domain there is now exactly one point

36The arrow notation is not generally used in your textbook, although you can see it at the bottomof [1, Figure 3, p. 12], and is introduced here so that, at least, you can see the kind of notation that is

used by mathematicians today. Sometimes we may modify the notation f : X → Y , writing Xf→ Y .

37R is the set of real numbers or, when interpreted geometrically, “the real line”; R

2 is the set of“ordered pairs” of real numbers, or, when interpreted geometrically, “the real plane” or “the Cartesianplane”.

38Some English dictionaries do not accept onto as a single word.


y such that (x, y) ∈ S, and so, if we define a function f by x 7→ y, then S is the graphof f .

Representations of Functions As mentioned above, a complete definition of a func-tion must specify

• the domain;

• the set where the function takes its values — we could call this the “target” or thecodomain);

• the rule that assigns to each point in the domain precisely one point in the target.

If there is even one point in the domain where the action of the function has not beenspecified precisely, the function is not “well defined”! The textbook proposes four waysof representing functions, some of which have restrictions in their applicability if therepresentation is to be interpreted as defining the function. Of these “numerically (bya table of values)” is possible only if the domain of the function is finite. Similarly, themethod described as “visually (by a graph)” is not adequate if by graph one thinks ofa sketch — a curve drawn on a piece of paper; if, however, the graph is specified in anunambiguous way, as, for example “the line through points (1,−2) and (4, 2)”, or “thecircle with centre (1,−2) and radius 5”, then the curve is precisely specified; the secondexample — the circle — would still not define a function, since it violates the “VerticalLine Test”. My objection is to the word visually , not to by a graph.

When a function is described by an algebraic formula, there is often a “natural”domain — the largest set for which the formula “makes sense”. So, for example, onemight be inclined to say that “the function 1

xhas domain all real numbers except 0”.

But, if the inventor wishes to specify a domain that is smaller than this set, and usethe same formula, that is completely “legal”. There are often good reasons to restrict afunction to a smaller domain than that for which a defining formula might be meaningful.So a formula is not enough to thoroughly specify a function: it is necessary to specifyits domain precisely: if that is not done, the reader will usually take the domain to bethe “natural” domain.

Piecewise Defined Functions Mathematicians use the term piecewise to describe adefinition of a function where the domain is dissected into (subsets of) mutually disjointsets, within each of which it may be possible to define the function in some unified way.(The reason for taking the sets to be mutually disjoint is to avoid the possibility thattwo parts of the definition would be in conflict.) The graph appears to have been createdby cutting and pasting pieces of various graphs. When we cut and paste intervals thatare disjoint, some of those intervals will contain their end-points, and others will not


(since we don’t want the intervals to overlap). This can lead to definitions where thereare abrupt changes in the behavior of a function as we pass through specific points ofthe domain. We will return to this issue when we discuss continuity in [1, §2.5].

Note that a definition can be given piecewise, and yet the function could admit amore pleasing definition. For example, we could define

f(x) =

{

−1 when x ≤ 5− sin2 x − cos2 x when x > 5

, (32)

but the function f is simply the constant function −1. There is nothing wrong withdefinition (32), but a definition

f(x) = −1 for all x (33)

is equivalent, and more pleasing.For an exercise on construction of functions in this way, see [1, Exercise 18, p. 79

(Review Exercises for Chapter 1)].


D.3 Supplementary Notes for the Lecture of September 11th,2006

Release Date: Monday, September 11th, 2006subject to correction

The first written assignment, W1, has been posted on WebCT. Thisassignment is expressed in a notation that will be defined when wediscuss [1, §1.3]. Until you understand the meaning of the notation,you cannot expect to be able to solve the problems. The problemsthemselves are concerned with domains of various functions. Some ofthe functions in the problems will be discussed in [1, §§1.5, 1.6].

The following provisional notes were posted in advance of the lecture. It is likely that,for want of sufficient time, some topics may not be explicitly discussed.

D.3.1 §1.1 Four Ways to Represent a Function (conclusion).

Symmetry This topic will not be discussed in depth at this time. You should have metsome of these concepts in your precalculus preparation. However, you are not expectedto know the Theorem given below.

Even and Odd Functions. Intuitively, a symmetry of a function f is a trans-formation of the graph that can be obtained by mapping the plane R

2 onto itself,in such a way that distances are not changed, and that the graph of f maps ontoitself. For example, the graph of the function sin is not changed when the wholeplane is shifted to the right a multiple of 2π units. A function f is even if it hasthe property that

f(−x) = f(x) (34)

for every point in the domain. So, if x is in the domain, condition (34) refers tof(−x); if it is permitted to mention f(−x), then −x must also be in the domain off . Not only must an even function have the property that its domain be symmetricaround the point 0; but, moreover, points which are symmetrically located withrespect to 0 must have the same function value. To prove that a function f is not

even, it suffices to show that there is a real number x such that either

1. f is defined at one of x, −x, but not at the other; or

2. f is defined at both x and −x, but the values of the function are different atthe two points.

Just one such pair is enough to permit the inference that f is not even, sinceevenness requires that no such pair exist.


A function f is odd if it has the property that

f(−x) = −f(x) (35)

Here, as for evenness, a function can be odd only if its domain is symmetric around0. The equation must hold for all pairs of points in the domain; if there is one realnumber x where either the equation fails or the function is defined at only one ofthe points x, −x, the function is not odd. The terminology is easily rememberedby thinking about the properties of even and odd powers of x.

Theorem D.2 Let f be a function whose domain is R. Then f may be expressed

as the sum of an even function fe and an odd function fo, both with domain R.

Proof: (This proof is called constructive; not only will we prove the existence ofthe functions postulated in the enunciation of the theorem, but we will show howto construct the functions.)

We can define, for all x ∈ R,

fe(x) =f(x) + f(−x)

2(36)

fo(x) =f(x) − f(−x)

2(37)

1. For all x ∈ R,

f(x) =1

2(f(x) + f(−x)) +

1

2(f(x) − f(−x))

= fe(x) + fo(x) (38)

= (fe + fo)(x)

according to the definition of the sum fe + fo, [1, p. 42]. Since functions fand fe +fo have the same domain — R — and have the same action on everypoint x in the domain, as shown in (38), the functions are said to be equal.39

2. The function fe is defined in (36). The proof that this function is even is leftto the student.

3. The function fo is defined in (37). The proof that this function is odd is leftto the student.

Exercise D.1 1. Show that the only function that is both even and odd is theconstant function 0.

39This is the definition of what we mean when we say two functions are equal: they have the samedomain, and they act in the same way on every point of that domain. The student who thinks thereis no need for a definition here, must remember that, hitherto, the = sign has been used only betweennumbers. We are now applying it to a different type of object; whenever we do that we must clarifyprecisely what we mean by this symbol and language in this new context.


2. Let r be a fixed real number. Is the constant function f given by f(x) = reven? Is it odd?

3. Give an example to show that there exist functions that are neither even norodd.

4. There is a function that is both even and odd. What is it?

Some examples of 2nd-degree equations whose graphs are symmetric about oneof the coordinate axes can be found in [1, Appendix C, pp. A16-A21].

Increasing and Decreasing Functions Here are the textbook’s definitions:

Definition D.1 [1, p. 21]

1. A function f is called increasing on an interval I if

f(x1) < f(x2) whenever x1 < x2 in I .

2. A function f is called decreasing on an interval I if

f(x1) > f(x2) whenever x1 < x2 in I .

There are variations of the terms increasing and decreasing whose meanings are obvious;you could see these variations when you use other textbooks, or look at old examinations.Among these are

• non-decreasing : f is non-decreasing on interval I, if, whenever x1 < x2 on I,f(x1) ≤ f(x2).

• non-increasing , defined analogously to the preceding,

• strictly decreasing , which is a term used by some authors for what Stewart callsdecreasing

• strictly increasing , which is a term used by some authors for what Stewart callsincreasing .

1.1 Exercises Remember, in problems like [1, Exercises 1.1.23-1.1.27, p. 23], whenthe textbook asks for the domain of the function, it always intends the largest possibleor maximal or natural domain for which the formula could be meaningful; the rangeor image in such a situation will be that corresponding to the largest possible domain.However, there will be situations where we will wish to prescribe for a function a domainwhich does not contain all points in the largest possible domain. This will happen, inparticular, in optimization problems, where the domain may be restricted by certainconditions specific to the problem under discussion.


D.3.2 §1.2 Mathematical Models: A Catalog of Essential Functions.

Linear Models The author defines a function f to be linear 40 if there exist realconstants m and b such that, for every x, f(x) = mx + b.

Exercise D.2 Is the squaring function x 7→ x2 linear?

Solution: NO! Here is a proof by reductio ad absurdum, i.e., by showing that, if thestatement were true, it would imply an absurd contradiction. Suppose that there existedreal numbers m and b such that

x2 = mx + b for all x . (39)

We can complete the proof in various ways.

1. One way to show that this constraint cannot be satisfied by all x is to chooseseveral values of x which lead to a contradiction. By specializing values of xwe can determine information about m and b, eventually obtaining contradictoryinformation. If we take x = 0, we find that 0 = 0 + b, so b = 0. If we take x = 1,we find that 1 = m + b = m + 0 = m. Then, if we take x = 2, we find that4 = m · 2 + 0 = 1(2) + 0 = 2, which is a contradiction.41

2. Another approach would be to determine precisely those values of x for which thegiven equation could be satisfied. The equation is quadratic, so, by completing thesquare, we obtain

(

x − m

2

)2

= b +m2

4,

implying that there couldn’t be any solutions at all unless b + m2

4≥ 0; and that,

even then, the only solutions are

x =m

2±√

b +m2

4.

Thus there are at most 2 solutions to (39), which contradicts the claim that theequation should hold for all real numbers in R.

40In some areas of mathematics, the word linear would be reserved for the case where b = 0, i.e.,where the graph of f passes through the origin, i.e. where the graph of the function is a line through

the origin. In such cases the word that is used to describe functions of the generality we want here isaffine. In this course we will usually follow the textbook’s use of language.

41This is an example of a proof by contradiction or by reductio ad absurdum: one assumes the falsity ofa statement, shows that the falsity would imply some nonsense, and then concludes that the statementmust have been true.


When we take the narrow definition of linear (i.e. the definition where b = 0), we findthat it is equivalent to the property

f(x1 + x2) = f(x1) + f(x2) .

Exercise D.3 Some students in MATH 140 appear to believe that both the square andthe square root functions have this property; see if you can prove that the square rootalso fails to be linear.

Polynomials A polynomial is a function of the form

P (x) = anxn + an−1xn−1 + · · · + a2x

2 + a1x1 + a0x

0

where n is a non-negative integer, and a0, a1, . . . , an are real numbers. We usually writejust x for x1, and a0 for a0x

0. If a0 = a1 = · · · = an = 0 the polynomial is the zeropolynomial. Otherwise there will be at least one non-zero coefficient among a0, a1, . . . , an,and we can assume that n is such that an 6= 0; we call n the degree of the (non-zero)polynomial.42 When a0 is the only non-zero coefficient, the polynomial is constant ,having degree 0, and may be identified with the real number a0: thus we may think ofthe real numbers as being contained in the set of polynomials.

As observed earlier, you are expected to know how to solve equations of the form

a2x2 + a1x

1 + a0 = 0 .

While you may be accustomed to doing so by using the “quadratic formula”, you areencouraged to use the (equivalent) method of completion of the square which will bediscussed in the lectures. You are not expected to know how to solve higher degreeequations, except where the polynomial has an obvious factorization. There will bespecific facts that you should know about polynomials; these will be discussed when wemeet examples during the term.

Power Functions The textbook calls any function of the form x 7→ xa, a powerfunction.43

Rational Functions A rational function maps the independent variable on to theratio of two polynomials, where the denominator is not the zero polynomial. Sincethe denominator could be a non-zero constant polynomial, e.g. 1, every polynomial is a

42Zero is also a polynomial, but there the term degree is difficult to define, if we wish it to haveexpected properties: some mathematicians refuse to use the term degree for the polynomial 0; othersdefine the degree to be −∞.

43Power function is not a “standard” term.


rational function. Other “degenerate” examples are1

x,x2 − 2x

x − 2: note that such functions

need not be defined for all real numbers x; for example1

xis not defined when x = 0,

andx2 − 2x

x − 2is undefined when x = 2 — in both cases the formula cannot be evaluated

because of a 0 denominator.

Algebraic Functions The textbook definition of algebraic function is not quite right,but in practice, the only algebraic functions we will be dealing with will indeed be“constructed using algebraic operations (such as addition, subtraction, multiplication,division, and taking roots) starting with polynomials”.

Trigonometric Functions The prerequisites for this course require a basic knowledgeof trigonometry. We will summarize here some of the definitions and facts that we willbe appealing to, in case students have forgotten some of them.

1. These definitions will involve an angle. We normally express angles in terms ofradians. A radian is the angle at the centre of a circle subtended by an arc whoselength is the radius of the circle. (This definition is somewhat premature, as wehave not yet defined what we mean by the length of a curve.) The number ofradians in a straight angle is denoted by the symbol π — the lower case of a letterof the Greek alphabet. Since a straight angle can also be defined to be 180 degreeswe have the relationship that

π radians = 180 degrees

which permits us to convert from degree measure to radian measure. In Calculus3 students see how to approximate π:

π ≈ 3.1415926535897932384626433832795028841971693993751...

Remember that we normally interpret the “argument” (“independent” variable)of a trigonometric function as representing an angle measured in radians; so, forexample, sin π = 0, but sin 180 is not equal to 0: 180 radians = (14 × 2π) +4.0708114... radians; sin 180 = sin 4.0708114... = 0.07098390....44 The distinctionis important because, when we determine formulæ for, for example, the derivative

44Occasionally we may wish to describe an angle in terms of degrees, and then we may write, forexample, sin 180◦ when we mean sin π. This is what we call an “abuse of notation”, since we are usinga well defined function — sin — to mean something other than its precise definition. If we wished tobe “formal”, one way out of this mess might be to introduce a new function, perhaps sindegrees. I willnot burden students with that type of formalism.


of the sine function, the formula will depend on the assumption that the argumentis to be thought of as an angle measured in radians; while we can develop formulæfor the case of angles measured in degrees, these would not be the familiar formulæfound in the textbook. To summarize: even when we may appear to write theargument of a trigonometric function in terms of degrees, the intention is that theangle is first expressed in terms of radians, and then the trigonometric function isapplied.45

2. We will have 6 trigonometric functions. As the name suggests, these functions arerelated to a trigon, i.e., to a triangle — to use a word of Latin origin instead of aword of Greek origin. We usually think of the two functions sine and cosine to bethe primary functions, and define the other four functions in terms of them. Forour purposes the sine and cosine will both have the entire real line as their domain!To define these functions for any x, think of a line segment of length 1 that is fixedat the origin O(0, 0) and is rotating in a counterclockwise direction — this is thedirection we call positive; call this rotating line segment the radius vector . Whenthe unit radius vector has rotated through an angle of x – expressed in radians— we define cos x and sin x respectively to be the x− and y− coordinates of theend P of the radius vector remote from the origin. Denote by Q the foot of theperpendicular dropped or erected from P to the x-axis. Then, in the right-angledtriangle, OPQ, where |OP | = 1,46 and the right angle is at vertex Q,

cos x = OQ =OQ

|OP | (40)

sin x = QP =QP

|OP | . (41)

We follow the usual sign conventions here:

Distances measured horizontally to the right are positive;distances measured horizontally to the left are negative;distances measured vertically upward are positive;distances measured vertically downward are negative.

When x lies between 0 and π2, these ratios are positive, so the functions can be

thought of as being defined as ratios of lengths of sides of the right-angled triangle.

45This distinction will be important when we come to discuss the derivatives of the functions.46We will try to follow the convention that the length of the line segment joining points P, Q is denoted

by |PQ|, while the directed distance from P to Q will be denoted by PQ.


3. The other four functions are defined in terms of these 2 functions as follows:

tanx =sin x

cos x=

QP

OQ(42)

cot x =cos x

sin x=

OQ

QP(43)

sec x =1

cos x=

|OP |OQ

(44)

csc x =1

sin x=

|OP |QP

(45)

if, for some x, the denominator is zero, the function is undefined; i.e., x is not inthe domain of the function.

4. By simple arguments based on similar triangles we may show that these definitionsmake sense — and yield the same values — even if the length of the radius vectoris a positive number different from 1. If we interpret the distances OQ and QPas directed distances, the definitions make sense for any real angle x, except forisolated cases where a denominator is equal to 0. For angles x other than positiveangles between 0 and 180◦ it is not recommended to think of the angle as beingpart of a triangle.

5. The full names of cot, sec and csc are cotangent , secant and cosecant ; the wordstangent and secant will also be used with other meanings.

6. Where the ratios are not defined, the corresponding functions are not defined.Thus, for example, the domain of the tangent function tanx consists of all realnumbers except the odd integer multiples of π

2.

7. Our definitions of sin x and cos x in terms of coordinates of a vertex of a right-angled triangle yield, after a simple application of the Theorem of Pythagoras, thebasic identity:

(sin x)2 + (cos x)2 = 1 . (46)

If we divide this identity by (sin x)2 or (cos x)2 we obtain the following identities:

1 + (cot x)2 = (csc x)2 , (47)

(tan x)2 + 1 = (sec x)2 . (48)

8. We are in the custom of writing exponents (i.e., powers) of trigonometric functionsas superscripts to the name of the functions. Thus, we usually write

sinn x for (sin x)n


etc. The one exception is that we will not use this convention when n = −1: sin−1 xwill have another meaning!

We also try to suppress parentheses wherever we can; so, for example, we will write

sin 2x instead of sin(2x) .

(Some of these notational conventions will not be followed by WeBWorK.)

9. Simple considerations of symmetry allow us to prove that the sine function isodd and the cosine function is even. These properties extend through our earlierdefinitions to the following identities :

sin(−x) = − sin x (49)

cos(−x) = cos x (50)

tan(−x) = − tanx (51)

sec(−x) = sec x (52)

csc(−x) = − csc x (53)

10. You should be able to quickly determine the values of these 6 functions for anyvalue of x which is an integer multiple — positive or negative — of any of thefollowing submultiples of π:

π

4,π

3,π

2where the respective functions are defined. In particular, see the values tabulatedin Table 9.

x 0 π6

π4

π3

π2

sin x 0 12

1√2

√3

21

cos x 1√

32

1√2

12

0

tanx 0 1√3

1√

3 undefined

cot xsec xcsc x

Table 9: Values of the Trigonometric Functions for simple, non-negative submultiples ofπ.

Exercise D.4 Some of the values in Table 9 have been left for you to computeyourself.


11. The following identities can be proved in several elementary ways: you will notbe expected to know how to prove them, but you are expected to remember theseidentities and to be able to use them:

sin(x + y) = (sin x)(cos y) + (cos x)(sin y) (54)

cos(x + y) = (cos x)(cos y) − (sin x)(sin y) (55)

tan(x + y) =tan x + tan y

1 − (tanx)(tan y)(56)

If we replace y by −y in the preceding, and use the facts that sin x and tanx areodd, while cos x is even, we obtain the related identities,

sin(x − y) = (sin x)(cos y) − (cos x)(sin y) (57)

cos(x − y) = (cos x)(cos y) + (sin x)(sin y) (58)

tan(x − y) =tanx − tan y

1 + (tanx)(tan y)(59)

12. In particular, substitutions into the preceding identities yield the following rela-tionships between the trigonometric functions:

sin x = cos(π

2− x)

(60)

cos x = sin(π

2− x)

(61)

tan x = cot(π

2− x)

(62)

cot x = tan(π

2− x)

(63)

sec x = csc(π

2− x)

(64)

csc x = sec(π

2− x)

(65)

13. Other substitutions yield the “double-angle” formulæ:

sin 2x = 2 sinx · cos x (66)

cos 2x = cos2 x − sin2 x (67)

= (1 − sin2 x) − sin2 x = 1 − 2 sin2 x (68)

= 2 cos2 x − 1 (69)

tan 2x =2 tanx

1 − tan2 x(70)


¿From identities (68) and (69) we can solve to obtain the “half-angle” formulæ

sin2 x =1 − cos 2x

2(71)

cos2 x =1 + cos 2x

2(72)

14. All of the trigonometric functions are periodic [1, p. 317]: there exists a positive realnumber p — called the period with the property that the behavior of the functionrepeats itself when the domain is shifted through that distance. More precisely, forall real numbers x,

sin(x + 2π) = sin(x) (73)

cos(x + 2π) = cos(x) (74)

tan(x + π) = tan(x) (75)

cot(x + π) = cot(x) (76)

sec(x + 2π) = sec(x) (77)

csc(x + 2π) = csc(x) (78)

Some authors, e.g., your text book, use the word period to mean the smallestpositive number with this property; other authors designate this smallest numberin some other way, e.g., by calling it the primitive period. We will not be concernedwith these distinctions in this course.

15. Graphs of trigonometric functions. Portions of the graphs of the 6 trigono-metric functions for −4π ≤ x ≤ 4π are shown in Figures 4, 5, 6, 7, 8, 9, respectivelyon pages 2019, 2020, 2021, 2022, 2023, 2024. You should be familiar with the

0100 5-10 -5

Figure 4: Graph of the Function sin x

approximate shapes of the graphs of the trigonometric functions [1, Figures 13, 14,pp. A30-A31]. No one expects you to draw a precise graph — just a crude sketchthat indicates things like the periodicity of the function, its domain and “range”,etc.

16. Law of Sines and Law of Cosines for the 3 angles of a Triangle


0100 5-10 -5

Figure 5: Graph of the Function cosx

(a) Law of Sines: If the (lengths of the) sides of a triangle are a, b, c, and theangles opposite them are A, B, C, then

sin A

a=

sin B

b=

sin C

c(79)

(b) Law of Cosines: [1, Exercise 83, p. A33] If the (lengths of the) sides of atriangle are a, b, c, and the angle between a, b is denoted by C, then

c2 = a2 + b2 − 2ab cos C (80)

(c) Area of a triangle: [1, Exercise 88, p. A33] The area of a triangle can beshown to be

1

2× length of base × height ,

(known to Euclid over two thousand years ago). By the Law of Sines this isequal to

1

2ab sin C =

1

2bc sin A =

1

2ca sin B

where we follow the convention that the lengths of the sides opposite anglesA, B, C are denoted by a, b, c.

17. What kinds of trigonometry problems should I be able to solve? Trigonometry isamong the prerequisites to this course, so you should be able to solve any typesof problems that normally appear in a precalculus course. Some of these are morelikely than others to appear in MATH 140. Usually a calculus problem will notbe primarily trigonometric; but, if you are unable to deal with the trigonometricissues that arise, you will be unable to solve the calculus problem. For example,

(a) In this course you are never expected to use a calculator or computer. If youare solving any of the problems mentioned in that way, your method is notwhat is intended!

(b) To solve an equation relating trigonometric functions of one or more variables.For example, [1, Example 6, p. A30] asks you to determine where the graphsof y = sin x and y = sin 2x intersect.


10

5

0

-5

-10

1050-5-10

Figure 6: Graph of the Function tanx

(c) Given the value of, for example, sec x, to determine the value of, for example,sin x. Usually this should not be solved by first attempting to determine x;rather, you need to be able to express one trigonometric function in terms ofthe others. This could involve a possible choice of two solutions; sometimesthere is additional information available to enable you to exclude one of thepossible solutions.

(d) Determination of some facts about side lengths and angle magnitudes of agiven triangle from other given facts; this is sometimes called “Solution of atriangle”. Where the triangle is right-angled, this is usually not complicated.Where the triangle is not right-angled, you could need to “drop” a perpen-


10

5

0

-5

-10

1050-5-10

Figure 7: Graph of the Function cot x

dicular from a vertex to the opposite side; or to use the Law of Sines or Lawof Cosines.

(e) More than being able to solve trigonometry problems, you need to be ableto work efficiently with the functions. Sometimes this is best done by apply-ing some identities involving the functions. For example, the formulæ whichexpress sin2 x and cos2 x in terms of cos 2x can be used to reduce the de-gree of a product of sines and cosines, a simplification which may make agreat difference in solving a problem where trigonometric functions happen toappear.

(f) While the derivation of trigonometric identities is not central to this course,


10

5

0

-5

-10

1050-5-10

Figure 8: Graph of the Function sec x

the skills learned from that part of a trigonometry course will be needed inthis first calculus course. Remember the procedures to follow in proving anidentity of the form

f(x) = g(x)

where f and g are functions that are known.

• Your proof must be completely general: you must not restrict the variablex in any way beyond the restrictions that are stated in the problem.

• One method of solution is to transform the two sides of the equation untileach of them can be shown to be equal to the same convenient functionof x.


10

5

0

-5

-10

1050-5-10

Figure 9: Graph of the Function csc x

• A more elegant solution would start on one side of the alleged equationand, through a sequence of reversible transformations, evolve it into theother side of the equation. This type of solution is more pleasing thanthe preceding, but does not always suggest itself immediately.



Release Date: Tuesday, September 19th, 2006

These notes were prepared in advance of the lecture; because of a family emergency Iwas unable to deliver that lecture, and Professor Sancho lectured in my place. His choiceof topics for this review lecture were certainly not the same as I had planned to discuss.This is review material — students in a calculus course are expected to be familiar withthese topics when they enter.

D.4.1 §1.2 Mathematical Models: A Catalog of Essential Functions (con-clusion).

Exponential Functions We will meet exponential functions next in [1, §1.6]; bothhere and there the discussion of these functions is not entirely rigorous. The intention isthat you should thoroughly understand what is meant by an integer power of a positivereal number: if n > 0, an is the product of n copies of a; if n < 0, an is the product of ncopies of 1

a; if a = 0, an is defined to be 1. You should then generalize this understanding

by taking roots: for integers r and s, where s > 0, ars is the (real) sth root of ar — and,

if s is an even integer, this sth root is chosen to be the positive root rather than thenegative one. Finally, if we wished to make this definition rigorous, we would need to beable to think of every real number as being a limit of a sequence of rational numbers.

Logarithmic Functions You are expected to be familiar with simple properties ofthe logarithm — to any base — from your precalculus prerequisite. We will be reviewingand proving some of these properties in connection with [1, §1.6].

Transcendental Functions You are not expected to be able to determine whethera function is transcendental . We will not be using this terminology — even though itappears in the title of the textbook.

Some review exercises in trigonometry, from the Trigonometry Supplementwhich was used in this course in 2005/2006.

[15, Exercise 62, p. 558] Prove the identity,

4(sin6 x + cos6 x) = 4 − 3 sin2 x .

Solution: By an identity we mean an equation involving variables, which is true forall values of those variables in a prescribed domain. Thus it’s not enough to know


the equation is true sometimes: it must be true for every value, and its failure forjust one value of x would be enough to prevent us from calling it an identity .

The most elegant solutions of problems of this type will begin with one side ofthe alleged equation, and, by applying various results that we know, progressivelytransform it into the other side. This strategy is not always easy to see at first. Itis best to begin with what you see as the “more complicated” of the two membersof the equation.

I plan to base my argument on the factorization of the left side of the equation.You may not know how to factorize a sum of 6th powers, but you should know howto factorize a sum of 3rd powers:

a3 + b3 = (a + b)(a2 − ab + b2) .

So let’s interpret the 6th powers as cubes of squares.

4(sin6 x + cos6 x) = 4(sin2 x + cos2 x)(sin4 x − sin2 x cos2 x + cos4 x)

= 4 · 1 · (sin4 x − sin2 x cos2 x + cos4 x)

by (46)

At this point we examine the expression within parentheses on the right, andobserve that it looks something like a square. If we compare it with

(sin2 x + cos2 x)2 = sin4 x + 2 sin2 x cos2 x + cos4 x ,

we see that we are short 3 sin2 x cos2 x. So we have

4(sin6 x + cos6 x) = 4(sin4 x − sin2 x cos2 x + cos4 x)

= 4(

(

sin2 x + cos2 x)2 − 3 sin2 x cos2 x

)

= 4(

12 − 3 sin2 x cos2 x)

= 4 − 3 (2 sin x cos x)2

And why have I expressed the subtracted term in this way? Because I wish tomake use of the double angle formula (66):

sin 2x = 2 sinx · cos x .

We obtain4(sin6 x + cos6 x) = 4 − 3 sin2 2x

as required.


[15, Exercise 48, p. 558] Find the value of the product 3 cos 37.5◦ · cos 7.5◦.

Solution: First note that does not contradict my earlier statement that I wish theargument of trigonometric functions to be expressed only in radians. I referred initem 1 on page 2015 of these notes that we should interpret these expressions asreferring to the radian equivalent of the angle given in angles. There will be noneed to actually compute that equivalent.

What is the point of this problem? The intention is that you should be ableto “simplify” products into sums. It is often — but not always — preferable towork with a sum of like trigonometric functions rather than a product. So theintention here is that you should convert this product of cosines into a sum ofsines or cosines. The tool we use is a family of identities which your textbook calls“product formulas”. These formulæ are consequences of the sum formulæ. If weadd the expansions of cosx + y and cos(x − y), we obtain

cos(x + y) + cos(x − y) = 2 cosx · cos y (81)

Taking x = 37.5◦ and y = 7.5◦, we obtain

3 cos 37.5◦ · cos 7.5◦ =3

2(cos 45◦ + cos 30◦)

=3

2

(

cosπ

4+ cos

π

3

)

=3

2

(

1√2

+

√3

2

)

Why, you might ask, is this answer an improvement over the original product —since one will still need to use a calculator to evaluate it? One needs to recognizethat numbers like square roots can be evaluated, albeit laboriously, by hand; whilethe evaluation of trigonometric functions is much more complicated47. The usesof the procedures to convert from products to sums are more convincing whenone considers, for example, a product of the cosines of variables rather than ofconstants, which is the case in the present, easy example.

[15, Exercise 52, p. 558] Find the value of the sum cos π12

+ cos 5π12

.

Solution: This is an example of the reverse of the operation considered in [15,Exercise 48, p. 558], solved above. There are situations where one finds a productmore amenable than a sum. In order to make this more convincing, I am goingmake the question more difficult. So let’s change it to

47although it can still be done by hand, as you will see it you take Calculus 3


Find the value of the sum cos(

x + π12

)

+ cos(

x + 5π12

)

.

(So the case that we are being asked about is x = 0.) To convert from a sum to aproduct we add identities (54) and (55), to obtain

cos(x + y) + cos(x − y) = 2 cosx cos y . (82)

This is an identity — true for all real numbers x and y. Suppose that we choose xand y so that

x + y = X

x − y = Y

These two equations are equivalent to those obtained by solving the system for Xand Y in terms of x, y:

x =X + Y

2

y =X − Y

2

so (82) can be rewritten as

cos X + cos Y = 2 cosX + Y

2· cos

X − Y

2(83)

For the general problem stated above take X = x + π12

, Y = x + 5π12

: thus

X + Y

2= x +

π

4X − Y

2= −π

6

so cos(

x +π

12

)

+ cos

(

x +5π

12

)

= 2 cos(

x +π

4

)

· cos(

−π

6

)

= 2 cos(

x +π

4

)

· cosπ

6

=√

3 cos(

x +π

4

)

.

The identity shows us that this sum of two cosines has a graph which is a scaledand translated sine curve. As for the special case, setting x = 0 yields

cosπ

12+ cos

5π

12=

√3 · 1√

2=

√

3

2.


Some “classical” exercises in trigonometry

Exercise D.5 The following problems are taken from a textbook which was once astandard high school source [30]. While the edition is over a century old, the problemsare still appropriate trigonometric background for a calculus course.

1. [30, Problem 13, p. 62] Show that tan2 x − sin2 x = sin2 x · sec2 x.

2. [30, Problem 26(2), p. 63] Solve the equation tan θ + sec π6

= cot θ.

3. [30, Problem 27, p. 63] If 5 tanx = 4, determine the value of5 sin x − 3 cos x

sin x + 2 cos xwithout using inverse trigonometric functions.

4. [30, Problem 10, p. 122] Determine the domain of the functionsin 3x

sin 2x − sin x, and

express the function in terms of cos x.

5. [30, Problem 4, p. 337] If sin x = 513

, find the value of tan x + sec x.

6. [30, Problem 45, p. 341] Find all x such that 0 ≤ x ≤ 2π and 2 cos2 x = 1 + sin x.

7. [30, Problem 63, p. 344] Let α and β be fixed real numbers, where β is not aninteger multiple of π. Determine the domain and image of the function

f(x) =sin(α + x) − sin(α − x)

cos(β − x) − cos(β + x).

D Exercises

[1, Exercise 72, p. A33] Find all values of x in the interval [0, 2π] that satisfy theequation 2 + cos 2x = 3 cos x.

Solution:

2 + cos 2x = 3 cos x ⇔ 2 + (2cos2x − 1) = 3 cosx

⇔ 2 (cos x)2 − 3 cos x + 1 = 0

⇔ (2 cos x − 1)(cos x − 1) = 0

⇔ cos x =1

2or cos x = 1

⇔ x =π

3,5π

3or x = 0, 2π.


Exercise 74, p. A33 [12]51 Find all values of x in the interval [0, 2π] that satisfy theinequality 2 cosx + 1 > 0.

Solution: The inequality is equivalent to cosx > −12

= cos 2π3

= cos 4π3

=. One wayto solve the problem is to make use of the fact that the cosine function is decreasingin the first and second quadrants, and increasing in the third and fourth. It followsthat either 0 ≤ x < 2π

3or 4π

3< x ≤ 2π.

D.4.2 Remedying deficiencies in your precalculus background

Some of the topics I have been discussing should be part of any reasonable high schoolprecalculus or trigonometry course — but they may not have been part of yours. Don’tdespair! Possibly you were short-changed by your high school, or by the bureaucracythat prescribed the syllabus for your high school — or you may simply not have paidattention when the topics were covered. Whichever was the case, you can probablyremedy deficiencies, as you may have to remedy other deficiencies in your training, butit will take some effort on your part, and you will have to manage that remediationon your own. Provided the gaps are not too extensive, if you were good enough to beadmitted to McGill, you should be able to accomplish this painlessly.

However if you never had a precalculus course, or have never studied any trigonom-etry, you might be advised to drop MATH 140 2006 09, and to take MATH 140 112 09— a precalculus course that we offer only in the fall term.



Release Date: Tuesday, September 19th, 2006

D.5.1 §1.3 New Functions from Old Functions.

Transformations of Functions Read this material. You will not be expected tocarry out “shifts”, “stretching”, “compression”, “reflection” on a given function, but youare certain to meet functions that have been obtained from others in these ways.

Combinations of Functions (Not discussed in the lecture). Given functions fand g, we can define the functions f + g, f − g, fg, f/g by a precise definition of thevalue of each of these functions at a general point x of its domain. The largest domainpossible is the set of points where both of the functions f and g are defined; except that,in the case of f/g, we have to exclude points where g is 0. That is why the domain isdescribed in terms of the intersection operator on pairs of sets.

Composition of Functions Remember that the definition of f ◦ g takes g as thefunction applied first, and f second [1, p. 44]. This definition is a practical consequenceof the notation we are using for functions, wherein we place the name of the function tothe left of the name of the point on which it acts. This means that the actions of thefunctions can be represented by a diagram like

xg7→ g(x)

f7→ f(g(x)) = (f ◦ g)(x) .

1.3 Exercises Try problems among [1, Problems 1.3.15, 1.3.31, 1.3.39 – pp. 46] anduse the Solution Manual or the TEC CD-ROM (cf. these notes, page 13) to check yourwork; look first at the worked examples in the textbook, like [1, Examples 1.3.7 and1.3.8, p. 44].

[1, Exercise 1.3.32, p. 47] (not discussed at the lecture) Find f + g, f − g, fg, andf/g, and state their domains:

f(x) =√

1 + x g(x) =√

1 − x .

Solution: Since the domains of the functions have not been stated explicitly, ourconvention is that the domains are to be the largest possible in each case. Thefunction f is the composition of two functions: first x is mapped on to 1 + x, afunction whose (maximal) domain is the whole real line R. Then the result of that


first operation is subjected to the square root function, which is defined for allnon-negative real numbers; thus that phase is well defined if and only if 1 + x ≥ 0,i.e., iff48 x ≥ −1. Thus the domain of f is x ≥ −1. In a similar way, we can showthat the domain of g is the set of all x such that 1 − x ≥ 0, i.e., x ≤ 1. When wecombine these two functions f and g, the domain will be the intersection of thesetwo domains — from which we will have to further delete the point x = −1 —the point where the g(x) = 0 — in the case of f/g. The intersection is the closedinterval −1 ≤ x ≤ 1. Thus we have

domain of f ± g, fg = [−1, 1]

domain of f/g = [−1, 1)

(As an exercise you might try to determine the range (or image) of each of the newfunctions.)

[1, Exercise 1.3.48, p. 47] Express the function G in the form f ◦ g: G(x) =1

x + 3Solution: The last operation performed in the evaluation of G is the taking of a

reciprocal. Define f(x) =1

x. Prior to that the first step was x 7→ x + 3, which we

can define to be the function g. Then G = f ◦ g.

(The domain of g is R; the domain of f is R − {0}. Hence the domain of G is49

“the set of all x in the domain of g such that g(x) is in the domain of f”, i.e., theset of all x such that x + 3 6= 0, i.e., R − {−3}.)The preceding is probably the solution that the textbook was seeking; but it notthe only solution. For example, if we define h(x) = x for all x; then G = G ◦ h;also, G = h ◦ G. Less trivially, if we define k(x) = − 1

x, and ℓx = −x − 3, then

G = k ◦ ℓ.

[1, Exercise 1.3.39, p. 47] Part of this problem was discussed in the lecture. A solu-tion can be found in the Student Solution Manual [3, p. 13].

D.5.2 §1.4 Graphing Calculators and Computers

This section may be safely omitted; we will not be using calculators or computers inthis course. Remember that, when I use technical words or symbols that are not in thetextbook, these are intended only to expose you to the current language of mathematics;you are not expected to retain such words or symbols for examination purposes, althoughyou may hear them in my lectures or see them in these notes.

48a mathematicians’ abbreviation for “if and only if”49[1, p. 44]


There will probably not be time to discuss §1.3 further in the lectures. This materialis a review of topics, most of which should have been seen in a pre-calculus or functionscourse. Try exercises and read the text when you find you cannot solve the problems.Ask the instructors or TA’s about problems that elude you.

D.5.3 §1.5 Exponential Functions

(cf., these notes, §D.4.1, p. 2025) In this “early transcendental” treatment of exponen-tials, the author’s introduction is intuitive, without proofs. He states the

Theorem D.3 (Laws of Exponents [1, p. 57]) Let a and b be any positive real num-bers, and x and y be any real numbers. Then

1. ax+y = axay .

2. ax−y =ax

ay

3. (ax)y = axy

4. (ab)x = axbx

Note that one consequence of these “Laws” is that a0 = 1, and that is true even whena = 0: 00 = 1. We will not attempt to prove Theorem D.3 in this course, and you willnot be expected even the cite the theorem by name whenever you need to use it. At thispoint in this course you are asked to assume these laws without proof, and to becomecomfortable using them. These are skills you should be bringing with you from yourpre-calculus and earlier courses.

Applications of Exponential Functions This is motivational material. Read it.

The Number e In MATH 141 you will see ways in which the constant e can be definedformally, and can be evaluated to any desired accuracy. For the present it is satisfactoryto think of e as that constant c — between 2 and 3 — for which the tangent to thegraph of the exponential function cx makes an angle of π

4radians with the y-axis where

it crosses that axis (at the point (x, y) = (0, 1)). An alternative definition will be foundon [1, p. 189, §5.1]. The “early transcendental” presentation of the calculus makes dowith these crude definitions as an expedient, in order that you can work early with theexponential and logarithm functions. Historically, the traditional definition would bebased on theory in [1, §11.9 et seq.], which is not even in the syllabus of our MATH 141.Most mathematicians today prefer to define the logarithm function first, where ln t is

defined to be the area under between the graph of y =1

xand the x-axis, the line y = 1


and the line y = t. This type of definition is not available to us yet, as we haven’t definedareas.

1.5 Exercises [1, Exercise 1.5.16, p. 62] (not discussed in the lecture) “Find the domainof each function: (a) g(t) = sin

(

e−t)

, (b) h(t) =√

1 − 2t .”Solution:

1. We first examine the “innermost” function in the composition. The function t 7→e−t is defined for all real numbers t, i.e., its domain is R. Likewise, the secondfunction in the composition, the sine function, is also defined for all values of itsargument. Consequently the domain of g is R.

2. While the first function applied, t 7→ 1 − 2t, is defined for all real numbers t, wehave to confine ourselves to those points in its domain which yield a non-negativevalue, in order that the square root may be taken. The inequality 1 − 2t ≥ 0is equivalent to 2t ≤ 1, hence to t ≤ 0. Thus the domain of h is the half-line50

(−∞, 0].

D.5.4 §1.6 Inverse Functions and Logarithms

Obtaining the inverse function of f from its graph The graph of a function fcan be interpreted as describing the action of the function: given any point a in R, thereis, by the “Vertical Line Test”, at most one point on the line x = a on the graph. Ifthere is no such point, then a is not in the domain of f . If there exists such a point,say (a, b), then f(a) = b. Thus the graph can be used to reconstitute the function. Thissame graph can sometimes be interpreted as describing a function that maps points inthe image of f , on the y-axis, on to points in the domain of f . We will call this test the“Horizontal Line Test”: any line y = b may intersect the graph in at most one point:

Horizontal Line Test: A function is one-to-one if andonly if no horizontal line intersects its graph more thanonce.

Once a function f is known to be injective, or one-to-one, we know that there exists aninverse function which acts on the image or range of f and takes its values in the domainof f . But, even if we have a simple formula for the action of f , this does not mean thatit will be easy or even possible to find a simple formula for f−1. When it is possible tofind a formula for the inverse, this can be done by following the boxed instructions on[1, p. 66]:

50We sometimes call a half-line a ray.


STEP 1 Write y = f(x).

STEP 2 Solve the equation y = f(x) for x in terms ofy. This gives a formula expressing y in terms ofx, i.e., a formula for x = f−1(y). This determinesthe function f−1 in a “non-standard” way, in thatthe action of the function is expressed in terms ofa variable named y.

STEP 3 If you wish to express your formula using thesymbol x for the independent variable, simply re-place the symbol y by x throughout your solution.

1.6 Exercises

[1, Exercise 24, p. 75] Find a formula for the inverse of the function f(x) = 4x−12x+3

.

Solution:

1. Denote the “dependent variable” by y:

y =4x − 1

2x + 3.

2. Solve for x in terms of y: y(2x+3) = 4x−4 implies that (2y−4)x = −4−3y;hence

x =−4 − 3y

2y − 4.

3. What we have found is the point in the target, i.e., on the x-axis, which isthe image of a point y on the y-axis. We can denote that fact by

f−1(y) =−4 − 3y

2y − 4. (84)

4. If we wish to represent the function f−1 as mapping part of the x-axis on topart of the y-axis, i.e., if we wish to name the “independent” variable x, thenwe can simply replace y by x in equation (84):

f−1(x) =−4 − 3x

2x − 4. (85)



Release Date: Wednesday, September 20th, 2006; updated on September 25th, 2006;subject to further revision

I am planning to shift my Friday office hour to 10-11 a.m. The changewill eventually be shown in the appropriate charts in these notes andon WebCT.

D.6.1 §1.6 Inverse Functions and Logarithms (continued)

Definition D.2 [1, Definition 1, p. 64] A function f is one-to-one or injective if distinctpoints are mapped on to distinct points; symbolically, if

x1 6= x2 implies that f(x1) 6= f(x2) .

Instead of writing the words implies that we will often use the symbol ⇒. To be precise,we should explain that the implication must be true whenever x1 and x2 are any pointsof the domain of the function f . We can write this using mathematical logic symbols as∀x1∀x2, or (∀x1)(∀x2); ∀ is called the “universal quantifier”, because it tells you that astatement must be true for all points in the “universe”, which, in this case, is understoodto be the domain of f . You won’t be expected to use either of these symbols, but youmay see them on the chalkboard or in the notes.

There is an equivalent way of defining one-to-one or injective which is often easier towork with:

Definition D.3 A function f is one-to-one or injective51 if

f(x1) = f(x2) ⇒ x1 = x2

or, more precisely,∀x1∀x2(f(x1) = f(x2) ⇒ x1 = x2) (86)

Logicians callf(x1) = f(x2) ⇒ x1 = x2

the contrapositive ofx1 6= x2 ⇒ f(x1) 6= f(x2) :

51The term one-to-one is older than injective, but is still in current use. A function that is notone-to-one might be described as “many-to-one”.


it is obtained by negating the statements connected by the ⇒ and reversing the di-rection of the arrow. It can be shown that any implication is true precisely when itscontrapositive is true.

The author provides a “geometric” test for a function being one-to-one (i.e. a test forinjectivity):

Horizontal Line Test: A function is one-to-one if andonly if no horizontal line intersects its graph more thanonce.

One way of inferring that a function is one-to-one is from the fact that a function ismonotone, i.e., it is always increasing or always decreasing. For, if the function isincreasing or decreasing, the graph can cross a horizontal line no more than once.

Exercise D.6 Can you construct an example of a function that is one-to-one but is notmonotone?

In the important cases below we will be naming and developing the properties of theinverse functions of some of the important functions that we need to have available tous.

Theorem D.4 Suppose that f : A → B has an inverse function f−1 : B → A. Then

1. The domain of f−1 is the image of f .

2. The image of f−1 is the domain of f .

3. (Cancellation Equation, cf. [1, p. 65])

f−1(f(x)) = x for all x ∈ A . (87)

4. (Cancellation Equation, cf. [1, p. 65])

f(f−1(y)) = y for all y ∈ B (88)

We can write the “Cancellation Equations” more compactly. For any set X, definea function 1X : X → X (called the identity function) by x 7→ x. Then we may rewriteequations (87), (88) as

(

f−1 ◦ f)

(x) = 1A(x) for all x ∈ A (89)(

f ◦ f−1)

(y) = 1B(y) for all y ∈ B (90)


or, even more compactly, as

f−1 ◦ f = 1A (91)

f ◦ f−1 = 1B . (92)

(Remember that two functions are equal when they have the same domain and the sameaction on every point of that domain.) This pair of equations will always hold whenthere exists a “true” inverse to a function, in particular, for logarithms and exponentials.But, when we come to trigonometric functions, the situation becomes murkier.52

Logarithmic Functions (Some of this subsection was not discussed in the lecture; Iwill return to it.)

Let a be a positive fixed real number (=positive constant). Once we have provedthat the function x 7→ ax has an inverse, we assign to the inverse function the name loga.We can then apply the preceding general theory of inverses to the pair of functions ax,loga x. Equations (87) and (88) become

loga (ax) = x (93)

alog y = y (94)

where the first equation is valid over the domain of the function ax, i.e., over the entireline R; and the second equation is valid over the domain of loga, i.e., over (0,∞). Wecan solve equations by applying either exponentiation or the taking of the logarithm (toany base) to both sides of the equation.

The basic property of the logarithm is a consequence of its being the inverse of theexponential:53

loga(xy) = loga x + loga y . (95)

This property can be shown to imply the other properties stated in the textbook:

loga

(

x

y

)

= loga x − loga y (96)

loga(xr) = r loga x (97)

first for integers r; then for rational numbers r; then for any real number r; theseproofs will not be given in the course, but you should remember the facts, and use themwhenever necessary.

52because, as we shall see, all of the 6 basic trigonometric functions fail the horizontal line test, andso none of them possesses an inverse!

53One can prove that this property is a consequence of properties (93), (94); I do not expect you tobe able to supply such a proof.


Natural Logarithms (Some of this subsection was not discussed in the lecture; I willreturn to it.)

Here we specialize the constant a of the previous item to a = e; instead of writingloge, the custom of calculus textbooks is to denote the function by ln. Mathematiciansoften persist in writing loge, or more likely just log, where the base of the logarithms isunderstood from the context54. The “change of base formula”

loga x =loga x

1=

loga x

loga a=

ln x

ln a

or, more generally,loga x

loga y=

logb x

logb y

for any positive real numbers a, b, x, y, permits statements in terms of logs to one baseto be converted to another base.

Example D.5 Use the “change of base formula” to express

log6 10 + log6 20 − 3 log6 2

in terms of logarithms to base 7 (cf. [1, Exercise 1.6.37(b), p. 76]). Simplify your answer.Solution:

log6 10 + log6 20 − 3 log6 2 = log6

(

10 × 20

2 × 2 × 2

)

= log6 25

=logb 25

logb 6to any base b

=log7 25

log7 6to the required base 7

I will return to this section next day, to discuss, among other things, the “inverse”functions of trigonometric functions. (The quotation marks have been used because theso-called inverses are not inverses of the full trigonometric functions.)

54In some situations mathematicians write log to denote a logarithm to a base other than e; you willnot likely encounter such situations in your calculus courses.


Textbook Chapter 2. LIMITS AND DERIVATIVES.

The essence of this chapter of the textbook is contained in [1, §2.4]. The concepts inthat section are difficult to grasp, and you should not expect to be comfortable with ituntil the end of the term. We have reasonable expectations of what you will be able toabsorb from this section, and, in practice, most of what you do will be discussed in thepreceding section, [1, §2.3], in which the abstract definitions of [1, §2.4] are applied tocreate workable “Laws”.

D.6.2 §2.1 The Tangent and Velocity Problems.

We will not spend much time on this section, which is mainly present at this point formotivational purposes. Here, and again in [1, §2.7], the author discusses two applicationswhich motivate the definition of the derivative, which will be defined in [1, §2.8]. Thatdefinition will apply the concept of limit , which we shall study in [1, §§2.2-2.4].

The Tangent Problem The determination of the slope of tangents is, indeed, onemotivation for the development of the calculus. But you should be cautioned that thesuggestion that one can infer the slope from a table of values is unwise, and even ob-jectionable. It is always possible to construct examples where a table of values can givetotally misleading information.

The Velocity Problem The textbook discusses two concepts:

average velocity =distance travelled

time elapsed

and

instantaneous velocity .

It is the latter that we will eventually call, simply, velocity . As for the former, youare free to use the 2-word name for the concept. But we cannot formally define whatwe mean by an average until [1, §6.5, p. 465], which we meet in MATH 141.55 Hereyou have an instance where mathematicians have appropriated words from the generalvocabulary, and given them “reasonable” meanings, but where the precise meaning stillrequires an explicit definition.

55This sounds unreasonable — after all, every school child knows what is meant by an average. Theproblem is that the concept is not being used for a finite list of numbers, but for a function defined overan interval.


D.6.3 §2.2 The Limit of a Function.

The discussion in this section is useful, motivational, and intuitive, but there are nomathematically valid proofs. Based on the ideas in this section we will justify results in[1, §2.3], which will then be used to evaluate limits. We will be discussing the meaningof the following statements (where a is a point in the domain of a function f , and L isa real number):

limx→a

f(x) = L limx→a

f(x) does not exist

limx→a+

f(x) = L limx→a+

f(x) does not exist

limx→a−

f(x) = L limx→a−

f(x) does not exist

limx→a

f(x) = +∞ limx→a

f(x) = −∞lim

x→a+f(x) = +∞ lim

x→a+f(x) = −∞

limx→a−

f(x) = +∞ limx→a−

f(x) = −∞

We shall then define what is meant by the statement.

“The line x = a is a vertical asymptote of the graph of the function f .”

To prove the validity of the results in [1, §2.3] we would need to apply the theory of[1, §2.4]. Students should not infer that we can evaluate lim

x→af(x) by computing and

tabulating values of f near a, or by using a graphing calculator. The tables might behelpful if we wish to guess the value of the limit, but are of no use if we wish to provethat our guess is correct. I reiterate: A table of values of a function has NO place ina proof that the limit of the function is a particular number at a particular point in itsdomain.

One property we will discuss when we reach the formal definitions in [1, §2.4] is thata function never has two different values for its limit at a point: either the limit existsand has a specific value, or the limit does not exist. While it’s not hard to prove, we willnot expect you to be able to prove this; but we will expect you to know the fact.

One-Sided Limits In denoting one-sided limits, using symbols + and − to the rightof the number being approached, your textbook follows the style of writing the symbolas a superscript, as in lim

x→3+f(x), or lim

x→3−f(x); some other authors write the symbols

“on the line”, as in limx→3+

f(x), or limx→3−

f(x). Both methods are in common use; your

instructor prefers the superscripted symbols.One-sided and two-sided limits are related by the following important theorem, which

we will not prove:


Theorem D.6 ([1, p. 98]) limx→a

f(x) = L if and only if both of the following statements

are true:lim

x→a−f(x) = L and lim

x→a+f(x) = L .

In words, the 2-sided limit exists only if the limits on both the left and the right existand are equal; its value is then their common value.

Infinite Limits When we first define limx→a

f(x) we will require both that a be a real

number (hence finite), and that the function values f(x) be also be (finite) real numbers.There are two ways in which we will wish to generalize this concept to the infinite:

• by allowing the function values to become arbitrarily large;

• by allowing a to become arbitrarily large.

2.2 Exercises Many of the problems in this set of exercises suggest the use of argu-ments that will not be acceptable once we have studied [1, §§2.3 – 2.5]. While you maywish to use figures and tables to suggest what the value of a limit might be, there is noplace for them in a proof of such a statement. Thus, for example, the wording of [1,Problem 2.2.12, p. 103] is not acceptable:

“Sketch the graph of the following function, and use it to determine the valuesof a for which lim

x→af(x) exists:

f(x) =

2 − x if x < −1x if − 1 ≤ x < 1

(x − 1)2 if x ≥ 1.”

We can make it acceptable by replacing the word determine by, for example suggest :when you are asked to determine a value, we usually expect you to be able to justifyyour work.

READ THIS AFTER WE HAVE STUDIED [1, §2.5]: To solve this problem

in a more acceptable way, use [1, Theorem 2.5.7, p. 129] to argue that 2 − x, x,

(x−1)2 are all polynomials, so they are continuous; so the limit of f is the function

value at all points on R except possibly for the points −1, 1. Those 2 points have

to be examined individually. At x = −1 we can argue, again from the continuity

of polynomial functions, that limx→−1−

(1 − x) = 2 − (−1) = 3, while limx→−1+

x = −1;

as the limits of f(x) from the right and left are different at x = −1, the function

has no limit there, by [1, Theorem 2.2.3, p. 98]. The same kind of reasoning shows

that f also lacks a limit at x = 1.


D.6.4 §2.3 Calculating Limits Using the Limit Laws.

We now know precisely what we mean by the limit. But the calculations appear to bevery complicated: is this what has to be done whenever we need to find a limit or provethat one does not exist? Fortunately not! We can prove some theorems that enable usto evaluate limits routinely for the types of functions we meet most often in practice.The author calls these theorems “Limit Laws”. You should remember the results andbe able to use them; you should also be able to identify situations where they cannot beapplied, i.e., where some condition of the theorem is not satisfied.

Theorem D.7 (Limit Laws [1, §2.3]) Let a, c be constant real numbers, n be a pos-itive integer, f and g be functions such that lim

x→af(x) and lim

x→ag(x) exist. Then

limx→a

c = c (98)

limx→a

x = a (99)

limx→a

[f(x) ± g(x)] = limx→a

f(x) ± limx→a

g(x) (100)

limx→a

c · f(x) = c · limx→a

f(x) (101)

limx→a

(f(x) · g(x)) = limx→a

f(x) · limx→a

g(x) (102)

limx→a

f(x)

g(x)=

limx→a

f(x)

limx→a

g(x)if lim

x→ag(x) 6= 0 (103)

limx→a

[f(x)]n =[

limx→a

f(x)]n

(104)

limx→a

xn = an (105)

limx→a

n√

x = n√

a

if n is even, we assume a > 0; (106)

limx→a

n√

f(x) = n

√

limx→a

f(x)

if n is even, we assume limx→a

f(x) > 0. (107)

2.3 Exercises

[1, Exercise 10, p. 112] 1. What is wrong with the following equation?

x2 + x − 6

x − 2= x + 3 ?


2. In view of part 1, explain why the equation

limx→2

x2 + x − 6

x − 2= lim

x→2(x + 3)

is correct.

Solution:

1. There is nothing wrong with the first equation, unless one wishes to interpretit as implicitly suggesting that the condition is to be true for all x. Anequation is a statement — a sentence. This sentence is, in fact, true for allbut one real number (since x2 + x − 6 = (x + 3)(x − 2) for all x). But, forthe value x = 2, the left side of the equation is undefined, while the right sideis still meaningful. It’s not that the equation is false for x = 2 — it is thatthe statement does not even make sense there, since the left side refers to adivision operation that is forbidden, as we have no meaning for the quotient0

0. This is what the textbook intends by “wrong”.

2. When we consider a limit as x → 2, we expressly avoid any consideration ofthe behavior of the function at x = 2. While the function x + 3 is defined atx = 2, that is completely irrelevant, as we don’t need to consider that value infinding the limit; the function on the left is undefined at x = 2, but that doesnot prevent us from showing that it has a limit as x → 2. Thus the equationbetween limits is meaningful. It is also true, since the value of the limit is, inboth cases, 5.

[1, Exercise 12, p. 112] Evaluate the limit, if it exists of limx→−4

x2 + 5x + 4

x2 + 3x − 4.

Solution: Since the limit of the denominator is 0, we cannot apply the QuotientLaw here. But both numerator and denominator factorize:

limx→−4

x2 + 5x + 4

x2 + 3x − 4= lim

x→−4

(x + 1)(x + 4)

(x − 1)(x + 4)

= limx→−4

x + 1

x − 1since x + 4 is a non-zero common factor away from x = 4

=lim

x→−4(x + 1)

limx→−4

(x − 1)

for the Quotient Law is applicable now

=−4 + 1

−4 − 1=

3

5.


[1, Exercise 22, p. 112] Evaluate, if it exists,

limh→0

√1 + h − 1

h.

Solution: Because the limit of the denominator is 0, we cannot apply the QuotientLaw...yet. Instead, we first perform an operation, sometimes called rationalizationwhich transforms the difference of unpleasant expressions in the numerator intoa sum in the denominator. This will permit a simplification, after which we canapply the Quotient Law.

The logic is one we will use repeatedly: we will multiply by a factor, therebydistorting the function, and then divide it out, thereby restoring the function toits original value.

limh→0

√1 + h − 1

h= lim

h→0

(√

1 + h − 1

h·√

1 + h + 1√1 + h + 1

)

= limh→0

(1 + h) − 1

h(√

1 + h + 1)

= limh→0

h

h(√

1 + h + 1)

= limh→0

1

1(√

1 + h + 1)

dividing numerator and denominator by h,

which we may assume is non-zero

= limh→0

1√1 + h + 1

=limh→0

1

limh→0

√1 + h + lim

h→01

by the Quotient and Sum Laws

=1

limh→0

√1 + h + 1

since we know the limit of a constant

=1

√

limh→0

(1 + h) + 1

=1√

1 + 1=

1

2.



Release Date: Monday, September 25th, 2006subject to correction

Since there appear to be temporary difficulties in accessing WebCT viathe usual route, I have sent to all students an e-mail message whichdescribes an alternative method of access.You don’t need to pass through WebCT to access WeBWork; instead,you may follow the instructions on page 4001 of these notes; namely,direct your browser to one of the sites

http://msr04.math.mcgill.ca/webwork/m140f06or

http://msr05.math.mcgill.ca/webwork/m140f06,

depending on the last digit of your student number: even last digits goto the first URL above, odd last digits go to the second; 0 is consideredto be even.

Some of this material was planned to be discussed during an earlier lecture; my planswere disrupted by my absence from one lecture.

D.7.1 §1.6 Inverse Functions and Logarithms (conclusion)

Review item: Change of base in logarithms In my notes for the last lecture onpage 2039 is included the following formula, for positive constants a and b:

loga x

loga y=

logb x

logb ywhich is equivalent to

loga x

logb x=

loga y

logb y(108)

Proof: If we take logarithms to base b on both sides of the equation

aloga x = x ,

we obtainloga x · logb a = logb x ,

implying thatlogb x

loga x= logb a, which is a constant, independent of x. This yields equation

(108), which is equivalent to the desired equation.


1.6 Exercises

[1, Exercise 1.6.18, p. 75] “Let f(x) = 3 + x2 + tan(πx

2

)

, where −1 < x < 1.

(a) Find f−1(3).

(b) Find f (f−1(5)).”

Solution: The first step in solving this problem is to show that there is an inverse.We cannot do so by following the boxed instructions given earlier and producingan explicit formula for the inverse, since we have no way of solving the definingequation for x in terms of f(x). However, we know that, for 0 ≤ x < 1, the functiontan πx

2is the sum of three functions, one of which is constant, and the other two are

increasing; hence, the sum is increasing, and, for x in that portion of the domain,the function possesses an inverse. For −1 < x ≤ 0 it is more difficult to prove thatthe function is increasing, although we will be able to do this using the methods of[1, §4.3, p. 296]. Thus this problem is premature, unless the author were to restrictthe domain to 0 ≤ x < 1.

(a) Once we know the inverse exists, we do not need to have an explicit formula,since we can see by inspection that the specific value of 3 for f is attained.Evidently f(0) = 3 + 02 + tan 0 = 3. Thus f−1(3) = 0.

(b) But we can’t use this approach for the second part of the problem, sinceinspection does not yield a convenient value for f−1(5). Here we may applythe “cancellation equation” (92) to conclude that

f(

f−1(5))

= 5 .

Since we can find the “answers” in both cases, does my objection about the proofof invertibility apply? Without having proved invertibility, we can’t be sure thatthe graph of f doesn’t cross the line y = 3 more than once: so the invertibility isneeded to prove the uniqueness of the answer.

[1, Exercise 1.6.28, p. 75] “Find a formula for the inverse of the function y = 1 + ex

1 − ex .”

Solution: Here the “boxed” procedure in the textbook [1, p. 66], given on page2035 of these notes, is applicable. Solving the given equation algebraically yields

ex =y − 1

y + 1.


We next apply the inverse function of ex to both sides of the equation, to obtain

ln (ex) = lny − 1

y + 1

⇒ x = lny − 1

y + 1

If we wish to express the inverse function in terms of an independent variablenamed x, we have to interchange the symbols: the inverse function is y = ln x−1

x+1or

x 7→ lnx − 1

x + 1.

The domain of the inverse function consists of those x for which the ratiox − 1

x + 1is

positive, which can be shown to be (−∞,−1)∪ (1, +∞), or R− [−1, 1], or |x| > 1(short for {x : |x| > 1}).

[1, Exercises 1.6.15–16, p. 75] I do not approve of these problems, which ask “Usea graph to decide whether f is one-to-one.” While the problems are marked withthe symbol the author uses for problems requiring the use of technology, and weare normally omitting such materials, that is not the reason for the objection.The objection is to the word “decide”: in this course we will often use a graph tosuggest a property, but will never use a graph to decide anything. The reason forthe objection is that, no matter how fine the technology is able to produce a graph,we could contrive functions whose important properties are not visible under thetechnology.

Inverse Trigonometric Functions None of the six trigonometric functions is one-to-one (injective), so none of them has an inverse! What we will be calling the inversesof trigonometric functions will actually be inverses of restrictions of these functions to asmaller subdomain. In the case of sin, cos, tan, cot, there is a “natural” choice for thisrestricted domain; these are not the only possible choices, but they are the ones that aremade in practice around the world at the present time.

The functions sec and csc also fail to be invertible, since their graphs cross somehorizontal lines more than once; i.e. they violate the “Horizontal Line Test” [1, p. 64].But we can find an inverse for a restriction of either function to a smaller domain. Unlikethe situations for the functions sin, cos, tan, cot, the author of the textbook can arguethat there is no “natural” choice for which restricted domains one should use. Someauthors select as the domains the same domains they used for the reciprocals of thesefunctions, except that they delete the points where cos and sin were zero. Such a selectionwould give us the restricted domains [0, π]−

{

π2

}

for the cosine, and[

−π2, π

2

]

−{0} for the


sine; within these respective domains the functions are one-to-one, and can be inverted.The author of your textbook finds these selections to be problematical. One reason isthat later, when we wish to determine the derivatives of the inverse functions, thesechoices would lead to unpleasant formulæ which involve the absolute value function.Possibly for this reason, the author selects a different part of the domain of the cosand sin functions. The resulting functions have graphs consisting of two branches whichare separated by an empty horizontal band; the possible choices mentioned earlier wouldalso have produced graphs with two branches, but the graphs would have been separatedonly by an empty horizontal line of 0 height. Since we have chosen to use the Stewarttextbook, we shall adopt the Stewart choice.56 For trigonometric functions, (and laterfor hyperbolic functions) the inverse functions are denoted either with a superscripted−1, or by prefixing the function name with the particle arc; thus the inverse functionassociated with the portion of the sine function defined for the variable between −π

2and

+π2

is often denoted by arcsin, and called the arcsine function.The discussion of inverse trigonometric functions in [1, §1.6] will be completed when

we reach [1, §3.6]. At that time we will explain further why the author of the textbookhas chosen to define the inverse secant and inverse cosecant functions in a way thatappears to be non-intuitive. Because these two definitions are difficult, we may avoidusing the two functions in question, except to illustrate the essential conditions neededfor an inverse function to exist.

The basic fact is that because all six trigonometric functions have graphswhich fail to satisfy the Horizontal Line Rule, none of them has an inverse!We cope with this difficulty in every case by restricting ourselves to a subset of thedomain of the original trigonometric function. There are infinitely many ways in whichsuch a restriction could be carried out. In the case of the sine, cosine, tangent, andcotangent we select a subdomain which one could argue is “natural”. Your textbook’sselected subdomain for the secant and cosecant is “unusual”, and I shall discuss thereasons for the author’s choice later in the term. In the meantime I will not expect youto be able to work with inverse secants and cosecants. Here are the restricted domainsthat we select for inversion of the sine, cosine, tangent, and cotangent:

56Other choices would have been possible, even for the inverses of the other trigonometric functions.For example, we could have chosen, for sin, to invert the portion of the graph of sin that is above thefollowing union of intervals:

[π

4,π

2

)

∪(

5π

4,3π

2

)

∪(

11π

4,13π

4

]

.

An inverse sine function determined by the restriction of the sine function to this union of intervalswould be just as useful as the function we usually denote by sin−1 or arcsin. The computations would,however, be hideous and nonintuitive.


function restricted domain for inversion

sin x −π2 ≤ x ≤ +π

2

cosx 0 ≤ x ≤ +π

tanx −π2 < x < +π

2

cotx 0 < x < π

The intervals shown in this table are also the images or “ranges” of the 4 respectiveinverse functions. I tabulate domains and ranges of the inverse functions, using the“arc-” names for these functions, instead of the “function−1” names:

function domain image=range

y = arcsinx −1 ≤ x ≤ 1 −π2≤ y ≤ +π

2

y = arccosx −1 ≤ x ≤ 1 0 ≤ y ≤ +π

y = arctanx −∞ < x < +∞ −π2 < y < +π

2

y = arccot x −∞ < x < +∞ 0 < y < π

(Note that in this last chart I have called the independent variable x and the dependentvariable y.) You should verify that the “cancellation equations”:

sin(arcsin x) = x for −1 ≤ x ≤ 1cos(arccos x) = x for −1 ≤ x ≤ 1tan(arctan x) = x for −∞ < x < ∞cot(arccot x) = x for −∞ < x < ∞arcsin(sin x) = x for −π

2≤ x ≤ π

2

arccos(cos x) = x for 0 ≤ x ≤ πarctan(tan x) = x for −π

2< x < π

2

arccot(cot x) = x for 0 < x < π

In the cases of the first four lines of the preceding table, the compositions have thedesired property wherever they “make sense”. But, in the case of the last four lines, thecompositions “make sense” for real numbers x outside of the designated intervals. Thenthe compositions do not have the stated properties!

You need to try to understand how the inverse trigonometric functions are con-structed. The construction always involves selecting a portion of the latter and reversingthe roles of the x- and the y-axes. We will have frequent occasions to need to rememberthe images of the inverse functions in the course of other calculations: typically whatwill be involved will be the choice of signs in some square root calculation.

Example D.8 Construction of the Inverse Sine Function, arcsinx or sin−1x.


1. Start with the sine function, (cf. Figure 4, page 2019).

2. Select a (maximal) portion of the domain for which the Horizontal Line Test issatisfied; usually we select the portion of the domain −π

2≤ x ≤ +π

2(cf. Figure

10). The graph of the inverse function, y = arcsin x = sin−1 x (cf. Figure 11,

1

-1 100 5-10 -5

Figure 10: Invertible restriction of the Function sin x

page 2052) can be obtained reflecting the selected portion of the graph of the sinefunction in the line y = x.

[1, Exercise 1.6.70, p. 77] Simplify the expression tan(arcsin x).

Solution:

Textbook SOLUTION 1: Let y = arcsin x. The first step is to try to expressthe tangent of y in terms of the sine:

tan y =sin y

cos y.

We know an identity that relates sines and cosines:

sin2 y + cos2 y = 1 ,

which could be solved to yield

cos y = ±(

1 − y2)

12 .

But which sign is correct? Are they both right? The answer is that they arenot both right! To resolve this dilemma we have to remember how the inversesince function was defined; and the resolution depends on the specific choicesthat we made when we constructed the inverse function. We decided to invertthe portion of the sine function defined for a variable between −π

2and +π

2.

So we know that y is in either the fourth or the first quadrant; and the cosine


-0.5

-1

10.50

1

0-1

Figure 11: Graph of the Inverse Sine Function

function is positive in these quadrants. Thus the correct sign here is +. Nowwe can complete the calculations:

tan(arcsin x) = tan y

=sin y

cos y

=sin y

+(

1 − sin2 y)1

2

=sin(arcsin x)

+(

1 − sin2(arcsin x))

12

=x√

1 − x2.


1

-1

1.50.5-1.5 10-0.5-1

Figure 12: Reflection (in red) of the restriction of sin x in the line y = x

NOTE: Your proof could be considered incomplete unless you explain clearlywhy you choose a particular sign when there was a choice of 2; it may not beenough to make the right choice without a clear explanation!

Textbook SOLUTION 2: (cf. [1, p. 74]) The textbook gives a second proofthat uses a diagram. As usual, I urge you to avoid proofs using diagrams, asthey tend to be incomplete. Here again it is simplest if we begin by definingy = arcsin x, and conclude that

sin y = x .

The author then constructs a right-angled triangle in which one angle is torepresent y. For its sine to be equal to x he denotes the side opposite y by


x, and the hypotenuse of the triangle as 1. Then, this being a right-angledtriangle, he applies the Theorem of Pythagoras to conclude that the 3rd sideof the triangle is equal to

√1 − x2. The tangent of y can then be read off

from the triangle.

What’s wrong with this proof? The essential difficulty is that the proof isimplicitly assuming that y ≥ 0; but the property we proved above holdsfor negative y as well — so this proof covers only “half” of the cases. Thisobjection could be removed by producing another diagram that holds for thenegative cases; this is delicate, since we would have to draw the triangle in acoordinatized plane.

In practice the diagram method “works”, but only because of the carefulway in which we have decided to invert our functions: it has its place in ahigh school, where the main interest is in positive, acute angles. I wouldrecommend that students in a science course like MATH 140 avoid this typeof solution.

(One student has observed that it would have been better form had the textbookrestricted the values of x to −π

2< x < π

2, since the tangent function is not defined

at ±∞.)

[1, Exercises 1.6.15–16, p. 75] I do not approve of these problems, which ask “Usea graph to decide whether f is one-to-one.” While the problems are marked withthe symbol the author uses for problems requiring the use of technology, and weare normally omitting such materials, that is not the reason for the objection.The objection is to the word “decide”: in this course we will often use a graph tosuggest a property, but will never use a graph to decide anything. The reason forthe objection is that, no matter how fine the technology is able to produce a graph,we could contrive functions whose important properties are not visible under thetechnology.

[1, Exercise 40, p. 112] Find limx→−4−

|x + 4|x + 4

, if it exists. If the limit does not exist,

explain why.

Solution: Numerator and denominator both approach 0, so the Quotient Law maynot be applied. However

|x + 4|x + 4

=

−x + 4

x + 4when x < −4

x + 4

x + 4when x > −4

=

{

−1 when x < −41 when x > −4.


We see that the limit from the right is +1, while the limit from the left is −1. Asthese one-sided limits are different, the “two-sided” limit does not exist.

Corollary D.9 (to Theorem D.7) If f is a polynomial or a rational function (=aratio of two polynomials) defined at a point a, then lim

x→af(x) = f(a).

[1, Theorem 2.3.1, p. 109] is the same as [1, Theorem 2.2.3, p. 98], which was introducedbefore we knew precisely what we intended limit to mean.

The “Squeeze” Theorem, [1, Theorem 2.3.3, p. 110] (The name of this theoremis not standard.) If you trap the values of a function g(x) between two other functionsf(x) and h(x), so that

f(x) ≤ g(x) ≤ h(x)

for all x in an interval of positive width around a real number a, and if the boundingfunctions f(x) and h(x) have the same limiting value as x → a, then g(x) also approachesa limit as x → a, and that limiting value is the common limiting value of f(x) and h(x).Symbolically, we can write

If

f(x) ≤ g(x) ≤ h(x)

limx→a

f(x) = limx→a

h(x)then lim

x→af(x) = lim

x→ag(x) = lim

x→ah(x) .

For this discussion the values of the functions at x = a are irrelevant.

The “greatest integer function” [[x]]. The function that your textbook denotesby [[x]] has also been denoted simply by [x]. Mathematicians, and computer scientistsnowadays usually denote this function by a different symbol, ⌊x⌋, and call it the “floor”function; an analogous function is denoted by ⌈x⌉, and called the “ceiling” function. Thevalue is defined to be the largest integer n whose value does not exceed x; for x ≥ 0[[x]] can be obtained by truncating the decimal expansion of x at the decimal point. Thegraph of [[x]] is a rising sequence of horizontal line segments of length 1, each containingthe left end-point but not the right. The function does not have a limit at any integerpoint, because the limits from the left and right are different; but, at any point a otherthan an integer point, the limit exists and is equal to [[a]].

2.3 Exercises (continued)

cf. [1, Exercise 50, p. 113] Let f(x) = x − [[x]].

(a) Make a crude graph of f .


(b) If n is an integer, evaluate limx→n−

f(x) and limx→n+

f(x).

(c) If α is not an integer, evaluate limx→α−

f(x) and limx→α+

f(x).

(d) For what values of a does limx→a

f(x) exist, and what is its value?

Solution:

(a) The graph is a sequence of line segments with slope 1, starting from an integerpoint (n, 0) on the x-axis (which point is included in the graph) and risinguntil the point (n + 1, 1), which is excluded from the graph.

(b) At any integer point the limit from the right is 0, which is also the functionvalue. But the limit from the left is +1.

(c) At any non-integer point α the limit from the left is equal to the limit fromthe right is equal to the function value, f(α).

(d) The limit exists at all points in R except at the integers.

[1, Exercise 36, p. 112] If 3x ≤ f(x) ≤ x3 + 2, for 0 ≤ x ≤ 2, evaluate limx→1

f(x).

Solution: The function f is “trapped” between the two given functions when 0 ≤x ≤ 2. Note that the 2 inequalities given imply that 3x ≤ x3+2, but we could provethat if we had to, by observing that this inequality is equivalent to x3−3x+2 ≥ 0,i.e., to (x − 1)2(x + 2) ≥ 0; since (x − 1)2 is a square, it is always non-negative,so the product with (x + 2) is positive for x > −2, which is where the interval0 ≤ x ≤ 2 is located. Both of the functions 3x and x3 + 2 have limit 3 as x → 1.The Squeeze Theorem implies that the function f , being trapped between thesetwo functions, also has limiting value 3 as x → 1.

Here we are introducing the first of these two generalizations. Students must understandthat this is a convenience of terminology, but that even when lim

x→af(x) = ±∞, we will

still say that the limit does not exist : we will speak of a limit existing only when it is a(finite) real number.

As for limits at a real number (as opposed to limits at ±∞), one can prove theanalogue of

limx→a

f(x) = L if and only if limx→a−

f(x) = L and limx→a+

f(x) = L

[1, p. 98] when we replace L by +∞ or −∞. Thus [1, Definition 6, p. 100] can besharpened:


Vertical Asymptotes

Definition D.4 The line x = a is called a vertical asymptote of the curve y = f(x) ifat least one of the following 4 statements is true:

limx→a−

f(x) = +∞ limx→a+

f(x) = +∞lim

x→a−f(x) = −∞ lim

x→a+f(x) = −∞

When does a limit exist? We say that limx→a

f(x) exists whenever there is a real

number L such that limx→a

f(x) = L. For a limit to exist it essential that L be a real

number; even iflimx→a

f(x) = ∞ or limx→a

f(x) = ∞ ,

we say that limx→a

f(x) does not exist : a fortiori we also say the limit does not exist when

left and right limits are different, or when one or other or both of them do not exist.Sometimes authors will include the word finite in a statement, as in Suppose that a finitelimit exists... to emphasize this distinction.

Whenever we have a statement of one of the forms limx→a−

f(x) = L, limx→a+

f(x) = L,

limx→a

f(x) = L, we can assert that the one- or two-sided limit referred to exists ; the

statement that it exists is weaker than a statement which actually asserts the value ofthe limit — but sometimes we may know that a limit exists without knowing its exactvalue!

Infinite Limits The definition for finite limits is modified to describe functions thatbecome arbitrarily large positively or arbitrarily large negatively as one approaches agiven finite value. The changes in the symbol just involve replacing a constant by ±∞,but the changes in the definition are more substantial. However, we shall see that thesekinds of limits have similar properties to finite limits; this justifies the use of a similarnotation to describe them.

“Turning the crank”. Students like to be able to treat calculus like a machine: dumpin the function, turn the crank, and out comes the limit, or the derivative, etc. We aregoing to discuss some Limit Laws which will enable you to evaluate limits for manycommon types of functions. But it is not hard to manufacture functions for which theserules together will not be enough to give you the value of a limit. Fortunately, few ofthese “pathological” functions will be studied in MATH 140. One type of case that wewill be considering frequently will involve functions that behave in one “nice” way onone side of a point x = a, and in another way on the other side. It will be easy to find


the limits from the left and right; if they are different, the function does not have a limitat the point.

D.7.2 §2.5 Continuity.

Definition of continuity The definition of

the function f is continuous at the point x = a

looks like a single condition:limx→a

f(x) = f(a) (109)

but is better remembered as three conditions:

1. f is defined at a, i.e. a is in the domain of f ;

2. lim f(x) exists as x → a; and

3. limx→a

f(x) = f(a).

Our limit notation includes all of this information in the single statement (109). Similarinterpretations can be made for the statements

limx→a−

f(x) = f(a) (110)

limx→a+

f(x) = f(a), (111)

which we read, respectively as

f is continuous from the left at aand

f is continuous from the right at a



Release Date: Wednesday, September 27th, 2006subject to correction

D.8.1 §2.4 The Precise Definition of a Limit (not examination material in2006-07).

As posted in the course outline, there will be no lecturein this section of MATH 140 on Monday, October 3rd,2006.

The order of doing mathematics For thousands of years mathematics has followeda logical order:

• First we define the terms that are to be used.

• Then we state the results we plan to prove.

• Then we prove what we claim.

• Finally, we may apply the new tools that we have forged.

Often there will be interruptions to this sequence where we may experiment with ideas, toexplore new possibilities, and to motivate the subsequent steps. Your textbook frequentlyincludes such motivational sections; §§2.1 and 2.2 are largely of this type. Where yousee, in these sections, statements that appear to be results you should interpret themas being advance hints of results that will become available after the formal definitionsare announced and theorems have been proved. That is the nature of, for example, [1,Theorem 2.2.3, p. 98]; the result is stated again as [1, Theorem 2.3.1, p. 109]. Some ofthe “definitions” in [1, §2.2] will become perfectly valid once the basic concept of limitis defined. But [1, Definition 2.2.1, p. 93] is purely motivational, and will be replacedby [1, Definition 2.4.2, p. 115]. In your first reading of the subject, these motivationalstatements help you to understand the purpose of the formal definitions.

The main formal definitions appear in [1, §2.4]. When we discuss [1, §2.3], studentsshould understand that we are actually assuming [1, §2.4]. Your responsibility will beto be able to be able to use the results in [1, §2.3] intelligently, without necessarilyunderstanding how they are proved. Most of our demands on the limit concept canbe satisfied with these results; we will return to it in [1, §2.6], where we generalizethe concept to lim

x→∞f(x) and lim

x→−∞f(x). In MATH 141 we will have several occasions

to return to the definition for further generalizations; but you will not be expected to


work extensively with the formal definitions in either of these courses: we are spendingsome time on the definition because we believe that, without exposure to it, you cannotunderstand how the calculus “works”. A logical order for reading these sections is

• Sections 2.1, 2.2 for motivation

• Section 2.4 — the formal definitions

• Section 2.3 — the basic theorems about limits

• Section 2.5, etc.

A game between x and y This part of the lecture will begin with a careful discussionof the formal definition of the statement lim

x→af(x) = L, where a and L are real numbers

[1, Definition 2.4.2, p. 115]. One way of understanding this definition is to think of a2-player game. The game is played with a function f . Player A chooses a real number, a,and player L responds with a choice of a real number L, the number that he suspects tobe the “limit” of f as x approaches a. Play proceeds as follows: A prescribes a tolerance,representing half of the width of a horizontal band that is centred at the line y = L;this half-width is traditionally denoted by the lower case Greek letter epsilon, ǫ. ThenL has to produce a positive real number, traditionally denoted by the lower case Greekletter delta, δ, with the property that all the points in the interval a − δ < x < a + δ(except possibly the point a) must have function value within a distance of ǫ from L —equivalently, that the graph of the function for the points in the given punctured intervalsurrounding x = a lies entirely in the vertical band of width 2ǫ centred at the line y = L;symbolically, L needs to produce δ such that

0 < |x − a| < δ implies that |f(x) − L| < ǫ .

Such a game would never end, since there are infinitely many values of ǫ that can bechosen. L wins the game if he can produce a procedure or formula that will always yielda value of δ, for any ǫ that A tries.57 When L wins we say that lim

x→af(x) exists and is

equal to L. A wins if she can show that there exists no procedure that can accomplishthe task; then we say that either lim

x→adoes not exist , or that it exists but is not equal to

L.How can A win? She has to show that there is some real number ǫ such that it is

impossible to find a real number δ with the property described above; that is, a positivereal number ǫ so that, no matter how close we get to the point a, there are points x forwhich f(x) is at least a distance of ǫ from L. We may discuss briefly examples of thissituation.

57This overstates the requirement. All L needs is to demonstrate that a value of δ exists — it isn’teven necessary to show how to find it.


Limits from the left and right. The preceding definition is modified in the obviousway to permit us to speak of “one-sided” limits. We say that lim

x→a−f(x) = L if, for every

positive real number ǫ, there exists a positive real number δ such that

a − δ < x < a implies that |f(x) − L| < ǫ ;

and that limx→a+

f(x) = L if, for every positive real number ǫ, there exists a positive real

number δ such that

a < x < a + δ implies that |f(x) − L| < ǫ .

Example D.10 Let f(x) =

3x + 1 when x < 02 when x = 0x2 when x > 0

.

1. Find limx→0−

f(x), or show that it does not exist.

2. Find limx→0+


3. Find limx→0


Solution:

1. Examination of the function f for x < 0 (using a graph or a table of values) suggeststhat, as x moves close to 0, the function value will also be close to 1; this suggeststhat the limit from the left is 1 — but this suggestion is not a proof. Given anypositive real number ǫ > 0, we find that we can confine the graph in a horizontalband of width 2ǫ centred at the line y = 1 by taking δ = 1

3ǫ: when

− ǫ

3< x < 0 ,

1 − ǫ < 3x + 1 < 1, so−ǫ < f(x) − 1 < 0 < ǫ .

As this argument holds for any positive real number ǫ, we conclude that

limx→0−

f(x) = 1 .

2. Examination of the function f for x > 0 (using a graph or a table of values) suggeststhat, as x gets close to 0, the function value gets close to 0; this suggests that thelimit from the right is 0 — but this suggestion also is not a proof. Given any


positive real number ǫ > 0, we find that we can confine the graph in a horizontalband of width 2ǫ centred at the line y = 0 by taking δ =

√ǫ: when 0 < x <

√ǫ,

0 < x2 < ǫ, so−ǫ < f(x) − 0 < ǫ .

As this argument holds for any positive real number ǫ, we conclude that

limx→0+

f(x) = 0 .

3. The limits from the left and right are different. To show that limx→0

f(x) does not

exist, we can consider the specific value ǫ = 14

(any positive number less than 12

will do here). To the right of x = 0 the graph of f is close to the x-axis; to theleft of x = 0 the graph is above the line y = 1. No horizontal line y = L is within14

of both of these parts of the graph, so there is no number L which could be thevalue of the limit. This argument could be made more precise by giving preciseinequalities. This proof needed the fact that we were working with ǫ “sufficientlysmall”: if we tried to give the same argument when ǫ = 1, it would not be correct;but any value less than 1

4would also work.

In future we will simply refer to the theorem [1, p. 98] (Theorem D.6 of thesenotes): a limit exists if and only if the limits from the left and right exist and areequal.

D.8.2 §2.5 Continuity (continued)

Classification of Discontinuities. We call a point where a function fails to be con-tinuous a discontinuity . If f has a discontinuity at x = a, but it is possible to extend thedefinition of the function so as to assign a value to the function at the offending point,or to change a value already assigned in such a way that the new function is continuousthere, then we say that the discontinuity is removable. This can happen in various ways:

* Sometimes a point a is not in the domain only because the inventor of the functiondid not choose to define the function there. For example, if we were to define

f1(x) = 6 for all x different from 1

and say nothing about the value of f1 at x = 1, that discontinuity may be removedby extending the definition to give the function the value 6 at x = 1. (Technicallywe should58 also change the name of the function now, since the extended function,having a changed domain, is not the same function as before.)

58but usually don’t


** Sometimes a point a is not in the domain because some operation on the expressionsthat were used to define the function may not be defined. For example, if we define

f(x) to be2(x − 1)

x − 1, then f is not defined at x = 1. Here the limit as x → 1 does

exist, but the graph of f has a hole in it over the point x = 1 on the x-axis. Wecan remove this discontinuity by assigning to the function the value 2 at x = 1.

*** Sometimes a point a is in the domain, but the inventor has assigned a functionvalue there that causes the discontinuity; it may be possible to change the functionvalue at x = a and thereby remove the discontinuity, as in the function

f2(x) =

{

2 if x 6= 17 if x = 1

where we can shift the function value at x = 1 from 7 to 2 and thereby remove thediscontinuity. Here again, the change in the function should remind us to changeits name.

Some discontinuities cannot be removed : there may be no way of repairing the functionat x = a to make it continuous. This can happen, for example

• if limx→a−

and limx→a+

both exist, but are different — we call this a jump discontinuity;

• if at least one of limx→a−

or limx→a+

is ∞ or −∞, i.e., if the line x = a is a vertical

asymptote to the graph — we call a an infinite discontinuity;

• more generally, if at least one of limx→a−

or limx→a+

does not exist, since continuity at a

requires the existence and equality of both one-sided limits

Continuity on an interval. A function is continuous on an open interval (c, d) if it iscontinuous at every point in the interval. We extend this concept to intervals includingeither or both of their left and right end-points, i.e. to [c, d), (c, d], [c, d], by requiringthat the appropriate one-sided continuity hold when an end-point is included. So, forexample, f is continuous on [a, b] if it is continuous at every point in (a, b) and if, inaddition, it is continuous from the right at a and from the left at b.

Consequences of the Limit Laws [1, Theorem 3, p. 127]. Functions that arecontinuous at the same point can be added, subtracted, multiplied, and divided (providedthe divisor has non-zero value). There are no surprises here.


Large families of continuous functions. In [1, Theorem 7, p. 129] the author sum-marizes his comments about various classes of functions, observing that all of the follow-ing types of functions are continuous wherever they are defined :

• polynomials; more generally,

• rational functions

• “root” functions

• trigonometric functions

• inverse trigonometric functions

• exponential functions

• logarithmic functions

(to be continued)

We can apply [1, Theorem 7, etc., §2.5] to evaluate limits at a point in the domain of afunction. Since continuity at x = a implies that the limit of f(x) as x → a is equal tothe function value f(a), we need only evaluate f at a to find the limit.59

Theorem D.11 1. If a function f(x) is continuous from the right at x = a, thenlim

x→a+f(x) = f(a).

2. If a function f(x) is continuous from the left at x = a, then limx→a−

f(x) = f(a).

3. If a function f(x) is continuous at x = a, then limx→a

f(x) = f(a).

Thus the only difficulties in finding limits of these functions will be at points not in theirdomains, or as x → ±∞.

The Intermediate Value Theorem. This is an important result, enabling us todetermine with any desired accuracy the points where graphs of continuous functionscross a horizontal line, in particular, the x-axis. As stated, the result is an existencetheorem, in that it asserts that a number exists, without telling you how to find it.However it may be shown in class that one can apply the result to obtain approximationsto the desired numbers. These approximations do not, in theory, require the use of acalculator; but, as the arithmetic becomes more and more complicated, the algorithm

59The fact that the limits of continuous functions can be found by substitution was seen earlier in the“Direct Substitution Property” stated for polynomials and rational functions in [1, p. 107].


we will describe is not practical for hand calculation. For that reason, we will not expectyou to carry out extended calculations at quizzes or the final examination.

The proof of this theorem is beyond the course, but you are expected to understandthe statement [1, Theorem 10, p. 131], and how to use it:

Theorem D.12 (Intermediate Value Theorem) Let f be continuous on the closedinterval [a, b], and let N be any number such that f(a) ≤ N ≤ f(b) or that f(b) ≤ N ≤f(a). Then there exists a number c in [a, b] such that f(c) = N .

Example D.13 While the Intermediate Value Theorem is an existence theorem, we canderive from its statement, a constructive procedure or algorithm to determine where acontinuous function assumes a particular value. Suppose, for example, we wished todetermine

√3 to any desired accuracy. We can consider the equation f(x) = x2. By

trial and error we observe two values of the function above and below 3: f(0) = 0 < 3,f(2) = 4 > 3. The theorem tells us that there will be a point in the interval [0, 2] atwhich f(x) = 3. If we cut the interval in half, and determine the value of f at themid-point x = 1, we find that f(1) = 12 = 1. As this is less than 3, we can now confineour attention to the half-interval [1, 2], and repeat the operation. We test the function atthe mid-point, 3

2, and find that f

(

32

)

= 94

< 3: thus we can now work with the interval[

32, 2]

. The mid-point of this interval is32+2

2= 7

4, at which point f

(

74

)

= 4916

> 3. Now we

work with the interval[

32, 7

4

]

. We evaluate f at the mid-point of this interval,32+ 7

4

2= 13

8:

f(

138

)

= 16964

< 3, so the next interval we work with is[

138, 7

4

]

. The mid-point is 2716

, etc.If we carry out this procedure 10 times, we reduce the size of the interval considered bya factor of 210 = 1024 > 1000, so every 10 applications improve our approximation by 3decimal places.

Several comments are needed:

• There are other, better ways of determining square roots — this example is pre-sented only to illustrate a consequence of the Intermediate Value Theorem.

• We can’t use an approach like this to approximate solutions to an equation wherethe functions involved are not continuous. For such functions there might not evenbe a solution. So, for example, the function [[x]] never takes the value 1

2, even

though it does have values 0 = [[0]] and 1 = [[1]] and 0 < 12

< 1. This exampleshows that failure of the hypotheses to hold, even at just one point, can cause thetheorem to be inapplicable. (The interval is [0, 2], and the point of discontinuity is1.)

We will return to this topic in connection with [1, §4.2 The Mean Value Theorem], whenwe will refine these investigations to determine maximum numbers of solutions in a giveninterval.


2.5 Exercises

[1, Exercise 2.5.41, p. 134] . For what value of the constant c is the function fcontinuous on (−∞, +∞)?

f(x) =

{

cx + 1 if x ≤ 3cx2 − 1 if x > 3

? (112)

Solution: The problem is solved in the Student Solution Manual [3, [p. 49], andhints are available on the CD-Rom “Tools for Enriching Calculus” (cf. §1.5.3).

[1, Exercise 2.5.43, p. 134] is a problem on removable discontinuities. Read the solu-tions in the manual and be sure you understand why certain of the discontinuitiesdescribed are removable and others are not.

Problems 44–54 are concerned with the Intermediate Value Theorem.

Definition D.5 A root or zero of a polynomial f(x) is a number a such that f(a) = 0;as solution to an equation f(x) = 0 is a number a such that f(a) = 0. The textbooksometimes uses the word root to describe what I call a solution, but there is no dangerof confusion.

[1, Exercise 42, p. 134] “Find the constant(s) c that make g continuous on (−∞,∞):

g(x) =

{

x2 − c2 if x < 4cx + 20 if x ≥ 4

”

Solution: First observe that, for x < 4, the function is a polynomial, and weknow that polynomials are continuous everywhere; thus g is continuous for x < 4.Similarly, for x > 4, the function is again a polynomial, and is again continuous. Ihave carefully avoided saying anything about the point x = 4; for all other pointsthe function is a polynomial in an interval surrounding the point, so the limits fromleft and right will exist and be equal. But, for the point x = 4, the description ofthe polynomial is different on the two sides of the point, and we cannot use thisargument in this simple a form.

To the left of x = 4 the function is a polynomial; as we know that polynomials arecontinuous everywhere, we can conclude that the limit of x2 − c2 as x → 4 will bethe value of this polynomial at x = 4, i.e., 42 − c2. Similarly

limx→4+

g(x) = limx→4+

(cx + 20) = c(4) + 20 .


For g to be continuous at x = 4 we need to know, in particular, that the limitexists. We have just seen that the limits from the left and right exist, so we needonly impose the condition that these two one-sided limits are equal:

42 − c2 = 4c + 20

which is equivalent to the equation

c2 + 4c + 4 = 0 , (113)

which has the single solution, c = −2. But we are not done yet. For continuity weneed to know that

• the limit exists as x → 4

• the function is defined at x = 4

• the limit is equal to the function value.

Now we return to the definition, and apply information that we haven’t used yet.That is the specific value of g(4), which is given in the second line of the definitionas c(4) + 20, which we now know must be (−2)(4) + 20 = 12. When c = 2 thecommon value of the limits from left and right is

42 − c2 = 42 − 4 = 12

which is equal to the function value at the point. Now we may conclude that thefunction is continuous everywhere when c = 2, and only for this value of c.

[1, Exercise 50, p. 134] (not discussed in the lectures) “Use the Intermediate ValueTheorem to show that there is a root of the given equation in the specified interval:ln x = e−x, on the interval (1, 2).”

Solution: Because students do not have access to calculators, I will provide60 thefollowing additional information:

2.7 < e < 2.8 (114)

0.69 < ln 2 < 0.7 . (115)

In order to apply the theorem we need to work with a single function. I will definef(x) = ln x − e−x; the same sort of reasoning would work if we took the functionto be, for example, − ln x + e−x, 5(lnx − e−x), (ln x − e−x)

2, ln x − e−x + 17.

60You could have determined (115) easily yourself by experimentally squaring numbers of the form1 + k

10 .



We calculate

f(1) = ln 1 − 1

e= 0 − 1

e< 0

f(2) = ln 2 − 1

e2> 0.69 − 1

(2.7)2= 0.69 − 1

7.29> 0

It is essential that we also observe that f is continuous on the interval [1, 2], since itis the sum of a logarithm and an exponential function, both known to be continuousby our observation on page 2064 of these notes, and since the sum of two continuousfunctions is also continuous. Then, as we have proved that

f(1) ≤ 0 ≤ f(2) ,

we may conclude that f assumes all values between f(1) and f(2), in particular,the value 0, say at a point c such that 1 < c < 2. The proof may now be completedby observing that f(c) = 0 is equivalent to ln c = e−c.

Continuity of the composition of two functions Stewart observes that the com-position of continuous functions is continuous, subject to conditions on the componentsat the appropriate points; more generally, he states that

limx→a

f(g(x)) = f(

limx→a

g(x))

,

provided limx→a

g(x) exists, and that f is continuous at that limit value limx→a

g(x). In par-

ticular this implies [1, Theorem 9, p. 130]:

Theorem D.14 If g is continuous at a, and f is continuous at g(a), then f ◦ g iscontinuous at a.

Example D.15 (Exercise 2.5.34, p. 134) “Use continuity to evaluate

limx→2

arctanx2 − 4

3x2 − 6x.”

Solution: The function in question may be interpreted as a composition (f ◦g)(x), where

g(x) =x2 − 4

3x2 − 6x


and f(x) = arctan x. Sincex2 − 4

3x2 − 6x=

x + 2

3xwhen x 6= 2,

limx→2

g(x) = limx→2

x2 − 4

3x2 − 6x

= limx→2

x + 2

3x

=limx→2

(x + 2)

limx→2

3x

by the Quotient Law

=2 + 2

3 · 2by the continuity of polynomials

=2

3

Then

limx→2

(f ◦ g)(x) = f(

limx→2

g(x))

by continuity of f at 23

= arctan2

3

The answer need not be reduced further, as 23

is not the tangent of a familiar angle, andis approximately 0.59 radians.

Existence of limits I reiterate the convention discussed in §D.8.1, page 2057 of thesenotes. We say that lim

x→af(x) exists only if there is a real number L such that lim

x→af(x) = L,

not if we know that limx→a

f(x) = +∞ or limx→a

f(x) = −∞; while those last two statements

do convey some information about the behavior of the function, we still say that thelimit does not exist in such cases. In practice many of the properties we know aboutlimits (e.g., the various laws) usually hold even when the limits are infinite, provided weavoid situations that cannot be given meaning — like subtracting ∞ from ∞. We couldhave allowed for this by “extending” the real number system to include the two objects,∞, and −∞; we have chosen not to do this because it replaces one set of inconvenienceswith another: we would then have a number system where certain operations would notbe defined, e.g. the multiplication of 0 and ∞, and the subtraction mentioned earlier.In this course we will continue to follow the prevalent convention stated in the secondsentence of the paragraph. “Infinity” and “Minus Infinity” are not real numbers, andare not the value of a function or a limit. When we write “= ∞” or “= −∞”, we aresimply using a notation, or rather an abuse of notation, that we find convenient.


D.8.3 §2.6 Limits at Infinity; Horizontal Asymptotes.

As soon as he has discovered a useful concept and proved a few theorems, a mathemati-cian tries to generalize it. After defining what is meant by lim

x→af(x) = L, we generalized

this definition by defining what we mean by limx→a

f(x) = ±∞. The generalization is a

new definition, but we adopt it because the newly defined “limits” behave similarly tofinite real limits. We can generalize the concept “to infinity” in another way also — bydefining what we mean by lim

x→∞f(x) = L and lim

x→−∞f(x) = L. When either or both of

these limits is equal to L, we say that the line y = L is a horizontal asymptote of thegraph y = f(x). This is intended to say that the vertical distance between the graph andthe line approaches 0 as one moves arbitrarily far to the right or arbitrarily far to theleft on the x axis. The either or both part of the definition is analogous to our definitionof vertical asymptotes, where we demanded that either a limit from the left or from theright or both approach ±∞.

Can a graph cross its asymptote? Some students have the impression that anasymptote is a line that is approached but never met. While this will be the casefor vertical asymptotes, because of the “Vertical Line Test”, it is not a restriction forhorizontal asymptotes: a graph can cross its asymptotes. For example, the graph of the

functionsin x

1 + x2has the x-axis as a horizontal asymptote, and it crosses this asymptote

infinitely often, at every integer multiple of π.

Limits of powers Consider the function f(x) = 1x

as x becomes large. If we want toensure that the graph of this function lies in a narrow horizontal band between the linesy = 0 and y = ǫ — where ǫ is any small, positive number — all we need to do is to takex greater than 1

ǫ; what this argument shows, using a definition that has not been stated

explicitly, is that

limx→∞

1

x= 0 .

Similar reasoning can be applied to any positive power of x: if r is any positive realnumber,

limx→∞

1

xr= 0 ,

or, equivalently,limx→∞

x−r = 0 .

(Your textbook restricts these results to rational real numbers r, since the author hasnot yet defined what he means by an irrational exponent.)


Infinite Limits at Infinity We can also combine the two generalizations of the originallimit definition, and define what we mean by lim

x→∞f(x) = ±∞, and by lim

x→−∞f(x) = ±∞.

The precise definitions are given in the subsection called “Precise Definitions”.

Limits of polynomials and reciprocals of polynomials We will illustrate thegeneral situation by considering some specific examples.

Example D.16 In each of the following cases evaluate the limit, or show that it doesnot exist:

1. limx→∞

(3x5 + 4x − 6)

2. limx→−∞

(3x5 + 4x − 6)

3. limx→∞

1

3x5 + 4x − 6

4. limx→−∞

(3x5 + 4x − 6)2

5. limx→∞

3x5 + 4x − 6

4x7 − 2x5 + 3x

6. limx→−∞

3x5 + 4x − 6

4x7 − 2x5 + 3x

7. limx→∞

4x7 − 2x5 + 3x

3x5 + 4x − 6

8. limx→∞

4x7 − 2x5 + 3x

3x7 + 4x3 − 6x2

Solution:

1. A recommended way of dealing with problems of this type is to factor out fromthe polynomial the leading power of x. Here

3x5 + 4x − 6 = x5

(

3 +4

x4− 6

x5

)

.

As x → ∞ or x → −∞, both4

x4and

6

x5approach 0, so

limx→∞

(

3 +4

x4− 6

x5

)

= limx→∞

3 + 4 limx→∞

1

x4− 6 lim

x→∞

1

x5

by the Sum Law

= 3 + 4 × 0 − 6 × 0 = 3 + 0 + 0 = 3


We see that as the variable becomes infinite, a polynomial behaves like its leadingterm: it is as though we simply disregarded the terms of lower powers. It nowfollows that

limx→∞

(

3x5 + 4x − 6)

= limx→∞

x5 · limx→∞

(

3 +4

x4− 6

x5

)

by a variant of the Product Law

= ∞ · 3 = ∞ .

It looks as though we have violated the instructions about ∞ stated earlier. But, infact, we can show that the various limit laws continue to hold with infinite limits,provided the various operations are defined. In the case of multiplication, a ProductLaw can be proved that would justify taking a product, even if one “factor” is ∞,provided the finite factor is not 0: we cannot give a meaning to a product ∞· 0 or

−∞ · 0. Notwithstanding this comment, we still say that limx→∞

1

3x5 + 4x − 6does

not exist .

2. A result analogous to the preceding holds as x → −∞:

limx→−∞

(

3x5 + 4x − 6)

= −∞ .

3. We may apply the Quotient Law: since the denominator approaches ∞, the re-ciprocal of the polynomial approaches 1

3. Or, we may factor out from the fraction

x−5, whose limit is 0, leaving a factor which approaches limit 13; so the limit of the

product is 0.

4. = ∞, left to the student.

5. We again factor out leading powers;

limx→∞

3x5 + 4x − 6

4x7 − 2x5 + 3x= lim

x→∞

x5 (3 + 4x−4 − 6x−5)

x7 (4 − 2x−2 + 3x−6)

= limx→∞

x5

x7· 3 + 4x−4 − 6x−5

4 − 2x−2 + 3x−6

= limx→∞

x5

x7· lim

x→∞

3 + 4x−4 − 6x−5

4 − 2x−2 + 3x−6

= 0 · 3

4= 0

6. Left to student.


7. This time the first factor has “limit” ∞, while the second has limit 43. The Limit

Law can be shown to applicable, and the limit is ∞.

8. Show that the limit in this case is 43.

Precise Definitions We are not expecting you to work with these precise definitionsin this course.

2.6 Exercises

[1, Exercise 2.6.24, p. 147] “Find limx→−∞

(

x +√

x2 + 2x)

.”

Solution: Since the limits of the two summands are, respectively −∞ and +∞, wecannot apply the Sum Law here; while we extended this and other laws to situationswhere limits are infinite, we cannot give a meaning to −∞ + ∞. Accordingly wemodify the function using a technique related to one we saw earlier:

limx→−∞

(

x +√

x2 + 2x)

= limx→−∞

(

(

x +√

x2 + 2x)

· x −√

x2 + 2x

x −√

x2 + 2x

)

= limx→−∞

x2 − (x2 + 2x)

x −√

x2 + 2x

= limx→−∞

−2x

x −√

x2 + 2x

Note that, superficially, this last function appears “more complicated” than the onewe started with. But there is an important difference: in the earlier function therewas a sum whose value was indeterminate, even if we were prepared to work with±∞; but, in the last quotient, both numerator and denominator are meaningful,provided we are still willing to work with ±∞. Unfortunately there is still a snag:we cannot evaluate the ratio of two functions that are both approaching ±∞.Fortunately, we can cope with that by dividing numerator and denominator by x:

limx→−∞

−2x

x −√

x2 + 2x= lim

x→−∞

−2

1 − 1x

√x2 + 2x

= limx→−∞

−2

1 − 1−|x|

√x2 + 2x

= limx→−∞

−2

1 − 1

−√

x2

√x2 + 2x

= limx→−∞

−2

1 +√

x2+2xx2


= limx→−∞

−2

1 +√

1 + 2x

=−2

limx→−∞

(

1 +

√

1 +2

x

)

by the Quotient Law

=−2

1 + limx→−∞

√

1 +2

xby the Sum Law

=−2

1 +

√

limx→−∞

(

1 +2

x

)

by the continuity of√

=−2

1 +√

1 + 0= −1

Suppose that we wished to determine the horizontal asymptotes to the graph ofx +

√x2 + 2x. The preceding shows that the line y = −1 is one asymptote.

To complete the investigation we would need to determine limx→+∞

(

x +√

x2 + 2x)

.

Since each of x and√

x2 + 2x approaches +∞ as x → +∞, we conclude that

limx→+∞

(

x +√

x2 + 2x)

= +∞ and so there are no horizontal asymptotes other

than the one already found.

[1, Exercise 2.6.38, p. 147] (Not discussed in the lecture) Find the horizontal and

vertical asymptotes of the graph of5x2 + 4

x2 − 1.

Solution: (The textbook suggests that this type of problem could use a calculator;but that is totally unnecessary.) Observe that numerator and denominator arepolynomials of the same degree. First observe that the denominator is the productof x − 1 and x + 1: it will be undefined at x = 1 and x = −1, and will approach±∞ as x → −1−, x → −1+, x → +1−, x → +1+, so each of the lines x = −1 andx = +1 is a vertical asymptote (in each case for more than one reason, because 2of these one-sided limits are infinite).

If we divide out from numerator and denominator the “leading” power of x, we


obtain

5x2 + 4

x2 − 1=

x2(

5 + 4x2

)

x2(

1 − 1x2

)

=5 + 4

x2

1 − 1x2

for x 6= ±1

→ 5 + 0

1 − 0= 5 as x → ∞ and as x → −∞ .

Thus y = 5 is the only horizontal asymptote.


D.9 Supplementary Notes for the Lecture of October 4th, 2006

Release Date: Wednesday, October 4th, 2006, subject to revision

I have posted the following message on WebCT:

Written Assignment W1 has been posted elsewhere on this site. Thesecond problem on this assignment involves the topic of MathematicalInduction, which we will not have time to discuss in the lectures ortutorials at this time. This topic will not be examination material.Mathematical Induction is discussed in the textbook on pages 81, 84.Students are expected to read the material up on their own, and thento work the assignment.

D.9.1 §2.6 Limits at Infinity; Horizontal Asymptotes (continued).

2.6 Exercises (continued)

[1, Exercise 2.6.49(a), p. 148] “Use the Squeeze Theorem to evaluate limx→∞

sin x

x.”

Solution: This limit should not be confused with the limits that we will be deter-mining in connection with [1, §3.4].

The sine function never exceeds 1 in absolute value; thus

−1 ≤ sin x ≤ 1 for all x .

We multiply these inequalities by1

xobtaining

−1

x≤ sin x

x≤ 1

xfor x > 0 . (116)

Similarly, for x < 0, we obtain

1

x≤ sin x

x≤ −1

xfor x < 0 , (117)

and the two inequalities can be combined into one:

− 1

|x| ≤sin x

x≤ 1

|x| for x 6= 0 . (118)

The extreme members of these inequalities approach 0 as x approaches ∞, trapping

the middle member; thus we may conclude that limx→∞

sin x

x= 0 . This is, as the

author suggests, another example of a function whose graph crosses a horizontalasymptote infinitely often.


[1, Exercise 2.6.44, p. 148] “Find a formula for a function (whose graph) has verticalasymptotes x = 1 and x = 3 and one horizontal asymptote, at y = 1.”

Solution: There are infinitely many solutions to problems of this type. We canconstruct a function by adding together functions having the properties indepen-

dently. One function whose graph has a vertical asymptote at x = 1 is1

x − 1;

similarly, one function whose graph has a vertical asymptote at x = 3 is1

x − 3.

Adding or subtracting these we obtain a function like

1

x − 1− 1

x − 3

whose graph has vertical asymptotes at precisely the designated places. The graphof the function we have found has a horizontal asymptote at y = 0. We can shiftthis asymptote to y = 1 by adding a constant:

1 +1

x − 1− 1

x − 3= 1 − 2

(x − 1)(x − 3)=

x2 − 4x + 1

(x − 1)(x − 3).

A “simpler” example would be

x2

(x − 1)(x − 3).

D.9.2 §2.7 Tangents, Velocities, and Other Rates of Change.

§2.7 continues the discussion begun in [1, §2.1], where the author proposed two applica-tions which would require the concept of limit.

Tangents For a given point P0 = (x0, y0) on the graph of the function f , a secantthrough P0 is a line passing through P0 and another point, P1 = (x1, y1), on the graph;since the points are on the graph, y0 = f(x0), y1 = f(x1). The slope of such a secant

isy1 − y0

x1 − x0=

f(x1) − f(x0)

x1 − x0. As we allow the point P1 to move along the curve towards

P0, the secant61 becomes a tangent62; its slope is

limx1→x0

f(x1) − f(x0)

x1 − x0.

61from the Latin, secare, to cut; do not confuse with the trigonometric function tangent whose namehas a similar etymology.

62from the Latin tangere, to touch; do not confuse with the trigonometric function tangent whosename has a similar etymology.


Note that, in the foregoing, x0 is the abscissa63 of the fixed point, and x1 is the abscissaof the variable point; we have avoided using the letter x here, as we may wish to writethe equation of a line, and wish to have the usual symbols free for that purpose. Whenwe reach [1, Chapter 3], we will be developing methods for routine computation oflimits of the preceding type; at present we will have to compute such limits “from firstprinciples”, without the use of the “Differentiation Rules”. Please remember that, evenif you know the “Rules”, you are expected to be able to carry out computations “fromfirst principles” when asked to do so. And, since the definition involves a limit, youcould even be expected to evaluate the limit according to the formal definition for that,although we don’t expect you to be able to work complicated limits that way.

Velocities Suppose a particle is moving along a straight line — which we convenientlycoordinatize as the x-axis — so that its position at time t is f(t). Between times t = t0and t = t1, the position of the particle has changed by a distance f(t1) − f(t0); thisis a signed distance. We often use the word displacement to describe the position of aparticle measured relative to a fixed point, which we often take to be the origin of thecoordinate system. The “average distance traversed per unit time between t = t0 and

t = t1” is defined to be the quotientf(t1) − f(t0)

t1 − t0, often called the average velocity .64

If, as we allow t1 to approach t0, we obtain a limit; we call that limit the velocity orinstantaneous velocity of the moving particle. We may denote the velocity at time t = t0by v(t), and define

v(t0) = limt1→t0

f(t1) − f(t0)

t1 − t0.

If we change our notation, and denote t1 by t0 + h, or by t0 + ∆t (so h or ∆t denotet1 − t0, the change in time t from t0 to t1), the equation becomes

v(t0) = limh→0

f(t1) − f(t0)

hor v(t0) = lim

∆t→0

f(t1) − f(t0)

∆t.

Other Rates of Change In the preceding discussion the symbol ∆t denotes a singlevariable with a two-letter name; it is not a product, and you must not pull the twocharacters ∆ and t apart. Historically, it was intended to denote the “increment” in t.

63x-coordinate64Later, in MATH 141, when we define the concept of average, we will be able to show that, in fact,

this is the average of the velocity function we are about to define. (The problem is that, at this stage,we haven’t defined what we mean by average of a function over an interval. We do know what we meanby the average of a finite list of numbers, and our eventual definition will generalize that definition.But being responsible mathematicians, we know that we can’t use a concept until it has been formallydefined. So, for the present, we can simply work with a concept having a 2-word name, average velocity.)


We also use the “operator” ∆ to describe the change in the function f , writing ∆y or∆f(t0) to denote f(t1)− f(t0), so that the instantaneous velocity may also be expressedas

v(t0) = lim∆t→0

∆f(t0)

∆t0= lim

∆t→0

∆y

∆t0.

Students may observe, quite correctly, that the meaning of the symbol ∆f is not alwayscompletely unambiguous, even when we write the variable, as ∆f(t0): how does oneknow what the second value of the argument65 is? The answer is that, in practice, thatinformation can be inferred from the context.

The author of the textbook again uses the word average in this section to describethe ratio of the increment of the function to the increment of the independent variable.We might call the first limit above the limit as h → 0 of the “average value” of f overthe interval [x, x + h] or the “average value” of f over the interval [x + h, x], dependingon whether h is taken to be positive or negative; but, as before, this is simply a 2-wordname for a concept, as we haven’t yet defined what we mean by an “average”.

2.7 Exercises

[1, Exercise 2.7.20, p. 156] (not discussed in the lecture) “The displacement (in me-ters) of a particle moving in a straight line is given by s = s(t) = t2 − 8t + 18,where t is measured in seconds.

(a) Find the average velocity over each time interval:

(i) [3, 4]

(ii) [3.5, 4]

(iii) [4, 5]

(iv) [4, 4.5]

(b) Find the instantaneous velocity when t = 4.

Solution:

(a) Each of the average velocities is defined to be the ratio of the displacementover the elapsed time.

s(3) = 32 − 8 · 3 + 18 = 3

s(3.5) = (3.5)2 − 8 · (3.5) + 18 = 2.25

s(4) = 42 − 8 · 4 + 18 = 2

s(4.5) = (4.5)2 − 8 · (4.5) + 18 = 2.25

s(5) = 52 − 8 · 5 + 18 = 3

65Argument is a synonym for variable; the argument of f(t) is t.


It follows that

s(4) − s(3) = −1

s(4) − s(3.5) = −0.25

s(5) − s(4) = 1

s(4.5) − s(4) = 0.25

and

s(4) − s(3)

4 − 3= −1

s(4) − s(3.5)

4 − 3.5= −1

2s(5) − s(4)

5 − 4= 1

s(4.5) − s(4)

4.5 − 4=

1

2

(b)

lim∆t→0

∆(s)

∆t= lim

∆t→0

s(t + ∆t) − s(t)

∆t

= lim∆t→0

((t + ∆t)2 − 8(t + ∆t) + 18) − (t2 − 8t + 18)

∆t

= lim∆t→0

2∆t · t + (∆t)2 − 8∆t

∆t= lim

∆t→0(2t + ∆t − 8)

= 2t − 8

for any value of t, in particular for t = 4, in which case the value of the limitis 0. Thus the particle is instantaneously at rest when t = 4.

D.9.3 §2.8 Derivatives.

The examples in the previous section, involving slopes of tangents and velocities ofmoving points, can be generalized in the following definition, applicable to any “wellbehaved” function of a real variable:

Definition D.6 The derivative of f at the point x = a, is defined to be any one of thefollowing limits, if they exist (as they are then all equal):

limh→0

f(a + h) − f(a)

h


lim∆x→0

f(a + ∆x) − f(a)

∆x

limb→a

f(b) − f(a)

b − a

A number of different notations are used for the value of the derivative at a; we may,from time to time, use any one of the following symbols, all of which have the same

meaning: f ′(a),df

dx(a),

df

dx

∣

∣

∣

∣

x=a

, Df(a), Df(x)|x=a. Other notations may be presented

later.

Intuitively a function f is differentiable at x = a when the graph of f does not have a“corner” at (x, y) = (a, f(a)).

Interpretation of the Derivative as the Slope of a Tangent

2.8 Exercises The intention is that all derivatives be computed “from first principles”,since we have not yet derived “Rules” for calculating derivatives of familiar functions.

[1, Exercises 2.8.8, p. 163] “If g(x) = 1−x3, find g′(0) and use it to find an equationof the tangent line to the curve y = 1 − x3 at the point (0, 1).”

Solution:

g′(0) = limx→0

g(x) − g(0)

x − 0

= limx→0

(1 − x3) − (1 − 03)

x

= limx→0

−x3

x= lim

x→0x2 = 0

so the tangent at (x, y) = (0, 1) has slope 0, i.e., is horizontal. Its equation isy − 1 = 0(x − 0), i.e., y = 1.

cf. [1, Exercises 2.8.17-18, p. 163] Find f ′(5) if f(x) =1

(√x + 4

)3 .

f ′(5) = limx→5

1

(√

x+4)3 − 1

(√

5+4)3

x − 5

= limx→5

33 −(√

x + 4)3

(x − 5) · 33 ·(√

x + 4)3


We can rationalize the numerator:

limx→5

33 −(√

x + 4)3

(x − 5) · 33 ·(√

x + 4)3 = lim

x→5

33 −(√

x + 4)3

(x − 5) · 33 ·(√

x + 4)3 · 33 +

(√x + 4

)3

33 +(√

x + 4)3

= limx→5

272 − (x + 4)3

(x − 5) · 33 ·(√

x + 4)3 ·(

33 +(√

x + 4)3)

= limx→5

1

27(

27 +(√

x + 4)3)

(√x + 4

)3

× limx→5

272 − (x + 4)3

x − 5

by the Product Law, in which the first limit is1

2 · 273. For the second limit we see

that the numerator is a polynomial which has value 0 when x = 5, which tells usthat x − 5 must divide the polynomial. By long division or otherwise we can seethat

272 − (x + 4)3 = −x3 − 12x2 − 48x + 665

= (x − 5)(

−x2 − 17x − 133)

.

Thus the second limit factor is

limx→5

272 − (x + 4)3

x − 5= lim

x→5

(

−x2 − 17x − 133)

= −25 − 85 − 133 = −243 .

We conclude that f ′(5) = − 1

162.

[1, Exercises 2.8.22, p. 163] “The limit limx→π

4

tanx − 1

x − π4

represents the derivative of

some function f at some number a. State such an f and a.”

Solution: (The solution is not unique: once we have one solution f , we can add aconstant to it and obtain another function, since the added constant will disappearwhen one value of the function is subtracted from another. We can also change thepoint at which the derivative is evaluated by an appropriate change of the formulafor the function.)

When we remember that tan π4

= 1, we see that the numerator is the the differencebetween values of the tangent function at a general point x and at the first pointπ4, precisely the same numbers whose difference is computed in the denominator.

Accordingly the given limit is the derivative of f(x) = tanx at x = π4. (We have

not yet determined the value of this derivative.)


[1, Exercises 2.8.35–36] are interesting, but it would be beyond the expectations ofthis course to expect you to be able to solve them without assistance. However, Imay discuss them in class, perhaps together with a third function:

f1(x) =

{

sin 1x

if x 6= 00 if x = 0

f2(x) =

{

x sin 1x

if x 6= 00 if x = 0

f3(x) =

{

x2 sin 1x

if x 6= 00 if x = 0

Here f1 is continuous everywhere except at x = 0, and has a derivative at everypoint; f2 is continuous everywhere, and has a derivative at every point exceptx = 0; and f3 is continuous everywhere and has a derivative everywhere.

Interpretation of the Derivative as a Rate of Change The derivative of f at apoint x in its domain was defined as any of the equal limits

limh→0

f(x + h) − f(x)

h

lim∆x→0

f(x + ∆x) − f(x)

∆x

limb→x

f(b) − f(x)

b − x

if they exist, and denoted by f ′(x).

D.9.4 §2.9 The Derivative as a Function

In the definitions given above, as we permit x to take values over the points in the domainof f — or rather, at those points where the limit exists — we define a function. Thisfunction is called the derivative of f , and denoted by f ′. If the limit in question is notdefined at a point a in the domain of f , then a is not in the domain of the derivative off .66 67

66When the limit in question is defined on only one side, we can speak of the left-hand derivative orright-hand derivative; these are discussed in [1, Exercise 2.9.46. p. 175].

67The domain of the derivative of f cannot include points that are not in the domain of f , since thedifference quotient for f ′(x) refers to f(x), which must exist before we can even discuss the existence ofa limit of the difference quotient.


Example D.17 The derivative of the function |x| is defined at all points in R except 0.Solution: The calculations were performed in the last lecture. At any point x > 0 thevalue of the derivative is f ′(x) = 1; at any point x < 0 the value of the derivative isf ′(x) = −1. But, at x = 0, the difference quotient has values +1 or −1 dependingon which side of 0 we take the second point of the secant. As the limits from the leftand right are not equal, the limit does not exist, and the function lacks a derivative atx = 0.68.

Other Notations We have already presented in Definition D.6, on page 2080 a numberof equivalent notations for the derivative evaluated at the value (point) a. We list

them again, and include a few others: f ′(a),df

dx(a),

df

dx

∣

∣

∣

∣

x=a

, Df(a), Df(x)|x=a, Dxf(a),[

df

dx

]

x=a

.

If a function is differentiable at every point in an open interval (a, b), i.e. at all x suchthat a < x < b, f is said to be differentiable on (a, b).

Differentiability implies Continuity. This result is proved in [1, Theorem 4, p. 171].You will not be responsible for proving the result, but a very short proof is possible. Firstobserve that the definition of f ′(a) involves f(a); that is, we can’t even write down thequotient whose limit we need unless we know that f ′(a) exists: this gives one of theconditions we need for continuity of f at x = a. Thus

limx→a

f(x) = limx→a

((f(x) − f(a)) + f(a))

= limx→a

((f(x) − f(a)) + limx→a

f(a) (by Sum Law)

= limx→a

(

f(x) − f(a)

x − a· (x − a)

)

+ limx→a

f(a)

(dividing and multiplying by x − a)

= limx→a

(

f(x) − f(a)

x − a

)

· limx→a

(x − a) + limx→a

f(a) (by Product Law)

= f ′(a) · limx→a

(x − a) + limx→a

f(a) (since f is differentiable at a)

= f ′(a) · 0 + limx→a

f(a) (as x → a, (x − a) → 0)

= 0 + limx→a

f(a)

= 0 + f(a) (limit of a constant, f(a))

= f(a)

68But the function has a left derivative and a right derivative at x = 0.


Note the author’s warning that the converse69 of the preceding theorem is not true! Youshould be able to supply examples of a function f and a point a in its domain such that fis continuous at a, but fails to be differentiable there. Graphs of the simplest examples aremade by piecing together at a point two line segments with different slopes; for example,the function |x| is of that type. You should be able to construct other examples, withadditional constraints on their construction.

How Can a Function Fail To Be Differentiable? The author describes severalsituations in which differentiability of a function f can fail at a point a.

• If limx→a−

f(x)−f(a)x−a

and limx→a+

f(x)−f(a)x−a

both exist, but do not have the same value, then,

by definition limx→a

f(x)−f(a)x−a

does not exist. The graph of the function will appear to

have a “corner” at x = a.

• If f is not continuous at x = a, then, a fortiori , by the result above, it cannot bedifferentiable.70

• If f is continuous, but has a vertical tangent line, we say that limx→a

f(x)−f(a)x−a

does

not exist, so f is not differentiable. An example of such a situation is at the pointx = 0 for the function

√

|x|. Note that the function is continuous at 0. We saythat its graph has a cusp at x = 0.

• Differentiability can fail for other reasons. For example, the function defined in [1,Problem 35, §2.8, p. 164] fails to have a limit because the slopes of the secantsthrough the origin oscillate between ±1 arbitrarily close to x = 0. In this courseyou are not expected to be able to construct “pathological” functions of this type.

2.9 Exercises Remember that, in these exercises, you are not permitted to use stan-dard formulæ for calculating derivatives, even though you may have learned them priorto taking this course. those formulæ will be developed in [1, Chapter 3].

[1, Exercise 26, p. 177] (not discussed in the lecturre) Find the derivative of the func-tion f(x) = x +

√x using the definition of derivative. State the domain of the

function and the domain of its derivative.

69The converse of a statement of the form “If A then B” is the statement “If B then A.” Can youthink of other implications which are true in one direction but not in the other?

70In the language of logic we are using the contrapositive of the result we proved: the contrapositiveis true if and only if the original statement is true. The contrapositive is obtained by reversing theimplication direction and also negating both statements. So the contrapositive of “A implies B” is “ifB is false, A is false”.


Solution: The square root is defined for all non-negative x; the polynomial x isdefined for all real numbers. The intersection of these sets is the domain of thefunction, [0,∞).

f ′(x) = limh→0

(x + h) +√

x + h − x −√x

x

= limh→0

(

1 +

√x + h −√

x

h

)

= 1 + limh→0

(√

x + h −√x

h·√

x + h +√

x√x + h +

√x

)

(rationalizing the numerator)

= 1 + limh→0

(

(√

x + h)2 − (√

x)2

h(√

x + h +√

x)

)

= 1 + limh→0

(

|x + h| − |x|h(√

x + h +√

x)

)

= 1 + limh→0

(

(x + h) − x

h(√

x + h +√

x)

)

since the domain of f is [0,∞)

= 1 + limh→0

(

h

h(√

x + h +√

x)

)

= 1 + limh→0

(

1√x + h +

√x

)

= 1 +limh→0

1

limh→0

√x + h + lim

h→0

√x

= 1 +1

√

limh→0

(x + h) +√

limh→0

x

by continuity of the square root function

= 1 +1√

x +√

x= 1 +

1

2√

x.

The domain of the derivative is (0,∞): the point 0, which is in the domain ofthe function, is not in the domain of the derivative. (As x → 0+ the slope of thetangent approaches = ∞.)

Suggested exercises:


1. Try to extend the method of rationalization shown above to find the derivativeof the derivative.

2. Replace√

x by√

|x|, so that the function is now defined over all of R, anddetermine the derivative.

D.9.5 2 Review

True-False Quiz (Good problems, but not discussed in the lecture) These are valuableexercises for determining whether you understand the material in the chapter. We can’tuse problems like this on assignments, since, as True/False problems, you can guess theanswer with an expectation of 50% correctness. But you should try them yourselves,and ask me or a TA if you are unsure of the correctness of your answer. Some of themcould be modified into good examination questions.

based on [1, True-False Question 6, p. 176] “Prove or disprove the following state-ment: ‘If lim

x→6f(x) · g(x) exists, then the limit must be f(6) · g(6).’”

Solution: The statement is false. A very simple example is

f(x) = g(x) =

{

1 if x 6= 6undefined if x = 6

,

where f and g are not even defined at x = 6, even though the limit of their productexists. In this case the discontinuity of the functions is removable, but we could

modify the example to make it unremovable, for example, by taking f(x) =1

x − 6and g(x) = x−6. Here g has limit 0, but f(6) is undefined and has an unremovablediscontinuity. Other variations can be constructed where f and/or g are more “illbehaved”, and one is the reciprocal of the other, so their product is “well behaved”.We can use the same device to construct functions whose sum is “well behaved”but where the individual functions are not, if we design the functions so that thesum cancels out the poor behavior of the individual summands.

based on [1, True-False Question 10, p. 176] “Prove or disprove the statement ‘Iff has domain [0,∞), and has no horizontal asymptote, then lim

x→∞f(x) = ∞ or

limx→∞

f(x) = −∞.’”

Solution: The statement is false. To disprove it I supply a counterexample. Tohave a horizontal asymptote we need the limit to exist (i.e., as a real number). Butthe non-existence of the limit does not imply that the limit is ±∞. For example,the function f(x) = [[x]]−x has no limit as x → ∞ or as x → −∞; another exampleis sin x.


based on [1, True-False Question 11, p. 176] “Prove or disprove the following state-ment. If you claim it is false, you are expected to supply a counterexample:

‘If the line x = 1 is a vertical asymptote of y = f(x), then f is notdefined at x = 1.’”

Solution: The statement is false. To be a vertical asymptote we need only thatone of lim

x→1−f(x), lim

x→1+f(x), is either −∞ or +∞; thus there are 4 possible reasons

why x = 1 might be a vertical asymptote, and we need at least one of them tohold; it could happen that 2 of them hold, but not more. These conditions areindependent of the question of whether f is defined at x = 1. For example, thefunction f defined by

f(x) =

1

x − 1if x 6= 1

5 if x = 1

is defined at x = 1 and has the line x = 1 as a vertical asymptote (for two differentreasons).

based on [1, True-False Question 17, p. 176] We could ask “Is the following state-ment true or false? If it is true simply state that. If it is false, you are expected toprovide an example — called a counterexample — to substantiate your claim:

‘If f is continuous at a, then f is differentiable at a.’”

Solution: The statement is false, as we have seen above. One counterexample isthe function f(x) = |x|, which is continuous at all points — you could be asked toprove that — but which fails to have a derivative at x = 0.

Exercises

[1, Exercise 22, p. 177] Find limx→∞

arctan (x3 − x).

Solution:

limx→∞

arctan(

x3 − x)

= arctan(

limx→∞

(

x3 − x)

)

by continuity of the arctangent function [1, Theorem 2.5.8, p. 129]. We cannotapply the Difference Law here, as it would lead to a difference of the type ∞−∞,whose behavior we cannot appraise. Instead we observe that x3 − x = x(x2 − 1).This expresses the argument of the arctangent as a product of two factors, both of


which become infinite as x → ∞. In this case the obvious extension of the ProductLaw does hold, and so lim

x→∞(x3 − x) = +∞. As u → ∞, arctan(u) → π

2.71

[1, Exercise 24, p. 177] Determine all horizontal and vertical asymptotes to the graphof f(x) =

√x2 + x + 1 −

√x2 − x.

Solution: By “completion of the square” we see that

x2 + x + 1 =

(

x +1

2

)2

+3

4> 0

so the domain of√

x2 + x + 1 is all of R. However,√

x2 − x =√

x(x − 1)

is defined only for x ≤ 0 and for x ≥ 1. The domain of the difference f(x) is theintersection of these two domains, i.e., again the union of the intervals (−∞, 0] and[1, +∞). But, at every point of this domain, the one-sided limits are finite; hencethere are no vertical asymptotes. Note that x = 0 is not a vertical asymptote, sincethe limit from the left is

√1 −

√0 = 1, not ±∞; the graph is, in fact, tangent to

the y-axis, and touches the axis, at the point (x, y) = (0, 1).

As x → ±∞ the two square roots both become infinite, so we cannot evaluate thelimit by using the Difference Law. We will use the usual device of rationalizing thesquare roots:

√x2 + x + 1 −

√x2 − x

=(√

x2 + x + 1 −√

x2 − x)

·√

x2 + x + 1 +√

x2 − x√x2 + x + 1 +

√x2 − x

=(x2 + x + 1) − (x2 − x)√

x2 + x + 1 +√

x2 − x

=2x + 1√

x2 + x + 1 −√

x2 − x,

provided x is not in the interval 0 < x < 1. Now what? We are planning to havex → ∞ and x → −∞. Both the numerator and the denominator approach ±∞,so we can’t use the Quotient Law.

2x + 1√x2 + x + 1 +

√x2 − x

=x(

2 + 1x

)

√

x2(

1 + 1x

+ 1x2

)

+√

x2(

1 − 1x

)

71Finally, we might ask precisely where the continuity of the arctangent function was required. Wecan see now it is the continuity of that function from the left at π

2 . This is a delicate observation, andI would normally not expect students in this course to be able to answer this question.


=x(

2 + 1x

)

√x2

√

1 + 1x

+ 1x2 +

√x2

√

1 − 1x

since√

ab =√

a ·√

b where everything is defined

=x(

2 + 1x

)

|x|√

1 + 1x

+ 1x2 + |x|

√

1 − 1x

by definition of |x|

=x

|x| ·2 + 1

x√

1 + 1x

+ 1x2 +

√

1 − 1x

As x → +∞ this product approaches 1 · 21+1

= 1, and as x → −∞ the product

approaches (−1) · 21+1

= −1. There are thus two horizontal asymptotes: y = ±1,(see Figure 13).

x

1050

y

-5

1

0-10

-1

y=f(x)

upper bound, y=f(1)

horizontal asymptote y=-1

horizontal asymptote y=1

Figure 13: Graph of the Function f(x) =√

x2 + x + 1 −√

x2 − x

Textbook Appendices.

Since the syllabus includes some topics in the appendices to the textbook, either directlyor as prerequisites, a few comments are in order; these comments have not been discussedin the lectures.

D.9.6 Appendix A. Numbers, Inequalities, and Absolute Values

When a decimal expansion of a real number is truncated — by ignoring all but a finitenumber of the digits, then, if any of the ignored digits is different from 0, an error isintroduced. This can happen in rational numbers, e.g. 1

3= 0.33333 . . . is not the same


as 0.3333333333333; and in irrational numbers, e.g. 3.14159265358979323846 is not thesame as π = 3.14159265358979323846 . . .72.

D.9.7 Appendix B. Coordinate Geometry and Lines

Familiarity with all of the material in this appendix is assumed.

D.9.8 Appendix C. Graphs of Second-Degree Equations

Not all students will be familiar with all the material in this appendix.

Circles You should be comfortable with all of the material in this subsection.

Parabolas We will be working with functions whose graphs are parabolæ, ellipses,hyperbolæ, but we do not expect you to know all of their properties; however, studentswho have not studied these curves before are urged to read the rest of the appendix.

Ellipses

Hyperbolas

Shifted Conics This is a useful topic, but we don’t expect you to know about it; somestudents will see the ideas in MATH 133.

D.9.9 Appendix D. Trigonometry

I have reviewed some portions of this appendix in the lectures. Students are assumedto be familiar with this material from their precalculus course. Any discussion in thelectures will be confined to isolated topics. However, problems on this material couldalready have appeared in WeBWorK Assignments or in quizzes. Students who findthat they are weak in this material could also consult [32].

72You may see a way of determining the exact value of π at the very end of MATH 141.


D.10 Supplementary Notes for the Lecture of October 10th,2006

Release Date: Tuesday, October 10th, 2006subject to correction

I began the lecture by explaining briefly the meaning of the concepts of MathematicalInduction, and the iterated derivative.

2.9 Exercises (conclusion)

[1, Exercise 36, p. 178] “Use the Intermediate Value Theorem to show that there is

a solution of the equation e−x2

= x in the interval (0, 1).”

Solution: To apply the given theorem we need to designate a continuous function.But the equation relates two functions! There are many ways to combine them,but the easiest is something like this. Define f(x) = e−x2 − x. We know thatthe polynomials −x2 and x are continuous everywhere; and we know that theexponential function ex is continuous everywhere. Hence the composition of thepolynomial function −x2 followed by the exponential function, i.e., e−x2

, is alsocontinuous everywhere. We also know that sums and differences of continuousfunctions are continuous where the summands and “differands” are continuous.That tells us that the function we have called f is continuous everywhere. Thetheorems we have applied are [1, Theorems 2.5.4, 2.5.5, 2.5.7, 2.5.8 in §2.5].

Now we have to show that the value of interest lies between the values of thefunction at the end points of the given interval. f(0) = e−02−0 = e0−0 = 1−0 = 1,and

f(1) = e−12 − 1 =1

e− 1 < 1 − 1 = 0 .

Thus f(0) > 0 > f(1). By the Intermediate Value Theorem [1, Theorem 2.5.10,p. 131] there exists a point c such that 0 < c < 1 and f(c) = 0, i.e., such that

e−c2 = c.

(One application of the theorem simply proves the existence of the point. But,if we were to bisect the interval (0, 1), and evaluate the function at the midpoint

x =1

2, we could trap a root in an interval of half the length; every 10 repetitions

of this bisecting reduces the size of the interval by a factor of 210 > 1000 = 103, sowe could determine a solution to any desired number of decimal places, providedwe had a way of evaluating the exponential function whenever we needed it.)


[1, Exercise 38, p. 178] “Find equations of the tangent lines to the curve y =2

1 − 3xat the points with x-coordinates 0 and −1.”

Solution: We will first determine the derivative in general from first principles,then take the specific values for the given points.

y′(x) = limh→0

y(x + h) − y(x)

h

= limh→0

21−3(x+h)

− 21−3x

h

= limh→0

2 · (1 − 3x) − (1 − 3x − 3h)

h(1 − 3x)(1 − 3(x + h))

= 2 limh→0

3h

h(1 − 3x)(1 − 3(x + h))

= 2 limh→0

3

(1 − 3x)(1 − 3(x + h))

= 2 limh→0

3

(1 − 3x)2= 6(1 − 3x)−2

Then y′(0) = 6, and y′(−1) = 38. The tangents at the points (0, 2) and

(

−1, 12

)

respectively have equations

y − 2 = 6(x − 0) or 6x − y = −2 and

y − 1

2=

3

8(x + 1) or 3x − 8y = −7.

[1, Exercise 14, p. 181] “Suppose that f is a function with the property that

|f(x)| ≤ x2 (119)

for all x. Show that f(0) = 0. Then show that f ′(0) = 0.

Solution: One reason I am discussing this problem is to remind you of how tointerpret an inequality where the left member is an absolute value.

First observe that when we set x = 0 in (119) we obtain |f(0)| ≤ 0; but, as weknow that |f(0)| ≥ 0 (being an absolute value), we conclude that

f(0) = 0 . (120)

The derivative of f at x = 0 is the limit of the quotientf(x) − f(0)

x − 0as x → 0.

From (119) we may conclude that

−x2 ≤ f(x) ≤ x2 (121)


for all x; hence, for x 6= 0, we may divide all members by x to obtain

−x ≤ f(x) − f(0)

x≤ x . (122)

Sincelimx→0

(−x) = 0 = limx→0

x ,

we conclude from the Squeeze Principle that limx→0

f(x) − f(0)

x= 0, i.e., that f ′(0) =

0.

Textbook Chapter 3. DIFFERENTIATION RULES.

D.10.1 §3.1 Derivatives of Polynomials and Exponential Functions

Beginning in this section we develop properties of the differentiation operation that willpermit the systematic computation of derivatives without the need for computations“from first principles”. While students are expected to master these “Rules”, you arealso expected to be able to carry out computations “from first principles”. Except forproblem functions that don’t fit into the categories to which the “Rules” apply, you willnormally not use “first principles” unless asked to do so.

Power Rules We can prove from first principles that

d

dxx = 1

This can be generalized tod

dxxn = nxn−1

in various ways. The textbook is careful to assert, at first that this generalized “PowerRule” is true for positive integers. But, in the course of several steps, it will eventually be“proved” for all real number n. You are not expected to be able to reproduce the proofs.Both of the proofs that the textbook gives of the Power Rule for integer exponents relyon this Principle.

Yet another proof could be based on applying to the resultd

dx(x) = 1 the Product

Rule (proved below in the discussion of [1, §3.2]). A proof would require using Mathe-matical Induction or an equivalent tool, [1, Exercise 3.2.43, p. 198].


New Derivatives from Old The author proves “rules” he calls the Constant MultipleRule, the Sum Rule, the Difference Rule (a consequence of the preceding two rules). Youare not expected to memorize these proofs. (But the level of difficulty of the proofs isno greater than could arise in a reasonable problem that could be set on the material ofthis section and some of the preceding sections.)

Exponential Functions The author considers, for any positive real number a, thevalue of the derivative of the function ax at any point x. He shows that

d

dxax = ax · lim

h→0

ah − 1

h.

Thus, if the limit

limh→0

ah − 1

h

exists — which he assumes without proof73 — the derivative is a constant multiple ofthe function; this can be shown to be a property that actually characterizes exponentialfunctions — they are the only functions that have this property. The textbook continuesto report without proof that the value of limit (D.10.1) ranges continuously over thepositive real numbers. Then Napier’s constant , denoted by e, is “defined” to be the

unique constant, lying between 2 and 3, with the property that limh→0

eh − 1

h= 1, and

consequentlyd

dxex = ex.

The differentiation rules developed in this section are summarized in Table D.10.1

3.1 Exercises Remember that the author’s intention is that you use only tools thatare mentioned, or, in any case, do not use tools that will be developed in subsequentsections. So, for example, [1, Exercise 3.1.22, p. 191] should not be solved using theproduct rule.

[1, Exercise 3.1.46, p. 191] (not solved in the lecture) “For what values of x does thegraph of f(x) = x3 + 3x2 + x + 3 have a horizontal tangent?”

Solution: To determine the points on a curve y = f(x) having a horizontal tangentone solves the equation f ′(x) = 0. In this problem that equation is quadratic, andleads to two points (with unpleasant coordinates). By the Rules developed thusfar

d

dx(x3 + 3x2 + x + 3) =

d

dx(x3) +

d

dx(3x2) +

d

dx(x) +

d

dx(3)

73Remember, this is the “Early Transcendentals” version of the textbook: the study of exponentialfunctions has been advanced to a point where some steps cannot be proved.


Derivative of a Constant:d

dx(c) = 0

Power Rule:d

dx(xn) = nxn−1

for any real number n

Constant Multiple Rule:d

dx(cf(x)) = c

d

dxf(x)

for any constant cand any differentiable function f

Sum and Difference Rules:d

dx(f(x) ± g(x)) =

d

dxf(x) ± d

dxg(x)

for any differentiable functions f , g

Derivative of Natural Exponential Function:d

dx(ex) = ex

Table 10: Differentiation Rules from [1, §3.1]

by the Sum Rule

=d

dx(x3) + 3

d

dx(x2) +

d

dx(x) +

d

dx(3)

by the Constant Multiple Rule

=d

dx(x3) + 3

d

dx(x2) +

d

dx(x) + 0

by the Derivative of a Constant Function Rule

= 3x2 + 3(

2x1)

+ 1(

x0)

+ 0

by the Power Rule

= 3x2 + 3 (2x) + 1 (1) + 0 = 3x2 + 6x + 1

Tangents are horizontal at points whose abscissa x is a root of the polynomial3x2 + 6x + 1. Students should be able to find the roots by using the “QuadraticFormula”. I prefer to use “Completion of the Square”, because it is a techniquethat we will find useful in the next course: 3x2 + 6x + 1 = 3(x2 + 2x) + 1 =


3((x + 1)2 − 1) + 1 = 3(

(x + 1)2 − 23

)

. The expression in parentheses may beinterpreted as a difference of squares, and we know how to factor such differences:

(x + 1)2 − 2

3= (x + 1)2 −

(

√

2

3

)2

=

(

x + 1 +

√

2

3

)(

x + 1 −√

2

3

)

so the derivative is 0 when either of these linear factors is 0, i.e., when

x = −1 ±√

2

3.

Some students may have been “taught” to avoid placing a surd in the denominator;if you prefer, we can write the solution as

x = −1 ± 1

3

√6 .

(Placing a surd in the denominator is not incorrect; in the days before calculatorswere commonly available, computation could be greatly simplified when denomi-nators were integers. Avoiding surds in the denominator may still be advisable,but it is not mandatory.)

[1, Exercise 3.1.50, p. 191] “Find equations of both lines through the point (2,−3)that are tangent to the parabola y = x2 + x.”

Solution: The point (x, x2 + x) on the curve has a tangent with slope

d

dx(x2 + x) = 2x + 1 .

As we wish to find an equation, and require the symbol x to denote the abscissaof the general point on that line, we will rephrase the preceding result: the point(a, a2 + a) on the curve has a tangent with slope 2a + 1. The tangent at the point(a, a2 + a) has equation

y − (a2 + a) = (2a + 1)(x − a) or y = (2a + 1)x − a2 .

We impose the condition that these tangent lines pass through the point withcoordinates (2,−3), i.e., that (x, y) = (2,−3) is a solution of the equation:

−3 = (2a + 1)2 − a2 or a2 − 4a − 5 = 0 .

Solving this quadratic equation yields a = −1, 5, so the two tangents through(2,−3) are

y = −x − 1 and y = 11x − 25 .

(There is a “classical” way of solving this problem without apparent use of thecalculus, but it will not be studied in this course.)


[1, Exercise 3.1.52, p. 192] “Where does the normal line to the parabola y = x − x2

at the point (1, 0) intersect the parabola a second time?”

Solution: (In this context the word normal means perpendicular : thus its slopewill be not the derivative but the negative reciprocal of the derivative:

− 1

dy

dx

.

This is by virtue of the property that

tan(

θ +π

2

)

=1

− tan θ, (123)

which is a consequence of the property that

tan(a + b) =tan a + tan b

1 − tan a · tan b. (124)

Of course, one can’t obtain (124) from (123) by substituting b = π2, since the tan-

gent is not defined at π2. But we can argue that the tangent function is continuous

from the left or right, and take either one-sided limit as b → π2:

limb→π

2−

tan(a + b) = limb→π

2−

tan a + tan b

1 − tan a · tan b

= limb→π

2−

tan a

tan b+ 1

1

tan b− tan a

=

limb→π

2−

(

tan a

tan b+ 1

)

limb→π

2−

(

1

tan b− tan a

)

=0 + 1

0 − tan a= − cot a .

An easier proof would be to show, again from the sum formulæ (or otherwise), that

sin(

a +π

2

)

= cos a

cos(

a +π

2

)

= − sin a.)


d

dx(x − x2) = 1 − 2x, so the tangent at the point (1, 0) has slope 1 − 2 = −1, and

the normal (line) has slope − 1

−1= 1; thus the equation of that normal is

y − 0 = 1(x − 1) .

If we solve this equation with the equation of the curve, we obtain two values for thex-coordinate of a point of intersection, x = −1 and x = 1, so there are two pointswhere the line and curve intersect: (1, 1 − 12) = (1, 0) and (−1,−1 − (−1)2) =(−1,−2). The first is the point of tangency; the second is where the tangent meetsthe curve a second time.

D.10.2 §3.2 The Product and Quotient Rules

The textbook continues the derivation of “Rules” which will permit the mechanicalcomputation of derivatives for common functions without the need to evaluate limits. Inpractice you will usually use these rules, but need to be able to find limits “from firstprinciples” when requested.

The Product Rule The proof of the Product Rule requires a familiar “trick”, similarto one we saw in the proof that differentiability implies continuity: we add a term, totransform a function to a more convenient form, then subtract the offending term sothat the value has not been changed. One point in the proof is delicate — we use thecontinuity of one of the functions at one point: we know this to be true since we areassuming that both functions are differentiable.

While we will not ask you to memorize this proof — or any proofs — we expectyou to understand the use of continuity, and could devise a way of calling on you todemonstrate that in an examination — without having to reproduce a memorized proof.

The Quotient Rule This “rule” could also be proved by first proving a ReciprocalRule [1, Exercise 3.2.44, p. 198]:

d

dx

(

1

g(x)

)

= −d

dxg(x)

(g(x))2,

and then applying the product rule to the product f(x) · 1

g(x). The differentiation rules

developed in this section are summarized in Table D.10.2.


Product Rule:d

dx(f(x) · g(x) =

d

dxf(x) · g(x) + f(x) · d

dxg(x)

for any differentiable functions f , g

Quotient Rule:

d

dx

(

f(x)

g(x)

)

=

d

dxf(x) · g(x) − f(x) · d

dxg(x)

(g(x))2

for any differentiable functions f , g,(where g(x) 6= 0)

Reciprocal Rule: [1, Exercise 44, p. 198]

d

dx

(

1

f(x)

)

= −d

dxf(x)

(f(x))2

for any differentiable function f , where f(x) 6= 0

Table 11: Differentiation Rules from [1, §3.2]

3.2 Exercises

[1, Exercise 3.2.43, p. 198] This problem is a special case of a generalization of theProduct Rule: If f(x) is the product of n differentiable functions,

f(x) = f1(x) · f2(x) · . . . · fn(x) ,

the derivative of f can be obtained by summing all products of the functions inwhich exactly one of the factors is differentiated, and the other n − 1 factors areunchanged. It is the general rule that I discuss here, as an opportunity to introducethe concept of Mathematical Induction.

We need to prove that

f ′(x) = f ′1(x) · f2(x) · . . . · fn(x)

+f1(x) · f ′2(x) · . . . · fn(x)

+ . . . + f1(x) · f2(x) · . . . · f ′n(x) (125)

This sequence of theorems can be proved by “Mathematical Induction” as follows:

I. First prove the case n = 1. We observe that, when n = 1, f = f1, and theclaimed result is just f ′ = f ′

1.


II. Assume that the theorem has been proved for n = N . (This is called the In-duction Hypothesis.)

III. Prove the case n = N + 1. If we know that

(f1(x) · f2(x) · . . . · fn(x))′

= f ′1(x) · f2(x) · . . . · fN(x)

+f1(x) · f ′2(x) · . . . · fN(x)

+ . . . + f1(x) · f2(x) · . . . · f ′N(x) (126)

we can differentiate the product

(f1(x) · f2(x) · . . . · fN(x)) fN+1(x)

by the Product Rule:

((f1(x) · f2(x) · . . . · fN(x)) fN+1(x))′

= (f1(x) · f2(x) · . . . · fN(x))′ fN+1(x)

+ (f1(x) · f2(x) · . . . · fN(x)) f ′N+1(x)

= (f ′1(x) · f2(x) · . . . · fN(x)) fN+1(x)

+ (f1(x) · f ′2(x) · . . . · fN(x)) fN+1(x)

+ . . . + (f1(x) · f2(x) · . . . · f ′N (x)) fN+1(x)

+ (f1(x) · f2(x) · . . . · fN(x)) f ′N+1(x)

which is just (125) when n = N + 1.

IV. Conclude that (125) is true for every natural number n.

The textbook problem is case n = 3; to prove it one does not need to use Mathe-matical Induction: simply apply the Product Rule twice.

[1, Exercise 3.2.42, p. 198] (not solved in the lecture) “Find equations of the tangent

lines to the curve y =x − 1

x + 1that are parallel to the line x − 2y = 2.”

Solution: One way of solving this problem would be similar to that given earlier for[1, Exercise 3.1.50, p. 191] above: one finds the slope of the tangent at any point onthe curve y = f(x), say with coordinates (a, f(a)), and imposes the condition thatthese tangents have the slope of a given line, thereby locating the points at whichthe tangents are parallel to the given line. There are two solutions. Problems ofthese types, involved specific techniques of using the calculus to solve geometricproblems, specifically to find tangents to a curve where the given information does


not include the point(s) of contact of the tangents, are well within the range ofproblems that students in Math 140 are expected to be able to solve.

Here is a slightly different approach. The derivative ofx − 1

x + 1can be shown to be

2(x + 1)−2. We wish to find the tangent lines that are parallel to x − 2y = 2, i.e.,those whose slope is 1

2. Imposing that condition on the derivative yields

2

(x + 1)2=

1

2

which reduces to x2 + 2x − 3 = 0 or (x + 3)(x − 1) = 0, satisfied by x = −3 andx = 1. We have thus shown that the tangents we seek are those which touch thecurve at the points

(

−3, −3−1−3+1

)

and(

1, 1−11+1

)

, i.e., at (−3, 2) and (1, 0). The lines

through these points with slope 12

are x − 2y = −7 and x − 2y = 1.

[1, Exercise 3.2.32, p. 197] (not solved in the lecture) “If f(3) = 4, g(3) = 2, f ′(3) =−6, and g′(3) = 5, find the following numbers:

(a) (f + g)′(3)

(b) (fg)′(3)

(c)

(

f

g

)′(3)

(d)

(

f

f − g

)′(3)”

These problems refer to the conventions that the textbook defined in [1, “Algebraof Functions”, p. 42].

Solution:

(a) By the Sum Rule, (f + g)′(x) = f ′(x) + g′(x) for all x. Here (f + g)′(3) =f ′(3) + g′(3) = (−6) + 5 = −1.

(b) By the Product Rule (fg)′(3) = f(3) · g′(3)+ f ′(3) · g(3) = 4 · 5+ (−6) · 2 = 8.

(c) By the Quotient Rule

(

f

g

)′(3) =

g(3) · f ′(3) − f(3) · g′(3)

(g(3))2=

2 · (−6) − 4 · 522

=

−8.

(d)

(

f

f − g

)′(3) =

(f − g)′(3) · f(3) − f(3) · (f − g)′(3)

((f − g)(3))2


=(f(3) − g(3)) · f ′(3) − f(3) · (f ′(3) − g′(3))

((f − g)(3))2

=(4 − 2) · (−6) − 4 · (−6 − 5)

(4 − 2)2= 8




D.11.1 §3.3 Rates of Change in the Natural and Social Sciences.

The purpose of this section in this course is to provide examples of the use of derivativesin problems stated verbally. Since some of the problems are stated in very technicallanguage, you should try to read those problems that are in an area that is close to yourinterest. The technical language is usually superficial, so that once a problem is statedin mathematical language, you should all be able to solve it, although you may not findthe language interesting. The differences in your backgrounds make it difficult to findproblems that will interest all of you. For that reason most of the material in this sectionis considered outside of the syllabus.

Physics Since one of the original motivations in the development of the calculus was tosolve problems of motion, and since some problems of that type are found in WeBWorK,I shall consider one such problem.

3.3 Exercises

[1, Exercise 3.3.10, p. 208] If a ball is thrown vertically upward with a velocity of 80feet per second, then its height after t seconds is

s(t) = 80t − 16t2 . (127)

1. What is the maximum height reached by the ball?

2. What is the velocity of the ball when it is 96 feet above the ground, on itsway up? ...on its way down?

Solution: 74

1. To determine the maximum height. One method would be to examine thefunction s(t). If we express the given formula as s(t) = −16(−5t + t2), we

can complete the square: s(t) = −16(

t2 − 5t +(

52

)2)

+ 16 · 254

= 100 −

74This problem could have been made more difficult by requiring that you determine the “equationof motion” from the initial velocity and the initial position. But the equation of motion is given toyou in (127). The determination of this equation would be premature, as it would depend on a resultin [1, §4.2] not yet proved; but the textbook occasionally does ask you to proved results of this typeprematurely.


(

4(

t − 52

))2= 100− (4t− 10)2. The subtracted term is a square, so it cannot

be negative; s(t) will be greatest when the subtracted term is as small aspossible, i.e. 0, i.e. where t = 5

2. When t = 5

2, s(t) = 100: this is the

maximum height. No calculus is needed!

2. The velocity is ddt

s(t) = 80 − 32t; but this is given in terms of t, not in termsof position s. When s = 96, t satisfies the equation 96 = 80t − 16t2, so t = 2or t = 3.

s′(2) = 80 − 64 = 16

s′(3) = 80 − 96 = −16

As the positive direction has been taken to be upward, height s = 96 isattained when t = 2, when the velocity is 16 ft/sec upward; and when t = 3,on the downward trip, when the velocity is again 16, but this time directedin the negative, i.e. downward, direction.

D.11.2 §3.4 Derivatives of Trigonometric Functions.

All the result in this section depend on the value of limx→0

sin x

x, through the application

of simple trigonometric identities.

Trigonometric functions defined (with isolated omitted points) over all of R.

(This is precalculus material, included in these notes only as a reminder tostudents who have to fill a gap in their preparation.)

Consider a point (x, y) moving on the unit circle centred at the origin, from a positionat rest at (1, 0). Suppose that the point has rotated through an angle of θ. Then wemay define

sin θ = y

cos θ = x

and then extend these definitions to the other 4 functions, defined as follows:

tan θ =sin θ

cos θ=

y

x

cot θ =1

tan θ=

cos θ

sin θ=

x

y

sec θ =1

cos θ=

1

x

csc θ =1

sin θ=

1

y


These definitions agree with definitions that should be familiar to all students for trigono-metric functions of acute positive angles. One can think of a triangle with one vertexat the origin, one at the point (x, y), and the right angle at the point (x, 0), which isthe foot of the perpendicular drawn from (x, y) to the x-axis. The definition has beensimplified by taking the hypotenuse to be of length 1. Students should be able to de-termine the sign of each of these functions from the position of the moving point in thevarious quadrants. All but the sine and cosine have points where they are not defined,and all of these are non-removable discontinuities of the functions, where the limit onone side of the discontinuity is infinite of the opposite sign from the limit on the otherside. Students should also be familiar with the approximate shapes of the graphs of thesefunctions [1, pp. A30–A31].

limθ→0

sin θ

θThe derivation we will discuss in the lectures of the value of the limit in

question will be somewhat different from that in the textbook, as we will consider areasrather than the lengths of arcs in its derivation.

Warning! The results of this section apply only if the variable of the trigonometricfunctions is considered to be in radians; if we wish to consider, for example, a functionf(θ) = the sine of an angle of θ degrees, the derivative will not be the cosine function ofthe angle measured in degrees. We can reconsider this question after we have completedthe next section of the textbook.

The determination of limθ→0

sin θ

θshown in the lectures differs from that in the textbook:

we base our inequalities on comparisons of areas, rather than of lengths. While studentsare not expected to memorize this proof, it is a valuable example of uses of the “Squeeze”Theorem, which student often find difficult. We first prove that

tan θ

2≥ θ

2≥ sin θ

2

for θ > 0; then we divide through bysin θ

2, to obtain

sec θ ≥ θ

sin θ≥ 1 ,

and then take reciprocals of the three members, which reverses both inequalities, yielding

cos θ ≤ sin θ

θ≤ 1 , (θ > 0) .

Our objective is to apply the Squeeze Theorem, taking the limit as θ → 0. For thatpurpose we need to have the pair of inequalities holding in a (possibly “punctured”)


interval surrounding 0, not only on one side. We obtain a corresponding result for θ < 0simply by renaming θ as −λ, and now requiring λ < 0; the inequalities become

cos(−λ) ≤ sin(−λ)

−λ≤ 1 (λ < 0) .

We observe that, since the sine function is odd , sin(−λ) = − sin λ, sosin(−λ)

−λ=

sin λ

λ

even when λ < 0; and cos(−λ) = cos λ: indeed, both cos λ andsin λ

λare even functions.

Thus we have shown that the same pair of inequalities hold on both sides of 0; of course,they fail to hold at 0 because the middle member is not defined there. There is no longerany reason to use two different symbols for the variable on the two sides of 0, so wecombine them into

cos θ ≤ sin θ

θ≤ 1 , (θ 6= 0) .

We may now apply the Squeeze Theorem, which does not require that the inequalitieshold at θ = 0, to obtain

limθ→0

cos θ ≤ limθ→0

sin θ

θ≤ lim

θ→01 ,

i.e., 1 ≤ limθ→0

sin θ

θ≤ 1

from which it follows that

limθ→0

sin θ

θ= 1 . (128)

Note that what we have found is the derivative of sin θ evaluated when θ = 0. We will

apply this result to find the value ofd

dθsin θ at any θ.

limθ→0

cos θ−1

θThere are a number of easy ways to extract this limit from (128). The

proof in the textbook proceeds as follows:

limθ→0

cos θ − 1

θ= lim

θ→0

(

cos θ − 1

θ· cos θ + 1

cos θ + 1

)

= limθ→0

cos2 θ − 1

θ(cos θ + 1)

= limθ→0

− sin2 θ

θ(cos θ + 1)

= limθ→0

sin θ

θ· lim

θ→0

− sin θ

cos θ + 1


= limθ→0

sin θ

θ·

− limθ→0

sin θ

limθ→0

cos θ + limθ→0

1

= 1 · 0

2= 0 (129)

This shows that the value of the derivative of cos θ at θ = 0 is 0.

Derivatives of trigonometric functions Thus far we have been able to determinethe values at θ = 0 of the derivatives of the sine and cosine function. As the other 4trigonometric functions can be expressed in terms of sines and cosines, it should not besurprising that we can now evaluate all of their derivatives at θ = 0. But, in fact, wenow have enough information to evaluate the derivatives of the these 6 functions at allpoints in their domains. (The functions are differentiable wherever they are defined, butall of them except the sine and cosine have points on R where they are not defined, anda fortiori there are no derivatives at those points.) Here are the derivatives:

d

dθsin θ = cos θ (130)

d

dθcos θ = − sin θ (131)

d

dθtan θ = sec2 θ (132)

d

dθcot θ = − csc2 θ (133)

d

dθsec θ = sec θ · tan θ (134)

d

dθcsc θ = − csc θ · cot θ (135)

I will prove at least one of these results in the lecture, but you should be able to deriveany of the others from one of them by using standard identities [1, pp. A28-A30] involvingthe trigonometric functions. Don’t memorize proofs! While you should remember howwe proved these formulæ, the first priority is that you have the formulæ committed to


memory, as several of them are in very frequent use. (This is one of the few times in thiscourse when I suggest memorization.)

Other limit problems involving trigonometric functions We can also apply the

limit results above to other limit problems. For example, to evaluate limx→0

sin 4x

tan 7x, we

would rewrite the given fraction as a product of two fractions of the form sin yy

wherey → 0:

limx→0

sin 4x

tan 7x= lim

x→0

(

sin 4x

4x· 7x

tan 7x· 4

7

)

= limx→0

sin 4x

4x· lim

x→0

7x

tan 7x· lim

x→0

4

7by the Product Law

= limx→0

sin 4x

4x· lim

x→0

cos 7xsin 7x

7x

· limx→0

4

7

= limx→0

sin 4x

4x·

limx→0

cos 7x

limx→0

sin 7x7x

· limx→0

4

7

by the Quotient Law

= 1 · 1

1· 4

7=

4

7

In the last line I am, on two occasions, using the result that limx→a

f(g(x)) = f(limx→a

g(x))

under assumptions of continuity, [1, p. 129]. I usually don’t expect students in Math 140to give this explanation, but it is the justification that is missing above.

3.4 Exercises

[1, Exercise 3.4.12, p. 216] “Differentiate y =tan x − 1

sec x.”

Solution: While it is possible to solve this problem correctly by applying the Quo-tient Rule naively, it is easier if the numerator and denominator are first bothmultiplied by cos x.

d

dx

(

tanx − 1

sec x

)

=d

dx(sin x − cos x)

= cos x + sin x

However, we must not forget that the original the is not defined at any odd multipleof π

2, where it has removable discontinuities.


[1, Exercise 3.4.42, p. 217] (not worked in the lecture) “Find limx→π

4

sin x − cos x

cos 2x”.

Solution:

limx→π

4

sin x − cos x

cos 2x= lim

x→π4

sin x − cos x

cos2 x − sin2 x

by a “double angle” formula

= limx→π

4

sin x − cos x

(cos x − sin x)(cos x + sin x)

= limx→π

4

1

(−1)(cos x + sin x)

since sin x − cos x 6= 0 near (but not at) x = π4

=1

−(

1√2

+ 1√2

) = −2−12 .

[1, Exercise 3.4.38, p. 217] (not worked in the lecture) “Find limcos θ − 1

sin θ.”

Solution: When you remember the simple trigonometric identities, there are oftenmany methods of solving problems of this type that will suggest themselves. Onesimple method that does not require such identities would involve first dividingnumerator and denominator by θ, in order to transform both into functions thathave been studied in this section:

limθ→0

cos θ − 1

sin θ= lim

θ→0

cos θ − 1

θsin θ

θ

=

limθ→0

(

cos θ − 1

θ

)

limθ→0

(

sin θ

θ

)

=0

1= 0.

Another attack could be as follows:

limθ→0

cos θ − 1

sin θ= lim

θ→0

(

1 − 2 sin2 θ2

)

− 1

2 sin θ2· cos θ

2

= limθ→0

tanθ

2= tan

0

2= tan 0 = 0



Release Date: Monday, October 16th, 2006, subject to revision

Today’s and next day’s lecture are being devoted to [1, §§3.5, 3.6].

D.12.1 §3.5 The Chain Rule.

The “Chain Rule” enables us to determine the derivative of the composition of severaldifferentiable functions. It can be described briefly using the “Leibniz notation”: Ify = f(u) and u = g(x), where f and g are both differentiable, then the derivative of thecomposite function f ◦ g is given by

dy

dx=

dy

du· du

dx,

which is easy to remember because it appears to be a statement about the product oftwo fractions. More precisely, we have

d

dxf(g(x)) = f ′(g(x)) · g′(x) .

The chain rule may need to be applied more than once when we differentiate a functionthat has been constructed by composing many functions. In general, it involves “peelingoff one layer after another” of the composed functions, at every stage multiplying bya factor which is a derivative evaluated at the point where the original function wasevaluated. When you are first learning the Chain Rule, you may find it convenientto name the various functions that compose. Eventually this intermediate step shouldbecome unnecessary.

The “Power Rule” combined with the Chain Rule Let u = g(x) be a differen-tiable function, and a be any non-zero real number. By applying the Chain Rule, wecan show that

d

dx(g(x)a) = a(g(x))a−1 · g′(x) . (136)

Note the presence of the final factor g′(x). If we apply these combined rules to a negativepower of a function

(

1

g(x)

)n

=1

(g(x))n= (g(x))−n

we obtain a generalization of the “Reciprocal Rule” [1, Exercise 3.2.44, p.198]:

d

dx

((

1

g(x)

)n)

= (−n)(g(x))−n−1 · g′(x) = (−n) ·ddx

g(x)

(g(x))n+1


= (−n) · g′(x)

(g(x))n+1 .

A special case is

The “Reciprocal Rule”

d

dx

(

1

g(x)

)

=d

dx

(

g(x)−1)

= (−1)(

g(x)−1−1)

· g′(x)

= − g′(x)

(g(x))2,

which can also be proved by the Quotient Rule.

Derivative of an exponential whose base is not e: Let a(x) be any positive,differentiable function of x. Then

d

dx((a(x))x) =

d

dx

(

(

eln a(x))x)

since exponential and logarithm are inverses

=d

dx

(

ex ln a(x))

by the exponent rules

= ex lna(x) · d

dx(x ln a(x))

= ex lna(x) ·(

1 · ln a(x) + x · d

dxln a(x)

)

by the Product Rule

= ex lna(x) ·(

ln a(x) + x · d

dxln a(x)

)

= a(x)x ·(

ln a(x) + x · d

dxln a(x)

)

(137)

which we cannot complete yet, because we don’t know the derivative of the logarithm.However, when a(x) is a constant, a, then ln a(x) is also constant, and its derivative is0; we obtain

d

dx(ax) = ax · ln a . (138)

In the next sections we shall show how our tools are now strong enough to determinethe derivative of the logarithm function; indeed, we are now equipped to differentiate allof the functions we are likely to meet in this course. We will also see another way of


determining the derivative of (a(x))x, and, more generally, of (a(x))b(x), where b is anydifferentiable function; the other method, which is no stronger than the method we haveshown here, is called logarithmic differentiation, [1, p. 246-248].

How to Prove the Chain Rule You are not expected to be able to prove thistheorem. The author includes this discussion because the earlier sketch of a proposedproof [1, pp. 218-219], while intuitively satisfying, is flawed.

3.5 Exercises

[1, Exercise 3.5.4, p. 224] “Write the composite function in the form y = f(g(x)). ...

Then find the derivativedy

dx: y = tan(sin x).”

Solution: The function that is applied first is sin: let’s call it g, so g(x) = sin x.The function applied next is tan; I’ll call it f , so f(x) = tan x. Then y = f(g(x)).By the Chain Rule,

dy

dx= f ′(g(x)) · g′(x)

= sec2(sin x) · cos x .

Alternatively, denoting g(x) by the single letter u, we could have written

dy

dx=

d

dxf(u)

= f ′(u) · du

dx= f ′(sin x) · g′(x)

= sec2(sin x) · cos x .

[1, Exercise 3.5.24, p. 225] “Find the derivative of the function y = 101−x2.”

Solution: Let u = 1 − x2. Then

d

dx

(

101−x2)

=d

dx10u

=d

du10u · du

dx

= 10u · ln 10 · d

dx(1 − x2)

by (138)

= 101−x2 · ln 10 · (−2x)


[1, Exercise 3.5.28, p. 224] “Find the derivative of the function y =e2u

eu + e−u.”

Solution: We could apply the Quotient Rule, using the Chain Rule where necessary:

d

du

(

e2u

eu + e−u

)

=(e2u · 2) · (eu + e−u) − e2u · (eu + e−u · (−1))

(eu + e−u)2

= e2u · eu + 3e−u

(eu + e−u)2

Another approach would be to first divide top and bottom by e2u, thereby makingthe Reciprocal Rule applicable:

d

du

(

e2u

eu + e−u

)

=d

du

(

1

e−u + e−3u

)

= −(

1

e−u + e−3u

)2

·(

e−u · (−1) + e−3u · (−3))

=e−u + 3e−3u

(e−u + e−3u).

Yet another approach would be to introduce an intermediate variable, t = eu:

d

du

(

e2u

eu + e−u

)

=d

dt

(

t2

t + 1t

)

· dt

du

=d

dt

(

t3

t2 + 1

)

· dt

du

(to simplify the rational function before differentiation)

=3t2(t2 + 1) − t3(2t)

(t2 + 1)2· dt

du

=t4 + 3t2

(t2 + 1)2· dt

du

=e4u + 3e2u

(e2u + 1)2 · eu .

[1, Exercise 3.5.30, p. 224] (not discussed in the lecture) “Find the derivative of the

function y =sin2 x

cos x.”

Solution: This problem could be solved by using the Quotient Rule, and by differ-entiating (sin x)2 by the Product Rule and the Chain Rule:

d

dx

(

sin2 x

cos x

)

=(2 sin x · cos x) · cos x − sin2 x(− sin x)

cos2 x


=sin x · (2 cos2 x + sin2 x)

cos2 x,

which, while correct, can certainly be simplified, since cos2 x + sin2 x = 1, tosin x · (cos2 x + 1)

cos2 x; some users would want to reduce it further to sin x ·(1+sec2 x).

However, it is easier to use a trigonometric identity earlier:

d

dx

(

sin2 x

cos x

)

=d

dx

(

1 − cos2 x

cos x

)

=d

dx(sec x − cos x)

=d

dxsec x − d

dxcos x

= sec x · tanx − (− sin x)

= sec x · tanx + sin x .

Another approach would be to replacesin x

cos xby tan x, and write the function as

sin x · tanx, whose derivative may be found by the Product Rule: cosx · tan x +sin x · sec2 x, which again admits simplification:

cos x · tan x + sin x · sec2 x = sin x + sin x · sec2 x = sin x · (1 + sec2 x) .

Additional Problems

1. (not discussed in the lecture) Differentiate y = sinx

2· cos 2x.

Solution:

(a) First Solution: There are no tricks. We apply the Product Rule, and,within it, the Chain Rule twice:

d

dx

(

sinx

2· cos 2x

)

= cosx

2· 1

2· cos 2x + sin

x

2· (− sin 2x) · 2

= cosx

2· 1

2· cos 2x − sin

x

2· (sin 2x) · 2 (139)

(b) [Second Solution]: We can apply the identity [1, 18.a p. A30] which thetextbook calls a “Product Formula”,

sin x · cos y =1

2[sin(x + y) + sin(x − y)].


(You are not expected to memorize this identity in this course.)

d

dx

(

sinx

2· cos 2x

)

=d

dx

(

1

2

(

sin

(

5

2· x)

+ sin

(

−3

2· x)))

=1

2

d

dx

(

sin

(

5

2· x)

+ sin

(

−3

2· x))

=1

2

(

cos

(

5

2· x)

· 5

2+ cos

(

−3

2· x)

·(

−3

2

))

=5

4cos

(

5

2· x)

− 3

4cos

(

−3

2· x)

=5

4cos

(

5

2· x)

− 3

4cos

(

3

2· x)

.

The answer obtained first could also be simplified by applying the sameidentity. The advantage of the second method is that the functions whichare eventually differentiated are “simpler” than those given, in that thetrigonometric functions appear only multiplied by a constant, rather thanin products.

2. (Only the part involving function y(x) was discussed in the lecture.) If y(x) =√

x + 1

x − 1, and z(x) =

√

x2 + 1

x2 − 1, find y′(2) and z′(2).

Solution:

(a)

dy

dx=

d

dx

√

x + 1

x − 1

=d

dx

(

(

x + 1

x − 1

)12

)

=1

2·(

x + 1

x − 1

)12−1

· d

dx

(

x + 1

x − 1

)

=1

2·(

x + 1

x − 1

)− 12

· 1 · (x − 1) − (x + 1) · 1(x − 1)2

= − 1

(x + 1)12 (x − 1)

32

When x = 2, y′ = − 1√3

= −√

3

3.


(b) Let u = x2. Then z(x) = y(u), so

d

dxz(x) =

d

duy(u) · d

dxu

= y′(u) · u′

= − 1

(u + 1)12 (u − 1)

32

· 2x

= − 1

(x2 + 1)12 (x2 − 1)

32

· 2x

so z′(2) = − 4

512 · 3 3

2

= − 4

45

√15.

D.12.2 §3.6 Implicit Differentiation.

The techniques we have developed in the preceding sections are sufficient for determiningthe derivatives of functions formed from the familiar classes of functions (polynomials,roots, rational functions, trigonometric functions, exponential functions, exponential andlogarithmic functions, etc.) by the methods seen in [1, §1.3, New Functions From Old].But sometimes we have to work with functions whose definitions are not explicit , thatis, where we do not have a precise formula for the function, and yet the function is stillfully determined. One way in which this happens is where the values of a function areknown to satisfy an equation, which we may not be able to solve explicitly . In such caseswe can think of the function as being implicitly defined by the equation. The ChainRule can often be used to find the derivative of such an implicitly defined function. It issomething of a misnomer to call the operation of differentiation in such a case implicitdifferentiation: there is nothing implicit in the differentiation — it is the definition thatis implicit. When we apply differentiation in cases like these, the result is often a formulawhich gives the values of the derivative of the function in terms of both the independentvariable and the function values; that is, we may have to express f ′(x) in terms of x andf(x), and may be unable to find an explicit formula for f ′(x) in terms of x alone.

The derivative of the logarithm function Since the logarithm and exponentialfunctions are inverses of each other, we know that

eln x = x for all x > 0 (140)

and ln ex = x for all x . (141)

Analogously to our earlier determination of the derivatives of the inverses of the trigono-metric functions, let us differentiate implicitly in (140) with respect to x:

eln x · d

dxln x = 1 ,


which equation we may solve for the derivative of ln x:

d

dxlnx =

1

eln x=

1

x.

Example D.18 [1, Exercise 10, §3.6, p. 233] Suppose we need to find y′, i.e.dy

dx, where

all we know is thaty5 + x2y3 = 1 + yex2

. (142)

We do not have practical methods for solving this equation for y explicitly in terms ofx. But we can apply the Chain Rule to the two sides of the defining equation, treatingeach term on each side as a function of x:

y5 : We differentiate by the Power Rule combined with the Chain Rule, to obtain

d

dxy5 = 5y4 · dy

dx,

and make no attempt to determinedy

dxat this time; indeed, this derivative is what

we are trying to find, and we are in the process of determining a constraint on it.

x2y3: This term is differentiated by the Product Rule, and then one of the resultingterms by the Power Rule combined with the Chain Rule:

d

dx(x2y3) = 2x · y3 + x2 · 3y2 · dy

dx.

1:d

dx1 = 0

In this case the definition is “explicit”.

yex2:

d

dx

(

yex2)

=dy

dx· ex2

+ y · ex2 · 2x .

Collecting the differentiated terms together, and equating the two expressions we havefound for the derivative of the function given by each of the sides of the equation, weobtain

5y4 · dy

dx+ 2x · y3 + x2 · 3y2 · dy

dx= 0 +

dy

dx· ex2

+ y · ex2 · 2x .

We shift all terms having a factordy

dxto one side of the equation:

5y4 · dy

dx+ x2 · 3y2 · dy

dx− dy

dx· ex2

= −2x · y3 + y · ex2 · 2x .


and solve fordy

dx:

dy

dx=

−2xy3 + yex22x

5y4 + x23y2 − ex2 =2xy

(

ex2 − y2)

5y4 + x23y2 − ex2 .

I normally do not work in the lecture problems which are solved in the textbook.However, the following example is related to a very interesting curve in the plane.

Example D.19 (Discussed very briefly at the end of the lecture; cf. [1, Example 2, p.230]) The Folium of Descartes has equation x3 + y3 = 6xy.

1. Find a formula for y′.

2. Find the tangent to the folium at the point (x, y) = (3, 3) on the curve.

3. Determine where the (tangent to the) curve is horizontal.

Solution: By implicit differentiation with respect to y we see that

3x2 + 3y2 · y′ = 6(1 · y + x · y′) ,

which we can solve for y′:

y′ =2y − x2

y2 − 2x.

This function can be evaluated at (x, y) = (3, 3) to yield y′ = −1; knowing a point onthe tangent line and its slope, we can find an equation for the line. Then, to determinethe locations of the horizontal tangents, we solve the equation y′ = 0, which is equivalentto

2y = x2 ,

with the equation of the curve, since the points (x, y) that we seek satisfy both of theseconditions. We obtain, after reduction, x3 = 0, 16, which imply that x = 0 or x = 3

√16.

When x = 0, y = 02

2= 0, so the point is the origin; when x = 3

√16 = 2

43 , y = 16

23

2= 2

53 .

Note that the curve has two tangents at (x, y) = (0, 0): one is horizontal, and the otheris vertical.





Today’s and last day’s lecture are being devoted mainly to [1, §§3.5, 3.6].

D.13.1 §3.5 The Chain Rule (conclusion)

§3.5 Exercises

[1, Exercise 3.5.52, p. 225] “Find the x-coordinates of all points on the curve y =sin 2x − 2 sinx at which the tangent line is horizontal.”

Solution: We have to find all solutions x of the equation y′ = 0, i.e.,

y′ = cos 2x · 2 − 2 cos x . (143)

We can solve this if we replace cos 2x by 2 cos2 x − 1, an identity that you areexpected to have available:

2(

2 cos2 x − 1)

− 2 cosx = 0 ⇔ 2 cos2 x − cos x − 1 = 0

⇔ (2 cosx + 1)(cosx − 1) = 0

which is satisfied when (and only when) cos x is equal to either 1 or −1

2.

cos x = 1 ⇔ x = 2nπ

cos x = −1

2⇔ x = ±2π

3+ 2mπ

⇔ x =

(

±2

3+ 2m

)

π

where m and n are any integers. We find that the points with horizontal tangentsare evenly spaced along the x-axis, at intervals of length 2π

3. (We could also have

solve equation (143) in a simpler way, by replacing the difference of cosines by aproduct of sines:

2(cos 2x − cos x) = −4 sin2x + x

2· sin 2x − x

2

= −4 sin3x

2· sin x

2.

The factor sin x2

vanishes precisely where x = 2nπ; the factor sin 3x2

vanishes pre-cisely where x = 2n

3π.)


D.13.2 §3.6 Implicit Differentiation (conclusion)

As explained in the last lecture, implicit differentiation means differentiation of functionswhich are defined implicitly (by an equation), rather than being defined explicitly. I beginby illustrating this technique with the very last problem in the Review Exercises for theChapter:

Example D.20 [1, Exercise 106, p. 273] Show that the length of the portion of any

tangent line to the astroid x23 + y

23 = a

23 cut off by the coordinate axes is constant.

Solution: Let’s differentiate all terms in the given equation with respect to x. Wedon’t have a formula for y as a function of x (although, in this case, we could find

one by solving the equation), so let’s just differentiate y23 by the Chain Rule without

attempting to obtain a formula. (We call this implicit differentiation, since we aredefining the function y(x) “implicitly” in terms of x, rather than attempting to obtainan explicit formula.) We obtain, as a result of that differentiation,

2

3x− 1

3 +2

3y− 1

3 · y′ = 0

so

y′ = − 3

√

y

x.

Suppose that we are considering the tangent to the curve at the point (x, y) = (u, v) on

the curve. Then the slope of the tangent is − 3

√

v

u, and the equation of that tangent is

y − v = − 3

√

v

u· (x − u) .

This line meets the x-axis in the point

(

u + v 3

√

u

v, 0

)

and the y-axis in the point(

0, v + u 3

√

v

u

)

The square of the distance between these two intercepts is

(

u + v 3

√

u

v

)2

+

(

v + u 3

√

v

u

)2

=(

u + u13 v

23

)2

+(

v + v13 u

23

)2


= u23

(

u23 + v

23

)2

+ v23

(

v23 + u

23

)2

=(

u23 + v

23

)3

.

But we assumed that (u, v) is a point on the curve, so

u23 + v

23 = a

23 (144)

and the square of the distance between the intercepts is(

a23

)3

= a2 , which is, indeed,

a constant. Hence the length of the portion of the tangent line intercepted by thecoordinate axes is the constant |a|.

Simplifications Sometimes it may happen that the formula we obtain fordy

dxmay

simplify further if we use the constraint we know to hold between x and y. In thepreceding example we saw such a constraint in equation (144).

Higher Derivatives In [1, §3.7] we shall consider problems where we will wish to findthe derivative of the derivative of a function defined explicitly. In those situations therewill be a variety of ways of applying implicit differentiation, and some methods may bemore efficient than others.

Orthogonal Trajectories Orthogonal means perpendicular . Sometimes we may wishto explore the angles between members of two families of curves, particularly in the casewhere they are mutually perpendicular. The determination of the slopes of members ofone or other of these families could require the use of implicit differentiation. Rememberthat two slopes are mutually perpendicular if their product is −1.

(This is a consequence of an application of the expansion of cot(θ1 − θ2) when θ1 − θ2 = π2 .

cot(θ1 − θ2) =cos(θ1 − θ2)

sin(θ1 − θ2)

=cos θ1 · cos θ2 + sin θ2 · sin θ1

sin θ1 · cos θ2 − sin θ2 · sin θ1

=1 + tan θ1 · tan θ2

tan θ1 − tan θ2

after division of numerator and denominator by cos θ1 · cos θ2.

If θ1− θ2 = ±π2 , the cotangent of the difference is 0, so the numerator of this last fraction must

be 0; equivalently,tan θ1 · tan θ2 = −1 .)


Derivatives of Inverse Trigonometric Functions Earlier in the term we investi-gated the definitions of inverse trigonometric functions, and found that, for each of the6 functions, we needed to restrict the domain in order that the graph would satisfy the“Horizontal Line Rule”, and that the function would have an inverse.

We can obtain, for example, the derivative of the inverse cosine function, as follows.We recall that the custom is to restrict the domain of the cosine to

0 ≤ x ≤ π , (145)

so thatarccos(cos x) = x (146)

for all x satisfying (145), andcos(arccos x) = x (147)

for all x in the domain of the arccosine function, i.e. for −1 ≤ x ≤ 1. Differentiating(147) “implicitly” with respect to x, we obtain

− sin(arccos x) · d

dxarccos x = 1 ,

sod

dxarccos x = − 1

sin(arccos x)= − 1

± (1 − cos2(arccos x))12

. (148)

But the values of the arccosine function are in the restricted domain of the cosine, i.e. in(145), where the sine function is positive. Thus the ambiguous sign ± may be replacedby +, and

d

dxarccos x = − 1

(1 − cos2(arccos x))12

= − 1√1 − x2

by (147). The derivatives of the other 5 inverse trigonometric functions may be deter-mined in a similar way. Here are the derivatives:


d

dxsin−1 x =

1√1 − x2

(149)

d

dxcos−1 x = − 1√

1 − x2(150)

d

dxtan−1 x =

1

1 + x2(151)

d

dxcot−1 x = − 1

1 + x2(152)

d

dxsec−1 x =

1

x√

x2 − 1(153)

d

dxcsc−1 x = − 1

x√

x2 − 1(154)

However, the textbook’s definitions of the inverse secant and inverse cosecant functionsare non-standard, and the resulting derivatives are also non-standard. If you need towork with either of these functions outside of this course, you must first check whatdefinition is in force.

Derivatives of inverse functions in general [1, Exercise 67, §3.6, p. 235] Thepreceding discussion generalizes. If we know that a function f has an inverse, so that wehave

y = f(x) (155)

x = f−1(y) (156)

f(

f−1(y))

= y (157)

f−1(f(x)) = x (158)

then we can differentiate implicitly to show that

dx

dy=

1

dy

dx

(159)


(wheredy

dx6= 0), so that the derivatives appear to behave like fractions. Sometimes we

need to determine the slope of a curve,dy

dx, but it is much easier to determine

dx

dy. We

can determine the latter first, then simply take its reciprocal.

3.6 Exercises

[1, Exercise 3.6.36, p. 234] extended: (not solved in the lecture)

1. “Show by implicit differentiation that the tangent to the ellipsex2

a2+

y2

b2= 1

(where a and b are non-zero constants) at the point (x0, y0) on the curve is

x0x

a2+

y0y

b2= 1 .”

2. Find an equation for the normal to the curve at the point

(

3a

5,4b

5

)

. (In this

context the word normal means perpendicular ; the normal to the curve at apoint is the line through the point which is perpendicular to the tangent line.)

Solution:

1. Differentiating the equation of the curve implicitly with respect to x yields

2x

a2+

2y

b2· y′ = 0 ,

so the slope of the tangent to the curve at the point on the curve with coor-

dinates (x, y) is y′ = − b2x

a2y. At the point with coordinates (x0, y0), the slope

of the tangent is, therefore, −b2x0

a2y0, and the equation of the tangent line is,

therefore,

y − y0 = −b2x0

a2y0

· (x − x0)

which may be rewritten as

x0x

a2+

y0y

b2=

x20

a2+

y20

b2. (160)

We haven’t yet used the fact that (x0, y0) lies on the curve; this implies thatthese coordinates satisfy the equation of the curve, so

x20

a2+

y20

b2= 1 .


When we substitute this information in equation (160), we obtain the desiredequation for the tangent line.

2. The slope of the tangent at

(

3a

5,4b

5

)

is y′ = −b2 · 3a

5

a2 · 4b

5

= −3b

4a. The slope of

the normal will, therefore, be the negative reciprocal of this number, i.e.4a

3b.

The equation of the line through the given point, with this slope, will be

y − 4b

5=

4a

3b·(

x − 3a

5

)

i.e. 5(4ax − 3by) = 12(a2 + b2).

[1, Exercise 3.6.46, p. 234] “Find and simplify the derivative of the function f(x) =arctan(x −

√1 + x2).”

Solution:

f ′(x) =1

1 + (x −√

1 + x2)2·(

1 − 1

2· (1 + x2)−

12 · 2x

)

=1

1 + x2 + (1 + x2) − 2x√

1 + x2·(

1 − x√1 + x2

)

=1

2· 1

1 + x2 − x√

1 + x2·(√

1 + x2 − x√1 + x2

)

=1

2· 1√

1 + x2(√

1 + x2 − x) ·(√

1 + x2 − x√1 + x2

)

=1

2· 1

1 + x2

[1, Exercise 3.6.50, p. 234] (not discussed in the lecture) “Find the derivative of thefunction

y = arctan cos θ ; (161)

simplify where possible.”

Solution:

1. First solution, using the Chain Rule:

d

dθarctan(cos θ) =

1

1 + cos2 θ· (− sin θ) .


2. Second solution, using implicit differentiation: Applying the tangent functionto both sides of (161) yields

tan y = cos θ .

We differentiate both sides with respect to θ:

sec2 y · dy

dθ= − sin θ

and solve fordy

dθ:

dy

dθ= − sin θ

sec2 y= − sin θ

1 + tan2 y= − sin θ

1 + cos2 θ.

[1, Exercise 3.6.56, p. 235] “Show that the given curves are orthogonal:

x2 − y2 = 5 (162)

4x2 + 9y2 = 72 .” (163)

Solution: Differentiating the first equation “implicitly” with respect to x yields

2x − 2y · dy

dx= 0 ⇒ dy

dx=

x

y;

while the second equation yields

8x + 18y · dy

dx= 0 ⇒ dy

dx= −4x

9y.

In the preceding, the derivativesdy

dxare not the same; they denote the slopes of

the two curves, even if we are considering a point which is simultaneously on bothcurves. That is precisely what we must do if we wish to test orthogonality, i.e.,perpendicularity: we must show that, at the points where the curves meet, theirtangents are perpendicular; or, equivalently, we must show that the products ofthe slopes of the two curves at a point where they meet is −1. The product is

(

x

y

)(

−4x

9y

)

= −4x2

9y2.

But this does not seem to be equal to −1; where have we gone wrong? Rememberthat we are considering this product of derivatives only at points that are on both


curves. So the only points we need to consider are those satisfying both of theequations. Consider a point (x, y) which is on both curves. If we solve equations(162), (163), interpreting them as equations in the variables x2 and y2, we obtain

(x2, y2) = (9, 4)

from which it follows that the product of the slopes of the tangents at the points(x, y) of intersection is

−4x2

9y2= −4 · 9

9 · 4 = −1

so the tangents at points of intersection are, indeed, perpendicular.

[1, Exercise 65, §3.6, p. 235] (not discussed in the lecture) Find all points on thecurve

x2y2 + xy = 2 (164)

where the slope of the tangent line is −1.

Solution: Implicit differentiation of (164) yields

2x · y2 + x2 · 2y · y′ + y + x · y′ = 0 . (165)

While we could solve for y′, we need only set y′ = −1 directly in (165), obtaining

2x · y2 − x2 · 2y + y − x = 0 , (166)

which factorizes as(1 + 2xy)(x − y) = 0 .

Thus the points we seek lie on at least one of the curves

y = − 1

2x(167)

y = x . (168)

They also lie on the given curve (164); we can find the points by solving each of thepreceding equations with (164). We see immediately that there are no points ofintersection on (167); but the coordinates of points of intersection on (168) satisfyx4 + x2 = 2, i.e. (x2 + 2)(x2 − 1) = 0, so x = ±1, and y = ±1.

(The equation of the given “curve” factorizes into

(xy + 2)(xy − 1) = 0

so the “curve” actually is the union of two curves, xy = −2 and xy = 1, indeed ithas 4 branches! We could have attacked the problem by explicit differentiation of


−2

xand −1

x, setting the derivatives equal to the given value of 1, etc. This is not

always the case, but often happens in textbook examples that are chosen to havesimplified algebra. Try to solve problems for the purpose they are intended, butthen look for facts like this one to verify your work or to permit more insight intothe particular example.)


D.14 Supplementary Notes for the Lecture of October 23rd,2006

Release Date: Monday, October 23rd, 2006(subject to revision or correction)


[1, Exercise 3.6.68, p. 235] 1. “Show that f(x) = 2x + cos x is one-to-one.

2. “What is the value of f−1(1)?

3. “... find (f−1)′(1)?”

Solution:

1. This question is actually premature, since a rigorous proof would need to usethe theory we will meet in [1, §4.2]. So let us look at the question intuitively.The slope of the graph is y′ = 2− sin x, which can never be less than 2− 1 =1 > 0; thus the tangents are always sloping upwards, so the graph is alwaysrising as one proceeds from left to right, and there cannot be two points onthe curve having the same y-value; what we have “proved” is that the graphsatisfies the Horizontal Line Test [1, p. 64]. This proof is a bit “fuzzy”, andwe will have a better way of expressing these ideas when we reach [1, §4.2]and consider the Mean Value Theorem.

2. We have to solve, for x, the equation 1 = y = 2x − cos x. While this wouldbe difficult if y were a general number, we can do it easily by inspection wheny = 1, as we can see that x = 0 has the desired property. Since the functionhas an inverse, x = 0 is the only solution.

3. The equation of the graph of the given function is

y = 2x + cos x,

and, as seen above, differentiation yields

dy

dx= 2 − sin x

We just saw that, when y = 1, x = 0. We know that the value ofdy

dxis

2 − sin 0 = 2 at this point; it follows from (159) thatdx

dy=

1

2.


D.14.1 §3.7 Higher Derivatives

Notation for higher derivatives In the notes for the earlier lectures (cf. §D.6) I havelisted some alternative notations for the (first) derivative. The “higher” derivatives, i.e.the iterated derivatives, may be denoted by variants of these notations, e.g., f ′′(a), f ′′′(a),

. . . , f (n)(a);d2f

dx2(a), . . . ,

dnf

dxn(a);

dnf

dxn

∣

∣

∣

∣

x=a

; Dnf(a), Dnf(x)|x=a; Dxx...xf(a);

[

dnf

dxn

]

x=a

.

Note that in the “Leibniz” (fractional) notation the symbol indicating the number ofdifferentiations is placed at a location in the “numerator” different from that in the“denominator” of the symbol.

Significance of the higher derivatives The second derivative represents the rate ofchange of the first derivative. In [1, §4.3] we shall see that, for a doubly differentiablefunction f(x), f ′′(x) has a significance in connection with the “concavity” of the graphof f .

When f(t) represents the position at time t of a particle moving on a line, the firstderivative represents the velocity of the moving particle, and y′′(t) represents the rate ofchange of velocity, which is called the acceleration of the moving particle. Many of thecommon equations of mathematical physics involve the second derivative, since Newton’s2nd Law of Motion [1, §6.4, p. 460] relates the time derivative of the product of massand f ′(t) to the applied force. Where mass is constant, this reduces to the equation

F = mass × f ′′(t) ,

which leads to a differential equation that must be solved to determine the equation ofmotion of the moving particle. Even where mass is not constant, as, for example, in arocket rising vertically under gravity, and gradually expending its fuel (and so reducingits mass), an equation can be obtained involving the second derivative:

F =d mass

dt· f ′(t) + mass × f ′′(t) .

Example D.21 Prove that the function y = e4x + 2e−x has the property that

y′′′ − 13y′ − 12y = 0 (169)

for all x. We say that y = e4x + 2e−x is a solution of the differential equation (169).Solution: We determine the derivatives by repeated applications of the rules of differen-tiation, including the Chain Rule:

y = e4x + 2e−x

y′ = 4e4x − 2e−x

y′′ = 16e4x + 2e−x

y′′′ = 64e4x − 2e−x


Hence

y′′′ − 13y′ − 12y =(

64e4x − 2e−x)

− 13(

4e4x − 2e−x)

− 12(

e4x + 2e−x)

= 0 · e4x + 0 · e−x = 0

Does the second derivative behave “like a fraction”? We have seen in §D.12.1of these notes that the first derivative “behaves like a fraction”, in the sense that, wherea function y(x) has an inverse,

dx

dy=

1

dy

dx

.

This symbolically simple situation does not hold for higher derivatives. Let’s attempt to

expressd2x

dy2in terms of the derivatives of y with respect to x. In the equation

dx

dy· dy

dx= 1

we interpret each of the factors on the left side as a function of x; but, in differentiating

the first factor we interpretdx

dyas a function of an intermediate variable y which is a

function of x; we obtain(

d2x

dy2

)

· dy

dx+

dx

dy· d2y

dx2= 0 ,

which we may solve ford2x

dy2:

d2x

dy2= −

d2x

dy2· dx

dy(

dy

dx

)2

= −d2y

dx2(

dy

dx

)3 .

(You are not expected to memorize this formula!)


Determination of higher derivatives by implicit differentiation:

Example D.22 Suppose that x and y are related by the equation ex+y = xy. Determined2y

dx2.

Solution: Implicit differentiation of the given equation with respect to x yields

ex+y · (1 + y′) = 1 · y + x · y′ .

We can now proceed in two ways.

Solve for y′: Solving for y′ yields

y′ =y − ex+y

ex+y − x. (170)

However, we could have simplified the equation before solving, since ex+y = xy:

xy · (1 + y′) = 1 · y + x · y′ . (171)

Now, when we solve, we obtain

y′ = −y(1 − x)

x(1 − y).

We then differentiate using the Quotient Rule:

y′′ =(y′(1 − x) + y(−1)) · x(y − 1) + y(1 − x) · (1(1 − y) + x(−y′))

x2(1 − y)2

Differentiate implicitly first: As in the preceding method, first simplify the equationto (171); then differentiate implicitly without solving for y′:

1 · y · (1 + y′) + xy′ · (1 + y′) + xy · (0 + y′′) = y′ + 1y′ + xy′′ (172)

At this point we could solve for y′′.

Neither of these solutions is fully simplified. For the 1st method we can further simplifyby using the information in (170):

y′′ =y

x2(y − 1)3·(

(x − 1)2 + (y − 1)2)

.


Example D.23 In [1, Example 3.6.2, p. 230] the textbook shows by implicit differenti-ation that, if

x3 + y3 = 6xy (173)

then

y′ =2y − x2

y2 − 2x. (174)

Determine the value ofd2y

dx2when (x, y) = (3, 3).

Solution:

y′′ =d

dx

(

2y − x2

y2 − 2x

)

=(2y′ − 2x)(y2 − 2x) − (2y − x2)(2yy′ − 2x)

(y2 − 2x)2

Substituting x = y = 3 yields y′′ =24y′(3) − 24

9, where the argument of y′ refers to

x = 3. Since substitution of (x, y) = (3, 3) in (174) yields y′(3) = −1, we have

y′′ = −−24 − 24

9= −16

3.

A high order derivative evaluated by Mathematical Induction (This subsec-tion is not examination material in 2006/07.) Consider [1, Exercise 40, p. 241]:to determine a formula for D1000 (xe−x).Solution: We begin by some experimentation. We evaluate Dn (xe−x) where n is a smallpositive integer, and try to guess what a general formula might involve.

D1(

xe−x)

= 1 · e−x + x · e−x · (−1) = (1 − x)e−x

D2(

xe−x)

= D(

D(

xe−x))

= D(

(1 − x)e−x)

= (−1) · e−x + (1 − x) · e−x · (−1) = (x − 2)e−x

D3(

xe−x)

= D(

D2(

xe−x))

= D(

(x − 2)e−x)

= 1 · e−x + (x − 2) · e−x · (−1) = (3 − x)e−x

By now we can guess what the general formula might be; if you don’t see how I arriveat the guess, differentiate a few more times until the likely pattern emerges. My guess is

Dn(

xe−x)

= (−1)n+1(n − x)e−x . (175)

We can observe that this formula “works” for n = 1, 2, 3; it also “works” for n = 0, wherewe interpret D0 as an operation that maps a function on to itself. Under Mathematical


Induction, having proved the formula valid at some starting value — here either n = 0or n = 1, we assume it true for a general value, n = N , and try to infer that it is alsotrue for the next value, n = N + 1. Indeed, if (175) is valid when n = N , then

DN+1(

xe−x)

= D(

DN(

xe−x))

by definition of DN+1

= D(

(−1)N+1(N − x)e−x)

which is our “induction hypothesis”

= (−1)N+1(

(−1) · e−x + (N − x) · e−x · (−1))

by the Product Rule

= (−1)N+1(

(x − (N + 1)) · e−x)

= (−1)(N+1)+1((N + 1) − x)e−x

which is precisely what we wanted to prove for the value of the (N + 1)st derivative.The Principle of Mathematical Induction permits us to infer that (175) is true for allnon-negative integers n, thereby proving that the guess was correct.

For a proof by Mathematical Induction to succeed, one needs to have the right guess,and to be able to infer each case from the preceding one, in a general way.

Let’s not lose sight of why we proved this result: it was to determine D1000(

xe−x)

.So we now may use the case n = 100 of equation (175).

3.7 Exercises

[1, Exercise 3.7.20, p. 240] “Find the first and second derivatives of the function y =tan−1(x2).”

Solution: By the Chain Rule,

d

dxtan−1(x2) =

1

1 + (x2)2· 2x =

2x

(1 + x4)2.

We then apply the Quotient Rule:

y′′ =2(1 + x4) − 2x · 4x3

(1 + x2)2=

2 (1 − 3x4)

(1 + x2)2.

[1, Exercise 3.7.36, p. 241] (This exercise may be omitted by students in 2006/07, asMathematical Induction, which is require in its proof, is not examination materialthis semester.)“Find a formula for f (n)(x) for all positive integers n: f(x) =

√x.”


Solution: This problem must be solved using Mathematical Induction, or an equiv-alent method. We first experiment to arrive at a general result, which we will thenhave to prove to be correct. Let’s start with the experimentation:

f (0)(x) =√

x = x12 (hypothesis)

f (1)(x) =1

2x

12−1 =

1

2x− 1

2

f (2)(x) =

(

−1

2

)(

1

2

)

x− 12−1 = − 1

22x− 3

2

f (3)(x) =

(

−3

2

)(

− 1

22

)

x− 32−1 =

1 · 323

x− 52

f (4)(x) =

(

−5

2

)(

1 · 323

)

x− 52−1 = −1 · 3 · 5

24x− 7

2

We can now guess that the general formula will be

f (n)(x) = (−1)n+11 · 3 · 5 · 7 · . . . · (2n − 3)

2nx

2n−12

for n ≥ 1. We need to formulate this conjecture75 in “closed form”, i.e., withoutusing the “. . .” symbol. The magnitude of the numerator is the product of the firstn − 1 odd integers. If we multiply by the even integers between them, and thendivide these out, we obtain

1 · 3 · 5 · 7 · . . . · (2n − 3) =1 · 2 · 3 · 4 · . . . · (2n − 4) · (2n − 3)

2 · 4 · 6 · . . . · (2n − 4)

=(2n − 3)!

2n−2(n − 2)!=

(2n − 2)!

2n−1(n − 1)!

where n! — read n factorial — is the product of the positive integers 1, 2, . . ., n,and where we define 0! to mean 1. We can now guess that the general formula is

f (n) = (−1)n+1 (2n − 2)!

22n−1(n − 1)!· x− 2n−1

2 (n ≥ 1) . (176)

All of the preceding work was just to prepare the guess, now the proof properbegins. In a proof by induction we must first prove a starting case. Usually westart from 0 or 1, but here I will start at n = 2. If we substitute n = 2 into theright-hand member of equation (176), we obtain

(−1)2+1 (2 · 2 − 2)!

22·2−1(2 − 1)!· x− 2·2−1

2 = (−1)3 2!

23(1)!· x− 3

2 = −1

4· x− 3

2 ,

75=guess


which agrees with the value computed earlier for the 2nd derivative. Thus the casen = 2 has been proved. Now suppose that we know that (176) is true for the casen = N > 1. Then we can differentiate both sides of the equation, to obtain

f (N+1) =d

dx

(

(−1)N+1 (2N − 2)!

22N−1(N − 1)!· x− 2N−1

2

)

= (−1)N+1 (2N − 2)!

22N−1(N − 1)!· d

dx

(

x− 2N−12

)

= (−1)N+1 (2N − 2)!

22N−1(N − 1)!·(

−2N − 1

2

)

x− 2N−12

−1

= (−1)N+1 (2N − 2)!

22N−1(N − 1)!·(

−2N − 1

2

)

x− 2(N+1)−12

= (−1)(N+1)+1 (2(N + 1) − 2)!

22(N+1)−1((N + 1) − 1)!· x− 2(N+1)−1

2

which is exactly the value that equation (176) gives in the case n = N + 1. Thuswe have shown that, starting from the value n = 2, the truth of every case ofthe equation implies the truth of its successor. By the Principle of MathematicalInduction we may conclude that the equation is true for all positive integers n ≥ 2.

[1, Exercise 3.7.50, p. 241] (Not solved in the lecture.) This problem asks that youconsider a particle moving along the x-axis, with position x at time t ≥ 0 given

by x(t) =t

1 + t2. You are asked to find a formula for the acceleration, and to

determine when the particle is speeding up and slowing down.

Solution:

x′(t) =1 · (1 + t2) − t · (0 + 2t)

(1 + t2)2

=1 − t2

(1 + t2)2(177)

x′′(t) =−2t · (1 + t2)2 − (1 − t2) · 2(1 + t2)2t

(1 + t2)4

=−2t(3 − t2)

(1 + t2)3(178)

To analyze when the particle is “speeding up”, we first need to ask what is meant

by speed . This concept was defined on [1, p. 161] to be

∣

∣

∣

∣

d

dtx(t)

∣

∣

∣

∣

, the absolute

value of the velocity. The particle is “speeding up” when the first derivative of


this function is increasing; equivalently, when the derivative of the first derivativeis positive; equivalently, when the second derivative is positive. We know from

(177) that x′ =(t + 1)(t − 1)

(1 + t2)2; as the denominator is always positive, the fraction

is positive when either 0 or 2 of the factors in the numerator are positive. Thisoccurs either when t < −1 or when t > 1. As we are told to consider only non-negative t, this yields

|x′| =

−(t + 1)(t − 1)

(1 + t2)2= x′ for 0 < t < 1

(t + 1)(t − 1)

(1 + t2)2= −x′ for 1 < t

(179)

We know from (178) that x′′ =−2(t +

√3)(t −

√3)

(1 + t2)3; knowing where x′(t) is posi-

tive and negative, we can infer from this statement that

d

dt|x′| =

x′′ =2t(t +

√3)(t −

√3)

(1 + t2)3for 0 < t < 1

−x′′ = −2t(t +√

3)(t −√

3)

(1 + t2)3for 1 < t

(180)

(We would have to examine the formulæ more carefully to show that the functionhas no derivative at t = 1.) The functions have a numerator which is a constantmultiple of t, t −

√3, and t +

√3. At time t = 0 the particle starts from x(0) = 0

and moves to the right (i.e., in the positive direction). The initial speed is 1,and this speed decreases as t → 1−. When t = 1, the particle has arrived atx(1) = 1

2, its speed has decreased to 0. As t increases from 1 to

√3, the velocity

becomes negative, and the particle moves to the left at an increasing speed, untiltime t =

√3, when it is at position

√3

4, and it is moving to the left at the rate of

18. At that point in time the speed stops increasing and starts to decrease again,

although the particle does not change its direction of motion. It continues to moveto the left, now at decreasing speed, approaching — but never reaching — theorigin, which is its limiting position as t → ∞.

D.14.2 §3.8 Derivatives of Logarithmic Functions

The derivative of the logarithm function Since the logarithm and exponentialfunctions are inverses of each other, we know that

eln x = x for all x > 0 (181)

and ln ex = x for all x . (182)


Analogously to our earlier determination of the derivatives of the inverses of the trigono-metric functions, let us differentiate implicitly in (181) with respect to x:

eln x · d

dxln x = 1 ,

which equation we may solve for the derivative of ln x:

d

dxlnx =

1

eln x=

1

x.

More generally, if a is any positive real number, and we differentiate implicitly in

aloga x = x ,

we obtain

aloga x ln a · d

dxloga x = 1 ,

which we may solve to obtain

d

dxloga x =

1

aloga x ln a=

1

x ln a.

The derivative of ln |x|: If we consider separately the domains 0 < x and x < 0, we

can show thatd

dxln |x| =

1

xfor x 6= 0. Read the proof in [1, Example 3.8.6, p. 246].

Fixed base and exponent vs. variable base and exponent. Let’s consider howwe differentiate functions of the form a(x)b(x).

1. When a(x) and b(x) are both constants, a(x)b(x) is also constant, and its derivativeis 0.

2. When b(x) is a constant b 6= 0, use the Chain Rule:

d

dx

(

a(x)b)

= b · a(x)b−1 · da(x)

dx;

of coursed

dx

(

a(x)0)

=d

dx1 = 0 .

3. When a(x) is a positive constant a, and b(x) = x, we saw in equation (138), that

d

dxax = ax ln a .


4. When a(x) is a positive constant a, and b(x) is unrestricted, we may apply theresult just preceding to obtain (by the Chain Rule with u = b(x)),

d

dx

(

ab(x))

=d

duau

∣

∣

∣

∣

u=b(x)

· d

dxu

= ab(x) · ln a · d

dxb(x)

5. But we are even able to differentiate the most general case, if a(x) > 0:

d

dx

(

a(x)b(x))

=d

dx

(

(

eln a(x))b(x)

)

=d

dx

(

e(lna(x))·b(x))

= e(ln a(x))·b(x) · ln e · d

dx((ln a(x)) · b(x))

= a(x)b(x) · ln e · d

dx((ln a(x)) · b(x))

= a(x)b(x) · ln e ·(

d

dx(ln a(x)) · b(x) + ln a(x) · d

dxb(x)

)

= a(x)b(x) · ln e ·(

1

a(x)· d

dxa(x) · b(x) + ln a(x) · d

dxb(x)

)

= a(x)b(x) ·(

1

a(x)· d

dxa(x) · b(x) + ln a(x) · d

dxb(x)

)

since ln e = 1.

Warning! Do not attempt to memorize these formulæ. If you are asked todifferentiate functions of these types, we will always expect you to show your work, so itwill not be acceptable to substitute in a memorized formula, even if you could rememberit perfectly.

Logarithmic Differentiation The last case discussed above can also be approachedas follows. First give a name to a(x)b(x), say y = a(x)b(x). Then take logarithms of bothsides of the equation:

ln y = b(x) · ln a(x) ,

and differentiate the resulting equation implicitly with respect to x:

1

y· dy

dx=

d

dxb(x) · ln a(x) + b(x) · 1

a(x)· d

dxa(x) .


Now solve the equation fordy

dxand express in terms of a(x) and b(x):

dy

dx= y

(

d

dxb(x) · ln a(x) + b(x) · 1

a(x)· d

dxa(x)

)

.

Then express this result in terms of the original function:

d

dx

(

a(x)b(x))

= a(x)b(x)

(

d

dxb(x) · ln a(x) + b(x) · 1

a(x)· d

dxa(x)

)

,

as before. You may use either the method given before, or this latter one, (which is oftencalled “logarithmic differentiation”).

The preceding formula should not be memorized! What you should rememberis the procedure:

• Name the function.

• Take logarithms of both sides of the naming equation.

• Differentiate both sides implicitly with respect to the original independent variable.

• Solve for the derivative.

• Simplify, and remove reference to the name given to the function.

3.8 Exercises

[1, Exercise 3.8.26, p. 249] To differentiate f(x) =1

1 + ln x, and to find the domain

of f .

Solution: The domain of f consists of all x where the denominator is defined, andwhere that denominator is non-zero. The denominator is defined wherever lnx isdefined, i.e. where x > 0. However, it is 0 when ln x = −1, i.e. when x = 1

e. So the

domain is(

0,1

e

)

∪(

1

e,∞)

.

d

dx

(

1

1 + ln x

)

=−1

(1 + ln x)2·(

0 +1

x

)

=−1

x(1 + ln x)2

Problems involving logarithms can often lead to several solutions that appear tobe different. For example, in this problem one can use the fact that 1 = ln e torewrite 1 + ln x as ln e + ln x, which is ln(ex).


[1, Exercise 3.8.38, p. 249] “Use logarithmic differentiation to find the derivative ofthe function

y =4

√

x2 + 1

x2 − 1.” (183)

Solution: Taking logarithms in equation (183) yields

ln y =1

4ln

x2 + 1

x2 − 1.

We observe that the numerator of the fraction is always positive; as we are takingthe 4th root of the fraction, it must be positive, so the denominator also must bepositive. We may, therefore, invoke the property of logarithms [1, 2nd Law, p. 68]that the log of a quotient is the difference of the logs:

ln y =1

4

(

ln(

x2 + 1)

− ln(

x2 − 1))

.

Differentiating implicitly with respect to x in the last equation yields

1

y· y′ =

1

4

(

1

x2 + 1· 2x − 1

x2 − 1· 2x)

=−x

(x2 + 1)(x2 − 1)

implying that

y′ = y · −x

(x2 + 1)(x2 − 1)

= −x(x2 + 1)−34 (x2 − 1)−

54 .

[1, Exercise 3.8.18, p. 249] “Differentiate the function

G(u) = ln

√

3u + 2

3u − 2.” (184)

Solution: In the Solutions Manual that the publisher provides instructors, theauthors write

G(u) =1

2[ln(3u + 2) − ln(3u − 2)] (185)

⇒G′(u) =

1

2

(

3

3u + 2− 3

3u − 2

)

=−6

9u2 − 4(186)


The value given for the derivative is correct; but the full solution is not alwayscorrect. Suppose that u < −2

3. Then both the numerator and the denominator of

3u + 2

3u − 2are negative: while the fraction is positive, and the square root function is

defined there, the logarithm function is not defined at either the numerator or thedenominator. One could modify the preceding incorrect derivation as follows:

When u >2

3, G(u) =

1

2[ln(3u + 2) − ln(3u − 2)] (187)

⇒G′(u) =

1

2

(

3

3u + 2− 3

3u − 2

)

=−6

9u2 − 4(188)

When u < −2

3, G(u) =

1

2[ln(−3u − 2) − ln(−3u + 2)] (189)

⇒G′(u) =

1

2

( −3

−3u − 2− −3

−3u + 2

)

=−6

9u2 − 4. (190)

Thus the expression given for the derivative was correct, even though the derivationwas incorrect when u < −3

2.

[1, Exercise 3.8.40, p. 249] To differentiate y = x1x .

Solution:

Using logarithmic differentiation:

y = x1x ⇒ ln y =

1

x· lnx =

ln x

x

⇒ 1

y· y′ =

1x· x − (ln x) · 1

x2

⇒ y′ = y · 1 − ln x

x2= x

1x · 1 − ln x

x2

Without using logarithmic differentiation:

y′ =d

dx

(

(

eln x)

1x

)

=d

dx

(

eln xx

)

=(

eln xx

)

· d

dx

(

ln x

x

)

=(

eln xx

)

·1x· x − (ln x) · 1

x2


=(

eln x)

1x ·

1x· x − (ln x) · 1

x2

= x1x ·

1

x· x − (ln x) · 1

x2

etc.



Release Date: Wednesday, October 25th, 2006subject to revision

The number e as a limit Let a be a real number. Then

limx→0

(1 + ax)1x = lim

x→0

(

eln(1+ax))

1x

= limx→0

(

eln(1+ax)

x

)

= elimx→0

ln(1+ax)x

by the continuity of the exponential function

= elimx→0

ln(1+ax)−ln(1+0(x))x−0

= ed

dx(ln(1+ax))|

x=0

= ea

1+ax |x=0

= ea .

In particular,limx→0

(1 + x)1x = e . (191)

We may wish to express this as a limit to +∞, by defining y = 1x:

limy→+∞

(

1 +a

y

)y

= ea . (192)

D.15.1 §3.9 Hyperbolic Functions

In this section we define functions that have a strong relationship to the 6 trigonometricfunctions; they are called the hyperbolic functions, and their names resemble those ofthe trigonometric functions, except for the letter “h” added at the end. Most of thereasons for this relationship cannot be proved in this course, however; in MATH 141,when we study [1, §10.1]76, we shall see that there is a certain analogy between two ofthese functions and the corresponding trigonometric functions.

Earlier in these notes I stated a theorem (D.2, p. ??) which asserted that any functioncan be decomposed into an “even” part and an “odd” part. The hyperbolic cosinefunction, cosh, is defined to be the “even” part of ex; formally, its definition is

cosh x =1

2

(

ex + e−x)

.

76see, for example [1, Exercise 17, p. 657]


By definition, it is even, (just like the cosine function), because

cosh x =1

2

(

ex + e−x)

=1

2

(

e−x + ex)

=1

2

(

e−x + e−(−x))

= cosh(−x) .

In a similar way, one can prove that the hyperbolic sine, defined by

sinh x =1

2

(

ex − e−x)

.

has the property thatsinh(−x) = − sinh x

which, by definition of odd , tells us that the function is odd, just like the sine function.The other four hyperbolic functions are defined by the same ratios used to define theanalogous trigonometric functions:

tanhx =sinh x

cosh x=

ex − e−x

ex + e−x

coth x =cosh x

sinh x=

ex + e−x

ex − e−x

sech x =1

cosh x=

2

ex + e−x

csch x =1

sinh x=

2

ex − e−x

The similarities with the trigonometric functions do not end here. We find that the hy-perbolic functions satisfy identities that resemble trigonometric identities. For example,you should be able to prove that

cosh2 x − sinh2 x = 1

sinh(x + y) = sinh x cosh y + cosh x sinh y

cosh(x + y) = cosh x cosh y + sinh x sinh y .

The similarities continue to the derivatives:

d

dxsinh x = cosh x


d

dxcosh x = sinh x

d

dxtanh x = sech2x

d

dxcoth x = −csch2x

d

dxsech x = −sech x · tanh x

d

dxcsch x = −csch x · coth x .

Inverse Hyperbolic Functions Hyperbolic functions sinh and tanh are invertible;as are csch and coth, but their domain excludes the origin. Functions cosh and sech areinvertible only if we confine ourselves to a portion of their domain, usually taken to bethe non-negative real numbers. The properties of the inverse functions can be discoveredusing similar methods to those we used to study the inverse trigonometric functions.However, unlike the inverse trigonometric functions, these inverse functions don’t enableus to express any new ideas, in that they are expressible in terms of known functionslike the natural logarithm.

Example D.24 Express the inverse hyperbolic cosecant in terms of familiar functions.Solution: Suppose that y = csch x; we wish to solve for x in terms of y.

y = csch x ⇔ sinhx =1

y

⇔ ex − e−x =2

y

⇔ e2x − 1 =2

y· ex

⇔ (ex)2 − 2

y· ex − 1 = 0

⇒ ex =1

y

(

1 ±√

1 + y2)

Case 1: Suppose y > 0. As exponentials are never negative, we can replace ± by +; thenwe take logarithms of both sides:

x = ln

(

1 +√

1 + y2

y

)

= ln(1 +√

y2 + 1) − ln y .

Since we are in the custom of calling the independent variable x, we will change thevariable’s name, to obtain

csch−1x = ln(1 +√

x2 + 1) − ln x .


Case 2: Suppose y < 0. Here the positivity of ex leads us to choosing the minus sign; weobtain, after reduction and the taking of logarithms,

x = ln

(

−y

1 +√

1 + y2

)

= − ln(1 +√

y2 + 1) + ln(−y) .

Since we are in the custom of calling the independent variable x, we will change thevariable’s name, to obtain

csch−1x = − ln(1 +√

x2 + 1) + ln(−x) .

This example demonstrates that anything we wish to “say” using the inverse hy-perbolic cosecant could be expressed in terms of polynomials and logarithms; the samephenomenon holds for the other inverse hyperbolic functions.

We sometimes use the alternative “arc” names: arcsinh, arccosh, . . .. The derivativesof these functions can be determined using implicit differentiation, similarly to what wedid for inverse trigonometric functions. The derivatives of the inverses may also be foundby explicit differentiation after we have found a formula for the inverse functions.

What should I remember? It is essential to remember the definitions of the functionsin terms of exponential functions. Then, even if you don’t remember some identities,you can reconstruct them by expressing the hyperbolic functions in those terms.

We will not expect students in this course to remember the explicit formulæ for theinverse hyperbolic functions, although you should know how to find those formulæ ifasked, using the method of the preceding example. You should also be comfortable withthe general question of finding the inverse function of any function whose graph satisfiesthe “Horizontal Line Test”, and know how to find its derivative.

3.9 Exercises

[1, Exercise 3.9.9, p. 255] “Prove that ex = cosh x+sinh x.” This is a decompositionof ex as a sum of an even function and an odd function.

[1, Exercise 3.9.20, p. 255] “If sinh x = 34, find the values of the other hyperbolic

functions at x.”

Solution: Since we know that the hyperbolic sine has an inverse, we could use thatfunction to determine x, then evaluate the other functions there.

x = arcsinh3

4


= ln

3

4+

√

(

3

4

)2

+ 1

= ln

(

3

4+

5

4

)

= ln 2

⇒ ex = 2

⇒ e−x =1

2.

It follows that cosh x = 54, tanh x = 3

5, etc.

Instead, we will proceed from first principles; in effect we are repeating some of thework that was done to determine a formula for the inverse function (which you arenot expected to memorize).

sinh x =3

4⇒ ex − e−x

2=

3

4⇒ 2(ex − e−x) = 3

⇒ 2 (ex)2 − 3ex − 2 = 0

⇒ ex =3 ±

√9 + 16

4=

3 ± 5

4= 2 or − 1

2,

But, of these 2 values, the second is extraneous, since an exponential cannot benegative. From the value of ex we can proceed as above to determine the values ofthe other hyperbolic functions.

[1, Exercise 3.9.38, p. 255] (not solved in the lecture) “Find the derivative: f(t) =ln(sinh t).”

Solution: We apply the Chain Rule:

f ′(t) =1

sinh t· cosh t

= coth t .

[1, Exercise 3.9.52, p. 256] Evaluate limx→∞

sinh x

ex.

Solution:

limx→∞

sinh x

ex= lim

x→∞

ex−e−x

2

ex

= limx→∞

1 − e−2x

2=

1

2.


[1, Exercise 3.9.54, p. 256] (not discussed in the lecture) “ If x = ln(sec θ + tan θ),show that sec θ = cosh x.”

Solution:

cosh x = cosh(ln(sec θ + tan θ))

=1

2

(

eln(sec θ+tan θ) + e− ln(sec θ+tan θ))

=1

2

(

eln(sec θ+tan θ) +1

eln(sec θ+tan θ)

)

=1

2

(

sec θ + tan θ +1

sec θ + tan θ

)

=sec2 θ + tan2 θ + 2 sec θ · tan θ + 1

2(sec θ + tan θ)

=2 sec2 θ + 2 sec θ tan θ

2(sec θ + tan θ)

= sec θ

D.15.2 §3.10 Related Rates.

In this section we consider problems where there are multiple functions, but usually oneindependent variable. Typically we need to relate the rates of change with respect tothis single independent variable. The problems are usually presented in a verbal form,and so the first step of solution requires translation into a mathematical statement towhich we may apply the tools we have available from the calculus. When a function isspoken about, it may be advisable to attempt to specify the domain; which will often —but not always — be the “maximal” domain. There are a number of “typical” questionsthat calculus books like to ask; of course, you can prepare yourself for those types byworking extensive examples. These “typical” questions involve the basic issues, but havebecome typical over the years because they entail a practical level of difficulty, or involveconcepts that students like you are able to identify with; however, the equations thatone obtains using the Chain Rule are normally “linear” — of a form where the variousderivatives are related by a sum of the form

function time derivative + function times derivative +...=0

and that means that it is normally possible to solve for the derivative that interests us interms of the other derivatives. Thus there is potentially a very wide range of problemsthat can be formulated, and, in theory, you should be able to solve all of them! Since it ispossible that you will be asked a problem different from any you have seen, so you shouldtry to understand the underlying issues. The textbook contains 5 worked examples of


different types, and we will not have time to work all of these types in the lectures. Donot assume that, because a particular type does not appear in the worked examples, ordoes not appear in a WeBWorK assignment, that it is not a reasonable question forMath 140.

TO BE CONTINUED



Release Date: Monday, October 30th, 2006, subject to further revision

D.16.1 §3.10 Related Rates (conclusion)

3.10 Exercises

[1, Exercises 3.10.3, 3.10.5, p. 260] Try these problems and look at the solutions inthe Solution Manual. These problems are already in purely mathematical language,so there is no verbiage to penetrate.

[1, Exercises 3.10.4, p. 260] If x2 + y2 = 25 anddy

dt= 6, find

dx

dtwhen y = 4.

Solution: Differentiating implicitly with respect to t in the equation of the circlegives

2x · dx

dt+ 2y · dy

dt= 0 .

Substituting y = 4 amd dydt

= 6 yields

2x · dx

dt+ 2(4)6 = 0 ,

so x · dx

dt= −24. However, the given value of y implies that x2 = 25 − 16 = 9,

so x = ±3. If follows that there are 2 different values of the derivative we seek:dx

dt= ∓8, according as x = ±3.

[1, Exercise 3.10.12, p. 260] “A spotlight on the ground shines on a wall 12 m away.If a man 2 m tall walks from the spotlight toward the wall, at a speed of 1.6 m/s,how fast is the length of his shadow on the building decreasing when he is 4 mfrom the building.”

Solution: It is usually good practice to draw a sketch of whatever situation is beingdescribed; (I shall not do that in these notes because the software I am using isvery cumbersome and time-consuming to create even simple figures).

Since I don’t have a figure to rely on, I am forced to do what you should alwaysdo: to define every symbol that is used precisely. I will place the spotlight at theorigin, and denote the foot of the wall by W ; I take the ground to be along thepositive x-axis, and so the coordinates of W are (12, 0). The top of the shadowwill be the point S, and I define its height as h, so its coordinates are (12, h). Let


the feet of the walking man be at A(x, 0), and the top of the man’s head B(x, 2).

The problem is to determine −dh

dt, when we know

dx

dt.

Triangles SOW and BOA are similar, and so the lengths of their sides must beproportional. This implies that

h

2=

12

x=

|OS||OB| .

From these equations, of which we are using only one, we infer that

h =24

x.

Differentiating all members of this equation with respect to t, we obtain

dh

dt=

d

dt

(

24

x

)

= −24

x2· dx

dt.

We now have the relationship we need in general; we apply it when |AW | = 4, i.e.,when x = 12 − 4 = 8.

dh

dt= −24

82· (1.6) = −0.6 m/s .

The negative sign of the answer indicates that the top of the shadow is movingin the negative direction, i.e., downwards. As the problem requested the rate atwhich the length is decreasing , that will be +0.6 m/s.

[1, Exercise 26, p. 261] “Two sides of a triangle have lengths 12 m and 15 m. Theangle between them is increasing at a rate of 2 degrees/minute. How fast is thelength of the third side increasing when the angle between the sides of fixed lengthis 60 degrees.”

Solution: To solve this problem you need to know a relationship between the angleand the lengths of the sides. This relationship has not been stated in the problem,but is in the elementary trigonometry that you are assumed to have brought to thecourse from your precalculus or earlier background. The simplest way to expressthis relationship is by the Law of Cosines , which is given in the textbook [1,Exercise 83, Appendix D, p. A33]:

If a triangle has sides with lengths a, b, c, and C is the angle between thesides with lengths a and b, then

c2 = a2 + b2 − 2ab cos C .


In the present problem, the relationship is given by

c2 = 122 + 152 − 2 · 12 · 15 cosC

so

cos C = −c2 − 144 − 225

360,

but we should precede the invocation of this equation by a statement definingwhat c and C are in our discussion, as we should never use a symbol that isn’tmentioned in the problem without first defining it. (This is a rule that we don’thave a chance to remind you about when you work problems on WeBWorK.)When we differentiate either of the last 2 equations with respect to time, we willobtain the relationship we need. Prior to doing that, if we wish to denote time bya symbol, we need to define it. So we begin with

Let t denote time, measured in seconds.77

Now, differentiating, say, the latter of the two equations, we obtain

− sin C · dC

dt= − 2c

360· dc

dt,

which we may solve to yield

dc

dt=

sin C2c360

· dC

dt,

which expresses the desired rate of change in terms of sin C,dC

dt, and c. But we

don’t know c yet; so we will have to apply the law of cosines to express c in termsof things we know, and substitute that formula to obtain

dc

dt=

sin C

2√

122 + 152 − 2 · 12 · 15 cosC360

· dC

dt

=sin π

3

2√

122+152−2·12·15 cos π3

360

· dC

dt

=30√

7· dC

dt

But there is still a snag! We are given that “The angle between (the sides) isincreasing at a rate of

77Strictly speaking, we should always indicate the orientation of the coordinate axis when we describea new variable; as time is normally measured from the past to the future, we will dispense with thatpedantic statement.


2 degrees/minute.”

Why can’t we simply substitute 2 as the value ofdC

dt? Because our convention is

that angles are always measured in radians! We have chosen to be consistent aboutthis, because the alternative would be to have 2 sets of trigonometric functions,with conversion tables. So we have to recall that

2◦ =2

360· 2π radians

sodC

dt=

π

90radians/minute; substituting this value, we obtain

dc

dt=

π

3√

7metres/minute.

While it is always possible for a nasty examiner to mischievously insert a trap in averbally stated problem, by an inconspicuous unit change, the distinction betweendegrees and radians is more serious than that, because the basic differentiationrules that we have developed depend on the use of radian measure. Be careful!

A word about notation. We often use letters like c and C to denote constants; but, when

you are in control, you are free to use any symbol for any purpose, and here both of these

letters represent variables, following an old convention in trigonometry that the angle is

named with a capital letter and the opposite side with the corresponding small letter.

Of course, it might be unwise to use a letter like d for a variable, since we might wish to

use it in representing a derivative.

[1, Exercise 3.10.38, p. 262] “The minute hand on a watch is 8 mm long, and thehour hand is 4 mm long. How fast is the distance between the tips of the two handschanging at 1 o’clock?”

Solution: This is the last problem in the set of exercises, and might be assumedfrom its location to be more challenging than the earlier problems. But, onceall the facts are expressed symbolically, it is only moderately difficult. It does,however, have one distinction: there are two different rates that are known, interms of which a third needs to be determined. There is quite a bit of work thatneeds to be done in managing definitions, etc., before the relevant mathematicalformulæ are available; once that has been done, the computations are simple andstraightforward.

Let’s first set up a coordinate system. It appears natural to locate the origin atthe centre of the watch, and to orient the watch so that 12 o’clock is on the y-axis,


that 3 o’clock is on the x-axis, and that the units are chosen to be mm, so that thetip of the minute hand rotates on the circle x2 +y2 = 82, and the hour hand on thecircle x2 + y2 = 42. The problem involves a rate with respect to time. I will denotetime by t, measured in hours. The hour hand on the clocks we use completes acycle in 12 hours; the minute hand in 1 hour. The problem is to determine therate of change of distance between the tips of the hands: let’s denote the positionof the tip of the hour hand by H , and of the minute hand by M ; since the motionof these hands is linked to a constant rate of rotation, it is convenient to namethe angles between the hands and the positive x-axis: call the angle made by HOh, measured in radians, and the angle made by MO m, also measured in radians;both h and m are functions of time — the hour hand moves at a constant rate,completing a revolution in 12 hours, while the minute hand, moving at a faster butconstant rate, completes a revolution in 1 hour. Note that we are not following theusual convention for clocks, where an angle is measured from the y-axis in whatwe would consider the negative direction — clockwise. Then we know that thecoordinates of H are (4 cosh, 4 sin h), and those of M are (8 cosm, 8 sin m). Thedistance between H and M is given by

|HM | =√

(4 cos h − 8 cosm)2 + (4 sin h − 8 sinm)2

However, we can use the Law of Cosines to express the distance in terms of thedifference of the respective angles:

|HM | =√

42 + 82 − 2 · 4 · 8 cos(h − m) .

It follows that

d|HM |dt

=1

2· 1√

42 + 82 − 2 · 4 · 8 cos(h − m)· d

dt

(

42 + 82 − 2 · 4 · 8 cos(h − m))

=1

2· 1√

42 + 82 − 2 · 4 · 8 cos(h − m)·(

−2 · 4 · 8 d

dt(cos(h − m))

)

=1

2· 1√

42 + 82 − 2 · 4 · 8 cos(h − m)·(

2 · 4 · 8 sin(h − m)d

dt(h − m)

)

=1

2· 1√

42 + 82 − 2 · 4 · 8 cos(h − m)·(

2 · 4 · 8 sin(h − m)

(

dh

dt− dm

dt

))

.

And what do we know? We know that

dh

dt= −2π

12,


dm

dt= −2π ,

h =π

3,

m =π

2.

Henced|HM |

dt= − 22π

3√

5 − 2√

3= −18.5896...

approximately. At 1 o’clock the distance between the tips of the hands is decreasingat the rate of 18.5896... mm/hour.

D.16.2 §3.11 Linear Approximations and Differentials

In this section we are interested in determining convenient approximations to a functionf . Our approximations will always be associated with a specific point in the domain off . For an approximation to be useful we need to know how close it is to the correctvalue; or, what is the worst that the error can be within a given interval around thepoint where the approximation is made.

0th degree approximations: Suppose that we approximate the function f near thepoint x = a in the domain by the constant function f(a). Then the error at x = a is0; but, without more information about the function, we can say little about the size ofthe error away from x = a. Nevertheless, approximation by a constant is often a useful,crude approximation.

1st degree approximations: A next stage of approximation is to approximate thefunction by one whose graph is still a straight line, but where we choose the line to bettermatch the graph of f . We can do this by taking the tangent line to the graph of f atx = a. This type of approximation is called a linear approximation, or a tangent lineapproximation, or the linearization of f . A linear function is one of the form kx + ℓ,78

and what we are doing is to choose the constants k and ℓ to “fit” the curve best. Tomake the approximating curve pass through the point (a, f(a)), we impose the conditionthat

ka + ℓ = f(a) ,

but this single equation is not enough to determine both of the constants. For a tangentline approximation the second constraint we impose is that the approximating line should

78The word linear is sometimes used in a more restrictive way, to describe a function of the form kx,i.e., where ℓ = 0; when an author uses the word linear in that sense, then she usually calls a functionkx + ℓ affine. You could meet this version of the terminology in your Linear Algebra course.


“touch” the curve at the point (a, f(a)); we require that its slope be the same as theslope of the tangent line; in fact, we are requiring that the approximating line be thetangent line. The constraint is

k = f ′(a) ,

which implies that ℓ = f(a) − af ′(a), and that the approximating function, which wemay denote by L(x) is

L(x) = f(a) + f ′(a) · (x − a) .

Higher order approximations: Could we improve the approximation by finding acurve that fits the curve better? If f is differentiable more than once, i.e. if f ′′ exists,we can approximate better with a polynomial of degree 2, i.e. a function of the formkx2 + ℓx + m. We will not pursue this possibility at this time, but may return to it atthe end of Math 141.

Why do we need approximations, anyhow? Sometimes an approximation is suchthat, when we replace the actual function by the approximation in an important equation,the problem changes from one that is very difficult to solve to one that is amenable. Yourtextbook gives examples of this in the subsection.

Applications to Physics One interesting example is in studying the motion of a“simple” pendulum. After determining the differential equation that describes the mo-tion of the pendulum, we simplify it by replacing a term sin θ by θ, where θ is a smallangle, measured in radians. In terms of linear approximation we are replacing the graphof the sine curve near θ = 0 by the graph of its tangent line: the slope is cos 0 = 1, sothe linearization is

sin θ ≈ sin 0 + cos 0 · θ = θ .

(Remember that the angle θ must be given in radians; if we wish to work in degrees,

then the limit ofsin x

xwill not be 1, and L(x) will take a different form.)

Differentials:

Increments: The notation here can be a little confusing. The prefixing of thesymbol for a variable with a Greek upper case (capital) Delta (∆), normally representsthe increment in the variable; when x is the independent variable — the variable overwhich we have control, the variable which is restricted only by the domain of the function— we may denote any change in its value by ∆x. In the days before the language offunctions was formalized, mathematicians used to talk of a dependent variable y, where


we today would speak of a function y(x). Then the change in y(x) corresponding to achange ∆x in x might be denoted by ∆y; in our notation, we would be defining

∆(y) = y(x + ∆x) − y(x) .

We give meanings to dx and dy, so that, with these definitions, the fractiondy

dxwill

actually be equal to f ′(x). We define dx to simply be any real number, although we arenot going to use this formalism for “large” numbers; the interpretation we wish to makeis that dx is to denote the increment: thus we are taking dx to be ∆x. Then we definedy to be a function of x and of dx also; more precisely, we define

dy = f ′ · dx

which depends on two variables — on x, and also on the value we assign to dx. We willnot dwell on this issue in Math 140, because the concept of functions of more than onevariable is not introduced in our calculus sequence until Calculus 3 (Math 222). HithertoI have been saying that the derivative “behaves like a fraction”; with these definitionsyou are free to actually consider it a fraction; it is not simply a convenient notation.

Using ingenuity to improve an approximation. Sometimes one can get “better”results with a mathematical tool by applying it more carefully.

Example D.25 Suppose we wish to approximate cosx for −π3

≤ x ≤ π3. A naive

application of the present theory would approximate with the linear function

cos 0 + (− sin 0)(x − 0) = cos 0 = 1

so the linear approximation is the same as the 0th degree approximation: we would beapproximating the graph of the cosine function by the horizontal line which is tangent

to the cosine graph at (0, 1). Now recall that we have proved that limx→0

cos x − 1

x= 0.

Let’s try to approximate the function cos x−1x

linearly. Since 0 is not in the domain of thisfunction, but since its limit as x → 0 exists, the discontinuity is removable. We define

f(x) =

cos x − 1

xif x 6= 0

0 if x = 0.

This function is continuous at all points x. In order to find a first order approximation,we will need its derivative. This is no problem for x 6= 0, since

f ′(x) =−(sin x)x − (cos x − 1)1

x2= −sin x

x+

1 − cos x

x2.


But, to approximate around x = 0, we will need to know f ′(0), and we don’t even knowwhether f is differentiable there. Consider the ratio

f(x) − f(0)

x − 0=

cos x − 1

x2=

1 − 2 sin2 x2− 1

x2= −1

2·(

sin x2

x2

)2

→ −1

2

as x → 0. Thus a first approximation of f around x = 0 is

f(x) ≈ 0 + x ·(

−1

2

)

which implies that

cos x ≈ 1 − x2

2

This is a much better approximation to the cosine function. In fact, we have found a2nd degree approximation. Indeed, we could continue this procedure through as manysteps as we like, and obtain the Taylor polynomial which approximate the cosine functionaround x = 0. This topic appears in Calculus 3, unless we succeed in introducing it atthe very end of Calculus 2.

Be careful when you choose variables! [1, Review Exercise 99, p. 273] “A windowhas the shape of a square surmounted by a semicircle. The base of the window is measuredas having width 60 cm, with a possible error in measurement of 0.1 cm. Use differentialsto estimate the maximum error possible in computing the error of the window.”Solution: We need some named variables. Let’s denote the side of the square by x. Then

the semicircle has radiusx

2. This makes the whole exercise involve fractions. So let’s

go back to the drawing board, and name the side of the square 2X; the radius of thesemicircle is X, and the area of the entire window, which we will denote by A(X) is

A(X) = (2X)2 +1

2· πX2 .

(Note that the remedy to eliminate fractions didn’t work — we would have to renamethe variable again if we wish to do that.) Differentiating the last formula yields

dA = 2(2X)2 dX + πX dX = (8 + π)X dX .

When dX = ±0.1, and X = 30, dA = . . .. WAIT! It was dx that was equal to ±0.1,dX = ±0.05; then dA = ±(8 + π)30(0.05) = ±

(

12 + 3π2

)

.


3.11 Exercises

[1, Exercise 32, p. 268] Use differentials (or, equivalently, a linear approximation) toestimate

√99.8.

Solution: Let f(x) =√

x; then f ′(x) =1

2√

x. We are going to approximate by a

tangent line. What should be the contact point of the tangent? If we take the point(a, f(a)) = (0, 0), the tangent line is vertical: it never crosses the line x = 99.8,and cannot be used to approximate this function at this point; or, alternatively, wecan say that the approximation is infinite. If we take the point (1, 1) on the graphof the function, we obtain a very, very bad approximation:

L(x) = f(1) + f ′(1) · (99.8 − 1) = 1 + 49.9 = 50.9 .

But, if we take as the contact point the closest convenient point to 99.8, which is(a, f(a)) = (100, 10), then the linearization is

L(x) = f(100) + f ′(100) · (−0.02) = 10 +1

20(−0.02) = 9.99 .

Laboratory Project: Taylor Polynomials (This is a topic which may be reachedat the end of Math 141; it is not in the syllabus of Math 140.)

D.16.3 3 Review

True-False Quiz

[1, True/False Exercise 10, p. 271] A discussion of this problem appears in thesenotes on page 2163 et seq.

Exercises

[1, Review Exercise 96, p. 273] Find the linear approximation to f(x) =√

25 − x2

near 3.

Solution: f ′(x) = 12(25 − x2)

12 · (−2x) = − x√

25 − x2; thus f ′(3) = −3

4. Since

f(3) =√

16 = 4, the linearization of f near x = 3 is

L(x) = f(3) + f ′(3) · (x − 3)

= 4 − 3

4· (x − 3) =

25 − 3x

4


[1, True/False Exercise 12, p. 272] “Determine whether the statement is true orfalse. If it is true, explain why. If it is false, explain why, or give an examplethat disproves the statement:

d2y

dx2

?=

(

dy

dx

)2

.” (193)

More precisely, the statement is asserting that the equality holds for all functionsf having derivatives of the given orders.

Solution: This statement is false, and most functions you know could serve ascounterexamples. For example, if f(x) = a + bx, then the second derivative is0, but the square of the first derivative is b2; so any linear function of the formf(x) = a + bx (b 6= 0) is a counterexample. We couldn’t take a constant function,as such functions do have the given property.

A more interesting question might be to characterize all counterexamples. Andit’s not hard to do. With a property we will meet in the next chapter, we can do

this! For convenience let’s denotedy

dxby v. Then (193) could be rewritten as

dv

dx= v2 .

If v 6= 0, this can be rewritten as

1

v2· dv

dx= 1 ,

or − d

dx

1

v= 1 .

We shall see in the next chapter that the most general function having 1 as aderivative is a linear function x + C, where C is any constant (real number). Thusthe most general solution of the “differential equation” is

v = − 1

x + C⇔ dy

dx= − 1

x + C.

But, again by the theory of the next chapter, the most general function having

− 1

x + Cas its derivative is − ln |x + C| + K, where K is another constant. So we

see that the most general non-zero function that possesses property (193) is

y = − ln |x + C| + K


where C and K are constants. We have seen earlier that the function 0 also hasthe property.

You are not expected to solve general differential equations per se in this course;this last paragraph is presented to show you an application of the theory we willbe meeting in [1, §4.2]. If you wish to read more about Differential Equations, youcould look at [1, Chapter 9, Differential Equations], which is not on the syllabus ofMATH 140 or MATH 141. The present application could, however, be formulatedas looking for an “antiderivative” of certain functions, and we shall be meetingthat topic later in this course in [1, §4.10].

Operations that “commute”. We say that two operations commute if the order inwhich they are performed does not affect the result. For example, on the set of functionsof a real variable x

• the operations of multiplication by a constant c and taking the limit as x → acommute (provided all the limits exist); this is the Constant Multiple Law;

• the operations of addition and taking the limit commute provided the limits exist;this is the Sum Law;

• the operations of multiplication and taking the limit commute provided the limitsexist; this is the Product Law;

• the operations of multiplication by a constant and differentiation commute (pro-vided all derivatives exist); this could be called the “Constant Multiple Rule” ofDifferentiation;

• the operations of addition and differentiation commute provided all derivativesexist; this is the Sum Rule of Differentiation.

But there are situations where operations do not commute; one that we have just metis the Product Rule of Differentiation, which gives a formula for the derivative of theproduct of two functions, but where you can see that the derivative of the product isnot usually equal to the product of the derivatives. The following problem investigatesa question about whether two given operations commute.

[1, True/False Exercise 10, p. 271]

“Determine whether the statement is true or false. If it is true, explain why. If it is false,explain why or give an example that disproves the statement:

d

dx

∣

∣x2 + x∣

∣ = |2x + 1| .”


More generally, the question could have asked: “Do the operations of taking the absolutevalue and differentiation commute; i.e., is it always true that

d

dx|f(x)| =

∣

∣

∣

∣

d

dxf(x)

∣

∣

∣

∣

?”

Solution: Your first reaction to the statement should be one of suspicion: while we haveseen that some operations “commute” with differentiation — operations like addition,multiplication by a constant — there is no good reason why a function that we alreadyknow to be “unruly” should commute with differentiation. But, until we investigate apossible proof, we can’t dismiss the possibility. The best way to approach this problemis to determine the value of the left and right sides of the proposed equation, and tocompare them carefully.

∣

∣x2 + x∣

∣ =

{

x(x + 1) if x(x + 1) ≥ 0−x(x + 1) if x(x + 1) < 0

=

{

x(x + 1) if x ≤ −1 or x ≥ 0−x(x + 1) if −1 < x < 0

Now we differentiate:

d

dx

∣

∣x2 + x∣

∣ =

d

dx[x(x + 1)] if x < −1 or x > 0

d

dx[−x(x + 1)] if −1 < x < 0

=

d

dx[x2 + x] if x < −1 or x > 0

d

dx[−x2 − x] if −1 < x < 0

=

{

2x + 1 if x < −1 or x > 0−2x − 1 if −1 < x < 0

(We had to exclude the points −1 and 0 because the function fails to be differentiableat those points.) We see both 2x + 1 and −(2x + 1) appearing in our formula, and it istempting to jump to the conclusion that the claimed equality holds. But let’s determinethe value of the right side of the proposed equation without using absolute value signs:

|2x + 1| =

{

2x + 1 if 2x + 1 ≥ 0−2x − 1 if 2x + 1 < 0

=

{

2x + 1 if x ≥ −12

−2x − 1 if x < −12


Now we see that significant points for analyzing this problem are −1, −12, and 0. These

divide the real line into 4 intervals. By examining the values of the two sides of the equa-tion for points in these four intervals, we should be able to find a counterexample to theclaimed equation. It’s hard work! Note first that, if x ≥ 0, the claimed equation holds;similarly, it can be seen to hold for −1 < x < −1

2. But we claim it fails throughout the

intervals (−∞,−1) and(

−12, 0)

. We don’t need to prove that much for a counterexample— we just need to provide one example. For example, take the point x = −1

4. Around

that point on the real line |x2 + x| = −x2 − x, so its derivative is −2x − 1: at x = −14

this is 12− 1 = −1

2. But, at x = −1

4, |2x + 1| is 1

2. (In fact, it is enough to observe that,

being an absolute value, it cannot be negative.) Thus we have disproved the claim.Most of this “solution” is superfluous, as I was showing you how I arrived at a

counterexample. All that is needed in the proof is something like the following:

The claim is false, because the derivative of x2+x at x = −14

is −2(

−14

)

−1 =−1

2, which is negative, but

∣

∣2(−14) + 1

∣

∣ is positive.

Another way to “discover” a counterexample in this case would be to sketch a graph ofleft and right sides. Of course, the graph has no place in the proof itself, but it could beused to discover the problem. The graph of y = |x2 + x| can be obtained from the graphof y = x2 + x by “folding” the portion that lies below the x-axis upward so that it nowappears above the x-axis. You can then see that there are 2 intervals where the functionis increasing, and two intervals where it is decreasing; but the graph of |2x + 1| has justone interval where the function is decreasing, and one interval where it is increasing.This shows that something is wrong, etc.


D.17 Supplementary Notes for the Lecture of November 1st,2006

Release Date: Wednesday, November 1st, 2006, subject to further revision

Textbook Chapter 4. APPLICATIONS OFDIFFERENTIATION.

D.17.1 §4.1 Maximum and Minimum Values

An “existence” theorem. A maximum point of a function is a point c in the domainof the function at which the function has a value that is not less than its value in somedesignated larger set which contains c. In particular, a global maximum point or absolutemaximum point is a point where the function has a value that is not less than its valueanywhere else in the domain. Note that we are permitting that the same value could beattained elsewhere, i.e. that there could be more than one maximum point. A minimumpoint, global minimum point or absolute minimum point are defined analogously —change less to greater throughout the definitions. We use the word extremum to mean“either a maximum or a minimum”; we use the Latin form of plural for maximum,minimum, extremum: maxima, minima, extrema.

Local or relative maximum, minimum, or extremum are defined analogously to thepreceding, except that we require that the designated larger set be an interval surround-ing the point c, where the interval could be as small as you like, so long as its ends havea positive distance from c. The values that the function assumes at extrema (also calledextreme points) are called the extreme values or the maximum values , etc. Note theterminology: a maximum or minimum must be attained. For example, we will say that

the function f defined by f(x) =1

1 + x2has no global minimum (value): if we make

x large enough, we can cause the value of f to be small than any positive number wetake. Why, then, don’t we say that the minimum is 0? Because the function is neverequal to 0, even though it gets as close to 0 as we wish. Fortunately we have79 a theoremthat ensures the existence of global extrema under certain conditions that are often notdifficult to satisfy:

Theorem D.26 (Extreme Value Theorem) [1, p. 281] A function f which is con-tinuous on a finite, closed interval [a, b] attains both a global maximum and a globalminimum at point(s) of the interval.

(Remember that continuity on a closed interval [a, b] entails continuity at every pointexcept a and b, as well as continuity from the right at a, and from the left at b.) This

79The proof of this theorem is beyond Math 140; it could be studied, for example, in Math 242.


theorem is “best possible”, in the sense that the conditions of continuity and the closedinterval are needed for it to be true. If the function is permitted to have a discontinuity,then it is possible to design a function that has no maximum or minimum; similarly, ifthe interval is permitted to be open at one or other of its ends, there again we can design

a function that lacks extrema. The domain of the function1

1 + x2discussed above is

not a finite closed interval, so the theorem does not apply there. Students should tryto construct “counterexamples” to extensions of the theorem, i.e. examples where somepart of the hypotheses is not available, and where the function fails to have extrema.

Note that the Extreme Value Theorem is only an existence theorem — it asserts thatextreme values exist, but gives no information about how to find them.

A “necessary condition” for an extremum. When a function is more than con-tinuous on an interval, we may have more information available that will enable us todetermine the extrema. For example, we know that differentiability is a stronger propertythan continuity. What if a function is known to be differentiable?

Theorem D.27 (Fermat [1, p. 282]) 80If g is differentiable at c and has a local ex-tremum there, then g′(c) = 0.

This theorem provides a condition that is necessary for the existence of an extremumat a point where g is differentiable; but it may happen that the condition is satisfied,and yet g still has no extremum at the point. That is, the condition is necessary for anextremum, but is not sufficient to ensure that the point is an extremum. As an example,consider the function g(x) = x3, and take as the domain either the whole real line R orany convenient interval around 0. To the right of 0 the function is positive, and to theleft of 0 it is negative. Thus it cannot have an extremum at x = 0, where g(x) = 0;but the derivative is equal to 0 at x = 0. However, the theorem does provide us a wayof narrowing our search for local extrema: just determine all the points in the domainwhere the derivative is 0, or where there is no derivative, and the local extreme points— if any — will be among them. Remember that a function which is being consideredon a closed interval [a, b] cannot be considered to be differentiable at x = a or x = b,because that property requires consideration of a limit on both sides of the point, andthe function is not defined on one of those sides; thus the theorem can never exclude thepossibility that the end point of a closed interval is an extremum. For that reason themethod we are developing will require explicit, separate consideration of the end pointsof intervals in the domain as candidates for extrema.

80This result is not always linked to P. Fermat; and there are other well known results that are oftencalled “Fermat’s Theorem”.


Critical Numbers or Critical Points. For ease in formulating a method for findingextrema, we define a critical number or critical point of the domain of f to be a numberor “point” c such that either

• f fails to be differentiable at c; or

• f is differentiable at c, and f ′(c) = 0.

The value of the function at a critical point is a critical value. Then Theorem D.27 maybe reformulated as follows:

Theorem D.28 [1, p. 283] The local extrema of f can occur only at critical points ofthe domain.

Is a global extremum always a local extremum? If a global extremum of f occursat a point a which is in the “interior” of the domain of f , in the sense that a is containedinside an interval (a−k, a+k) in the domain of f (where k is some positive number), thenthis global extremum is also a local extremum. But if, for example, the extremum occursat the end point of an interval [a, b] of the domain, where the domain does not containthe points immediately to the left of a, then a cannot be called a local extremum.81 Wecan formalize the situation for a function being considered on one closed interval [a, b]:

Theorem D.29 (The Closed Interval Method) [1, p. 283] The following procedureserves to narrow the search for absolute extrema of a continuous function f on a closedinterval [a, b]:

1. Determine all critical points of f in (a, b), and the critical values of f at each ofthese critical points.

2. Determine the values of f at the end points, viz., f(a) and f(b).

3. The global maximum of f on [a, b] will be the maximum of the set of values de-termined in the preceding two steps; analogously, the global minimum will be theminimum of the set.

It can happen that the preceding method is still not finite, as the set of critical points.

81There are other ways of defining the use of the term local , and some authors may follow a practiceunder which the end points can still be considered local extrema, if they are “one-sided” local extrema.These variations in definition will not affect you in this course, but you may see slightly differentformulations if you read other textbooks.


Exercise D.7 The function given by

f(x) =

{

x sin 1x

when x 6= 00 when x = 0

when considered in the interval −1 ≤ x ≤ 1 has infinitely many critical points. (Whatare they?)Solution: The derivative vanishes at all points of the form 1

x= nπ, where n is any integer,

i.e., where x = 1nπ

. In addition there is one point — just one — where the function isnot differentiable. That point is x = 0, since the limit of the slope of the line segmentjoining (0, f(0)) to a nearby point on the curve does not exist, as the slopes oscillatebetween ±1.

Are end-points of an interval domain critical? If a function f is defined onlyon [a, b], it can’t have a derivative at either a or b. Technically, such points satisfy thetextbook’s definition of critical . Some authors call such points critical, and some do not.You will not be expected to make such a decision; in any case, Part 1 of Theorem D.29,speaks only of critical points in (a, b); then Part 2 requires you consider the end pointswithout reference to any question about differentiability. To summarize it doesn’t matterwhether you wish to consider end-points as critical, you will not need to use that conceptat end-points.

Example D.30 Find the global extrema of the function |x| on the interval −1 ≤ x ≤ 2.Solution: This function has no points where the derivative vanishes, but it does haveone point where it is not differentiable; that is x = 0, and is, by definition, a criticalpoint. The global extrema will occur among the points −1, 0, 2: the global minimum isat x = 0, of value 0; and the global maximum is at x = 2, of value 2.

4.1 Exercises

[1, Exercise 4.1.50, p. 287] “Find the absolute (=global) maximum and absolute(=global) minimum values of f on the given interval”: f(x) = x3 − 6x2 + 9x + 2,on the interval [−1, 4]

Solution: The function is a polynomial, which is differentiable everywhere. Thusthe only critical numbers possessed by f1 are points in the domain where thederivative vanishes. f(x) = 3x2 − 12x + 9 = 3(x − 3)(x − 1). Thus the criticalpoints (=critical numbers) are 3 and 1. Following the “Closed Interval Method”,we have to consider the function’s values at the critical points and at the end pointsof the interval; we calculate that

f(3) = 27 − 54 + 27 + 2 = 2


f(1) = 1 − 6 + 9 + 2 = 6

f(−1) = −1 − 6 − 9 + 2 = −14

f(4) = 64 − 96 + 36 + 2 = 6

We see that the maximum value is 6, and it is attained at 2 maximum points, 1 and4; the minimum value is −14, and it is attained uniquely at the minimum point−1. Point 3 is neither a global maximum, nor a global minimum. (Using methodswe have still to develop, we could prove that it is [the only] local minimum.)

[1, Exercise 4.1.54, p. 287] Same instructions as the preceding, with function f(x) =x2 − 4

x2 + 4and interval [−4, 4].

Solution: Before applying the Closed Interval Method, I will simplify the function,so that its behavior will become obvious.

f(x) =x2 − 4

x2 + 4=

x2 + 4 − 8

x2 + 4= 1 − 8

x2 + 4.

The subtracted term has numerator and denominator which are both positive.Since the numerator is constant, the fraction increases as the denominator de-creased, and the denominator cannot be less than 02 + 4 = 4, a value which isattained at x = 0. Thus we see — without even appealing to the calculus — thatf attains its minimum when x = 0. But the denominator has no maximum, so thesubtracted term may be made arbitrarily small, This shows that f approaches 1from below, but never attains this value: f does not have a maximum value. Butwait — we were not asked to consider the function over the maximal domain, onlyover the domain −4 ≤ x ≤ 4. The same reasoning we have used shows that thevalue of the function increases as we move from 0 to either +4 or −4, where its

value is, in either case, 1 − 8

20=

3

5.

Now let’s use the calculus.

f ′(x) = 0 − 8

(x2 + 4)2 · 2x =16x

(x2 + 4)2

for all x in −4 < x < 4; this function is 0 only when x = 0, so x = 0 is a criticalnumber of f . Because we haven’t been authorized to consider f to the left of −4or to the right of +4, we really can’t claim to know f ′(±4), so, technically, ±4could be considered to be critical numbers of f . Some textbook authors definetheir terms slightly differently from yours, and, under their definitions, the endpoints ±4 might not be critical points. It doesn’t matter which definitions we use,however, because we are going to meet the end-points in the next step of the Closed


Interval Method anyhow, as this step requires us to determine the values of the

function at the end-points of the interval of definition. As seen earlier, f(±4) =3

5.

To complete the application of the Closed Interval Method we consider the valuesof the function at the three points — −4, 0, +4:

x −4 0 +4

f(x) 35

−1 35

and conclude that the absolute minimum of −1 on the interval is attained at x = 0,and the absolute maximum of 3

5on the interval is attained at x = ±4.

[1, Exercise 4.1.58, p. 287] Same instructions as the preceding, with function f(x) =x − 2 cosx and interval [−π, π].

Solution: This function, being a sum of a polynomial and a cosine, is differentiableeverywhere. f ′(x) = 1 + 2 sinx, which vanishes when the sine is −1

2, i.e. at x =

(

2n − 16

)

π, and at x =(

2n − 56

)

π, n being any integer. In the interval given, thereare only the points −π

6; −5π

6.

f(

−π

6

)

= −π

6− 2 ·

√3

2

f

(

−5π

6

)

= −5π

6+ 2 ·

√3

2

We must also check the end-points of the prescribed domain.82

f(−π) = −π + 2

f(π) = π + 2

Of the four values, we find, using a very crude approximation for π, that the largestis π+2, attained when x = π, and the minimum is −π

6−√

3, attained when x = −π6.

[1, Exercise 4.1.62, p. 287] (not discussed in the lecture) Same instructions, f(x) =e−x − e−2x, interval =[0, 1].

Solution: f ′(x) = −e−x + 2e−2x = e−x (−1 + 2e−x). In this product, the firstfactor, e−x cannot be zero, as no exponential is zero. Hence f ′ can vanish only

when e−x = 12, i.e. when x = ln 2; we note that f(ln 2) = 1

2−(

12

)2= 1

4. Checking

82Note that, again, this is not the “natural” domain of the function, but a restricted domain prescribedby the author of the problem.


the function at the end points, we find that f(0) = 1 − 1 = 0, and f(1) = 1e− 1

e2 .Comparing the values, again using a very crude approximation for e, we find thatthe global maximum is at x = ln 2, with value 1

4, and the global minimum is at 0,

with value 0.

[1, Exercise 4.1.68(b), p. 287] (not discussed in the lecture) “Use calculus to find

the exact maximum and minimum values: f(x) =cos x

2 + sin x.”

Solution: (In the (a) part of this problem the textbook asks you to graph thefunction; but that graphing is not at all necessary to finding the extrema.)

f(x) =− sin x · (2 + sin x) − cos x · cos x

(2 + sin x)2= − 2 sin x + 1

(2 + sin x)2

which vanishes when sin x = −12, i.e., when x = −π6

+ 2mπ and x = 7π6

+ 2nπ,where m and n are any integers. We are confined by the conditions of the problemto the closed interval 0 ≤ x ≤ 2π, so the critical points are

−π

6+ 2π =

11π

6

and7π

6. We must also consider the function at the end-points of the given closed

interval. We tabulate the values:

x 0 7π6

11π6

2π

end point critical point critical point end pointf(x) 1

2− 1√

31√3

12

Since the function√

x is an increasing function,

3 < 4 ⇒√

3 <√

4 = 2 ⇒ 1√3

>1

2

so the absolute maximum on the given interval is 1√3, and the absolute minimum is

− 1√3; in fact, because the function is periodic with period 2π, these are the absolute

extrema for the natural domain of the function, R.

[1, Exercise 4.1.74, p. 288] (not discussed in the lecture) “Show that 5 is a criticalnumber of the function g(x) = 2 + (x − 5)3, but g does not have a local extremevalue at 5.”

Solution: g(x) = 3(x − 5)2, which vanishes only at x = 5; thus x = 5 is the onlycritical point. This function resembles the function x3 considered on page 2167.


If we take x < 5, the function value is less than 2, while, for x > 5, g(x) > 2.It follows that 5 cannot be either a local maximum or a local minimum, and, afortiori it cannot be a global maximum or a global minimum. (The only way inwhich 5 could be a local extremum is if we were to consider the function on aninterval ending at 5.)

For the final examination, do we need to know how to prove theorems? Math140 is mainly a “problem-oriented” course — we usually don’t expect students to mem-orize proofs of theorems; however if a theorem can be formulated as a “reasonable”problem, then it could be posed as such, and you could be expected to be able to solveit.

While you aren’t expected to memorize proofs, you are certainly expected to know

• the statements of important theorems

• precisely what the theorem assumes, and what it applies to

• how and when to use the theorem

You could certainly be presented with questions which test that knowledge. So, forthe Mean Value Theorem, discussed below, [1, Exercise 4.2.15, p. 295] is a perfectlyreasonable question for students in Math 140. Few, if any, questions that probe your un-derstanding of a theorem are to be found on WeBWorK. We have adopted WeBWorKbecause of its great value in monitoring your progress in solving numerical problems, butit does not “drive” the course: the best aid for learning the material in the course is theexercises in the book, monitored with the Student Solutions Manual. Excellent problemsfor testing your understanding of the concepts are in the Review exercises entitled “Con-cept Check” and “True-False Quiz” at the end of each chapter of the textbook. However,these problems are not necessarily formulated in a form that would be practical for anexamination.

Summary of §4.1 In §4.1 you were introduced to the concepts of maximum, min-imum, and extremum, and to the terms global (=absolute) and local (=relative); theExtreme Value Theorem asserts that a function which is continuous on a closed intervalhas a global maximum and a global minimum in the interval. While our goal is often tofind the global extremum, we saw that any global extremum not at an end-point is alsoa local extremum; we were introduced to Fermat’s Theorem, which provides a necessarycondition for a local extremum: At a local extremum either the derivative does not exist,or it does exist and is 0. This theorem sees application in the Closed Interval Methodfor finding the global extrema of a continuous function on a closed interval.


D.18 Supplementary Notes for the Lecture of November 6th,2006

Release Date: Monday, November 6th, 2006, subject to further revision

D.18.1 §4.2 The Mean Value Theorem

We interrupt our study of extrema to consider two “existence” theorems, whose impor-tance is often misunderstood because of a type of problems that often appear in calculustextbooks. The theorems assert the existence of a number with a particular property,but make no attempt to locate that number; problems in other textbooks that ask youto find the numbers in question are misleading, since, in applications of these theorems,we are interested not in the numbers themselves, but only in the “worst possible” valuesthey may have. Most of the problems in your textbook do adopt a “correct” stance; but,for example, [1, Exercises 4.2.1-4.2.4, p. 295] and the part of [1, Exercises 4.2.11-4.2.14,p. 295] which states “Then find all numbers c that satisfy the conclusion of the MeanValue Theorem” are alien to the spirit of the theorem, unless one interprets them asverifying that there is at least one point with the desired property.

The theorems There are two main theorems in this section, and, while it appearsfrom the enunciations that one of them is “stronger” than the other, they are equivalent,in the sense that either of them can be used to prove the other.

Theorem D.31 (Rolle’s Theorem) Let [a, b] be a given closed interval. Let f be afunction such that

1. f is continuous on [a, b];

2. f is differentiable on (a, b); and

3. f(a) = f(b).

Then there exists a number c such that a < c < b and f ′(c) = 0.

Theorem D.32 (Mean Value Theorem) Let [a, b] be a given closed interval. Let fbe a function such that

1. f is continuous on [a, b];

2. f is differentiable on (a, b).

Then there exists a number c such that a < c < b and f ′(c) =f(b) − f(a)

b − a.


It appears that the Mean Value Theorem should be “stronger” than Rolle’s Theorem,since Rolle’s is the special case of the Mean Value Theorem where f(a) = f(b). But, infact, the theorems are equivalent, since the Mean Value Theorem can be proved fromRolle’s Theorem.

Geometric interpretation of these theorems The Mean Value Theorem says that,if we consider the secant to the graph of f obtained by joining the points (b, f(b)), and(a, f(a)), there will always be at least one point c between a and b where the tangent tothe graph of f is parallel to that secant.

Doesn’t one hypothesis imply another? Since you know that differentiability im-plies continuity, you may ask why we need both 1. and 2. Indeed, the portion of 1.that speaks of continuity at the points of (a, b) is totally redundant: continuity at thesepoints is implied by differentiability. Remember what continuity f on [a, b] means83:

• continuity of f on the interval (a, b),

• continuity from the right at x = a, and

• continuity from the left at x = b.

Thus we could weaken the first hypothesis of the theorem, to

1′. f is continuous from the right at x = a, and continuous from the left at x = b.

The conclusion of Rolle’s Theorem. If the hypotheses of the theorem are satisfied,the theorem asserts the existence of a point (c, f(c)) on the graph of f such that a < c < b,where the tangent is horizontal.

Proof of the Mean Value Theorem from Rolle’s Theorem Given a function fwhich satisfies the conditions of the Mean Value Theorem, we will consider, instead, thefunction h defined by

h(x) = f(x) − f(a) − f(b) − f(a)

b − a· (x − a) .

The continuity and differentiability properties of f imply that h has the same properties,so h satisfies the first 2 conditions of Rolle’s Theorem. Moreover, we can see that h(a) =

83This is a definition, not a theorem.


h(b), so the 3rd condition of Rolle’s Theorem is also satisfied. Hence the conclusion ofRolle’s Theorem holds: there exists a point c such that a < c < b and

h′(c) = f ′(c) − f(b) − f(a)

b − a,

is zero, and so f ′(c) has the claimed property.

Corollary D.33 (to the Mean Value Theorem) Rolle’s Theorem is the special casewhere f(a) = f(b).

Corollary D.34 (to the Mean Value Theorem) If f ′(x) = 0 for all x in an interval(a, b), then f is constant on the interval.

Proof: If we take any two distinct points x1, x2 in the interval, we know that betweenthem there is a point c such that f(x2)−f(x1) = f ′(c) ·(x2−x1) = 0, since the derivativemust be 0 at c. Since f ′(c) = 0, f(x2) = f(x1), proving that, at any point in the interval,the function has the same value as at any other point; or, equivalently, that the functionhas the same value at every point.

Corollary D.35 (to the Mean Value Theorem) If f ′(x) = g′(x) for every point xin an interval (a, b), then there exists a constant k such that, for all x in the interval(a, b),

g(x) = f(x) + k .

Proof: To the function h defined by h(x) = g(x) − f(x) we apply the corollary immedi-ately preceding. Since h′(x) = g′(x) − f ′(x) = 0 for all x, there exists a constant k suchthat h(x) = k for all x; equivalently, such that g′(x) − f ′(x) = k for all x.

This is the result I used earlier in these notes, when I solved a differential equation to find

all functions f that had the property that f ′′ = (f ′)2.

Comparison with Tangential Approximations In a linear or tangential approxi-mation to a function f near a point a we approximate

f(x) ≈ f(a) + f ′(a) · (x − a) ,

where the derivative is evaluated at the point a, but where an error is introduced — anerror that we have made no attempt to quantify. If we apply the Mean Value Theoremon an interval [a, b], then, for x in that interval,

f(x) = f(a) + f ′(c) · (x − a) ,

where the exact location of c in the interval (a, x) is not known, but where there is noerror in the equation.


Example D.36 Prove that, for any x in [−1, 1],

arcsin x + arccos x =π

2

Solution:

Proof using the calculus. Define f(x) = arcsin x + arccos x; then

f ′(x) =1√

1 − x2+

−1√1 − x2

= 0

so, by one of the Corollaries above, f is a constant function over its domain. Wecan determine the value of that constant by selecting a convenient value of x inthe domain; we might be tempted to choose x = 1,

arcsin 1 + arccos 1 =π

2+ 0 =

π

2

and to conclude that π2

must be the value of the sum at any point in [−1, 1]. But,if we were to make this choice, we would be overextending ourselves: our corollaryapplies to an open interval, and one of the two functions, arcsin, is not defined inan open interval containing x = 1. It would be possible to “repair” the corollaryto one that would permit this choice, but the version we have available now doesnot permit it: neither of the inverse functions has a derivative at the end-pointof its domain. So, instead, let’s choose a convenient point in the interior of thedomain, where both functions have a derivative. Among the values that shouldbe convenient to students knowing just a little trigonometry are x = 0, x = ± 1√

2,

x = ±12, x = ± 1√

3: for all of these values you should know the arcsine and arccosine.

I shall take x = 0:arcsin 0 + arccos 0 = 0 +

π

2=

π

2,

which is the same result we obtained earlier by a proof of doubtful validity.

Trigonometric proof. Define θ = arcsin x, so sin θ = x. Moreover, the image (=range)of the arcsin function is

[

−π2, +π

2

]

. Then

cos(π

2− θ)

= cosπ

2· cos θ + sin

π

2· sin θ

= 0 · cos θ + 1 · sin θ

= sin θ = x

Since −π2≤ θ ≤ π

2, 0 ≤ π

2− θ ≤ π. By definition of the inverse cosine function,

π2− θ = arccos x, so arcsin x + arccos x = π

2.


Warning! When a mathematical theorem requires that a function possess a certainproperty for all x in some “universe”, then the failure of that property at even onepoint can cause the result of the theorem to fail to apply. So, for example, the function|x| is differentiable at all but one point. If that point, x = 0, is in the domain we areconsidering, then we have no right to claim that the results of the two theorems apply.For example, on the interval [−1, +1] there is no point where the derivative is equal to0. In mathematics even one exception can destroy the rule.

A moderately difficult examination problem made much more difficult.

Example D.37 The following problem appeared on the Final Examination in thiscourse in December, 2002:

1. [8 MARKS] Determine the derivative of the function

h(x) = arctan x + arctan 1 − arctan1 + x

1 − x.

for x 6= 1 . You are expected to simplify your answer.

2. [7 MARKS] Use your solution to (the preceding) question to determine the valueof h(−5) . Only a solution using the previous result will be accepted. Reduce youranswer as much as possible; the examiners are aware that you do not have the useof a calculator.

3. [THIS IS DIFFICULT, AND WAS NOT PART OF THE EXAMINATION QUES-TION] Use your solution to (the preceding) question to determine the value ofh(+5) .

Solution:

1. For all points x in the domain of h,

h′(x) =d

dx

(

arctan x + arctan 1 − arctan1 + x

1 − x

)

=1

1 + x2+ 0 − 1

1 +

(

1 + x

1 − x

)2 · 2

(1 − x)2

= 0

2. We have shown that h has a derivative, and that the value of that derivative is 0for all x. By the corollary above, that would appear to imply that the function h


is constant. But we must be careful. The corollary refers to an interval, and wemust also refer to an interval here. We will take the largest possible interval, toobtain the strongest result.

The domain of the function h is the set of all x for which both arctan x and

arctan1 + x

1 − xare defined. We know that the arctangent function is defined every-

where. But we need to restrict x so that1 + x

1 − xis defined, and this requires the

exclusion of the point x = 1. Thus the domain of h is a “punctured” real line,R − {1}. If we wish to apply the corollary, we will need to choose an interval thatavoids the excluded point x = 1. For x < 1 the corollary tells us that the value ofh is constant; in order to determine that constant, we need only evaluate h at a“convenient” point. For example, take x = 0:

h(x) = h(0)

= arctan 0 + arctan 1 − arctan1 + 0

1 − 0= arctan 0 + arctan 1 − arctan 1

= 0 +π

4− π

4= 0

for all x < 1. In particular, this tells us that h(−5) = 0.

3. But the preceding computation does not apply for x > 1; all we know is that h isconstant over the interval. It’s not so easy to find a convenient value of x in thisinterval, and it will turn out that the constant value of the function for this intervalis not the same as for the interval containing −5.84 Try, for example x =

√3; this

is a reasonable choice, since we know that tan π3

=√

3. Then we have

h(x) = h(√

3)

= arctan√

3 + arctan 1 − arctan1 +

√3

1 −√

3

=π

3+

π

4− arctan

1 +√

3

1 −√

3

=7π

12− arctan

1 +√

3

1 −√

3

84This is not a contradiction to the Corollary to the Mean Value Theorem: the corollary does notapply to the entire domain of h because there is one point where the domain is broken, and where thefunction fails to satisfy the conditions of the theorem; so the most we can do is to apply the corollaryseparately on the two halves of the domain; of course, it could have happened that the constant valuesfor the two halves were the same — but it didn’t happen, and the constants are, indeed, different!


But what is the last term? You may recall an identity found on one of the endpapersof your textbook:

tan(x + y) =tan x + tan y

1 − tanx tan y.

Apply this identity with x = π4

and y = π3, to obtain

tan7π

12=

1 +√

3

1 − 1 ·√

3.

As we defined the arctangent function using the branch(

−π2, π

2

)

, our inverse tan-

gent function takes its values on the interval(

−π

2,π

2

)

; we restate the last result

in a form that we can use:

tan−5π

12=

1 +√

3

1 − 1 ·√

3,

and infer that

arctan

(

1 +√

3

1 − 1 ·√

3

)

= −5π

12,

from which it follows that

h(x) =7π

12−(−5

12

)

= π

for all x > 1.

4.2 Exercises

[1, Exercise 4.2.21, p. 295] “Show that a polynomial of degree 3 has at most 3 realroots.”

Solution: Let the polynomial be named f(x), and suppose that it had 4 or moredistinct roots:

r1 < r2 < r3 < r4 . (194)

Then, by definition of a root , f = 0 at each of these 4 locations; by Rolle’s Theoremf has a horizontal tangent between each successive pair of these roots. That is,there exist real numbers c1, c2, c3 such that

r1 < c1 < r2 < c2 < r3 < c3 < r4 ,

f ′(c1) = f ′(c2) = f ′(c3) .


But the derivative of f is a quadratic polynomial, and we are able to calculate itsroots: we know that there cannot exist more than 2 points where f ′ = 0. Fromthis contradiction we can infer that our assumption that there were 4 distinct rootswas unjustified. By reductio ad absurdum we have shown that the number of rootscannot exceed 3.

This theorem can also be proved using purely algebraic methods, without anymention of the calculus.

[1, Exercise 4.2.18, p. 295] “Show that the equation 2x − 1 − sin x = 0 has exactlyone real (solution).” The problem could also have been stated in the following,equivalent way: “Show that the graphs y = 2x − 1 and y = sin x cross at exactlyone point.”

Solution:

1. We will show that the difference in height between where a line x = u crossesthe graph is equal to zero exactly once. That difference is 2u− 1− sin u, andwe will denote it by f(u).

2. We use two theorems in this type of proof: the Intermediate Value Theoremto prove that the function is 0 at least once; and the Mean Value Theorem orRolle’s Theorem to show that it is not 0 more than once.

3. Since f(u) is the difference between a polynomial and sin u — two differen-tiable functions — it is differentiable everywhere;

f ′(u) = 2 − cos u ≥ 2 − 1 = 1 > 0 .

Since it is differentiable everywhere, it is surely continuous everywhere.

4. When u gets large, 2u also becomes large; but the subtracted sine functionremains between ±1; thus we can certainly find a value of u for which f(u) > 0.For example, f(π) = 2π − 1 > 0. Similarly, when u becomes large negatively,f becomes large negatively; we need only one convenient point: f

(

−π2

)

=−π − 1 + 1 = −π < 0.

5. We now have an interval[

−π2, π]

such that f is negative at one end-point,positive at the other, and continuous throughout. By the Intermediate ValueTheorem, f is zero somewhere in the interval.

6. Suppose that f was 0 twice — not necessarily in the interval, but twice any-where on the real line — say at points u = a and u = b. Then we could applyRolle’s theorem to conclude that there is a point c such that a < c < b andf ′(c) = 0. But we saw earlier that f ′(u) > 0 for all u.


7. From this contradiction we conclude that it was not correct to assume therewere two points where f = 0; as we have already shown that there is at leastone such point, we can now conclude there is exactly one such point.

[1, Exercise 4.2.23, p. 295] “If f(1) = 10 and f ′(x) ≥ 2 for 1 ≤ x ≤ 4, how small canf(4) possibly be?”

Solution: For a solution, please consult the Student Solutions Manual [3, p. 140].The solution given is not quite complete: the author shows that f(4) ≥ 16, butdoes not show that there could exist a function f for which this value of 16 wouldbe attained. One such function is 2x + 8 is a function that satisfies the conditionsof the problem for which this inequality is “best possible”.

[1, Exercise 4.2.29, p. 296] “Use the Mean Value Theorem to prove the inequality

| sin a − sin b| ≤ |a − b|

for all a and b.

Solution: For a solution, please consult the Student Solutions Manual [3, p. 140].



Release Date: Wednesday, November 8th, 2006

This lecture was poorly attended, possibly because of competition from atest in another section. Students who are diverting calculus time to othersubjects have already been reminded of the risks.

D.19.1 §4.3 How Derivatives Affect the Shape of a Graph

What does f ′ say about f? The first theorem is another corollary to the Mean ValueTheorem:

Corollary D.38 (to the Mean Value Theorem:) (“Increasing/Decreasing Test” or“I/D Test”)

1. If f ′(x) > 0 at every point of an interval [a, b], then f is increasing on the interval.

2. If f ′(x) < 0 at every point of an interval [a, b], then f is decreasing on the interval.

Remember that the textbook’s definitions of increasing and decreasing [1, p. 21] do notpermit equality; these are concepts that some mathematicians call strictly increasing andstrictly decreasing . There are variations of these results that can be proved to cover caseswhen equality may hold. Based on the above results, we can prove the following “Test”.This “test” describes conclusions that hold when f ′ exists in an interval on both sides ofc — even if it does not exist at c itself. When we speak of the left “side” of c, we meanan interval of the form (c − ǫ, c), where ǫ is a positive number — any positive number;the right “side” is defined analogously. The theorem requires that certain conditionsoccur on the “sides”: if they do not hold, then you cannot use this theorem to draw theconclusions stated.

Theorem D.39 (First Derivative Test) Suppose that c is a critical number of a con-tinuous function f .

1. If f ′ changes from positive to negative at c, then f has a local maximum at c.

2. If f ′ changes from negative to positive at c, then f has a local minimum at c.

3. If f ′ is defined on the two sides of c, but does not change sign at c then f hasneither a local maximum nor a local minimum at c.


What does f ′′ say about f?

Definition D.7 1. If the graph of f lies above all of its tangents on an interval [a, b],the graph is said to be concave upward on the interval.

2. If the graph of f lies below all of its tangents on an interval [a, b], the graph is saidto be concave downward on the interval.

3. A point P = (c, f(c)) is an inflection point of the graph of the function f if

(a) f is continuous at x = c;

(b) the curve is concave upward on one side of P , and concave downward on theother side.

Theorem D.40 (Concavity Test) 1. If f ′′(x) > 0 for all x in I, then the graph off is concave upward on I.

2. If f ′′(x) < 0 for all x in I, then the graph of f is concave downward on I.

Theorem D.41 (Second Derivative Test) Suppose that f ′′ is continuous in a neigh-bourhood of x = c, i.e., in some interval (c−ǫ, c+ǫ), where ǫ is some positive real number.

1. If f ′ = 0 and f ′′(c) > 0, then f has a local minimum at c.

2. If f ′ = 0 and f ′′(c) < 0, then f has a local maximum at c.

3. If f ′ = 0 and f ′′(c) = 0, this theorem gives no information.

Note that the Second Derivative Test is inconclusive when f ′′(c) = 0. Consider thefunctions f1(x) = x3 and f2(x) = ±x4 at x = 0. All of these functions have a zero firstderivative at x = 0, but the second derivatives also vanish. In the case of f1 there isno extremum at x = 0; in the case of x4 there is a local minimum; while −x4 has alocal maximum. It is possible to refine the Second Derivative Test by considering higherderivatives, but that is beyond Math 140.

What should I retain from this section? This section contains definitions and teststhat are required in the analysis of functions: all of these need to be retained and drilledfor examination purposes. However, you will not be expected to be able to reproduceproofs of these results, although they are all simple consequences of the results in thepreceding two sections of the textbook.


Tabulation of function data. The textbook suggests that routine problems can bebest approached by tabulating the necessary data. If these methods appeal to you, usethem — they tend to keep your information orderly, and to reduce the chance that youforget something. The rationale for the methods is that we are often interested in thesign of a derivative or second derivative; since the sign of a product is related to the signsof its factors, a table can be created where the signs of the factors are tabulated andthe sign of the product deduced. Not all functions can be factored into simple factors,and for them these tabular methods are of limited value. The best way to prepare forexamination on these topics — after you have thoroughly absorbed the definitions andtheorems is by extensive drill, using the Student Solutions Manual to correct your workand to guide you as to what constitutes a good solution.

4.3 Exercises As stated earlier, you are advised to work multiple problems and toverify your solutions in the Student Solutions Manual.

[1, Exercise 4.3.14, p. 305] You are asked to

1. Find the intervals on which f is increasing or decreasing.

2. Find the local maximum and minimum values of f , (if any).

3. Find the intervals of concavity and the inflection points (if any).

Here f(x) =x2

x2 + 3.

Solution: This is a straightforward example, and the exposition is certainly im-proved by tabulation of the data. We will be needing the derivatives. The firstderivative may be computed by the Quotient Rule and Chain Rule, but the compu-tations are even easier if you notice that f(x) = 1− 3

x2+3. Thus, by the Reciprocal

Rule and the Chain Rule,

f ′(x) = 0 +3

(x2 + 3)2· 2x =

6x

(x2 + 3)2.

The denominator is a power of a factor which is always positive; the numerator isa constant multiple of the polynomial x, which changes sign at x = 0. Extendingthe author’s style, I will tabulate, but will also include information about the endpoints of the intervals.:

Interval x x2 + 3 f ′(x) fx < 0 − + − decreasing on (−∞, 0)x = 0 0 + 0 critical pointx > 0 + + + increasing on (0,∞)


Note that f(−x) = f(x): this is an even function, and its graph is symmetricunder reflection in the y-axis. There is only one critical point. As f is differentiableeverywhere, we know that the local extrema will occur only at points where f ′ = 0;so the only candidate is x = 0. The change of sign of f ′ tabulated above shows, bythe First Derivative Test, that f has a local minimum at x = 0; there are no localmaxima.

The second derivative may be computed by the Quotient Rule:

f ′′(x) = 6 · 1(x2 + 3)2 − x · 2(x2 + 3)(2x)

(x2 + 3)4=

18(1 − x)(1 + x)

(x2 + 3)3.

To analyze its behaviour, we may again tabulate:

1

0.8

0.6

0.4

0.2

0

x

20100-10-20

Figure 14: Graph of the Functionx2

x2 + 3and its horizontal asymptote, y = 1

Interval 1 − x 1 + x x2 + 3 f ′′(x) fx < −1 + − + − concave downward on(−∞,−1)x = −1 + 0 + 0 inflection point

(because concavity changes)−1 < x < 1 + + + + concave upward on(−1,−1)

x = +1 0 + + 0 inflection point(because concavity changes)

x > 1 − + + − concave downward on(1,∞)

At the points x = ±1 f ′′ is continuous, and changes sign. Consequently both ofthese are inflection points of f .

We have determined the nature of the critical point x = 0 by using the FirstDerivative Test; we could also have applied the Second Derivative Test there: since


f ′′(0) > 0, the critical point is a local minimum; the minimum value is 0. A sketchof the graph with its horizontal asymptote is given in Figure 14, page 2186).

[1, Exercise 4.3.16, p. 305] (This example was not worked completely in the lecture.)The instructions are the same as for the preceding example, but the function is

f(x) = cos2 x − 2 sin x ,

no longer a pure polynomial; however, it may be possible to treat it as a polynomialin a trigonometric function, and to still use the methods used above. This functionis periodic with period 2π, since both the sine and cosine functions have periods2π; thus it suffices to study its behavior on an interval of length 2π: the authorhas selected the interval 0 ≤ x ≤ 2π.

While it has not been requested, we may observe by completion of the square that

f(x) = cos2 x − 2 sin x

= (1 − sin2 x) − 2 sinx = 2 − (sin x + 1)2

= (√

2 − 1 − sin x)(√

2 + 1 + sin x)

This information can also be analyzed in a tabular form, and used to locate thegraph above and below the x-axis: the second factor is always positive, but thefirst changes sign at x = arcsin(

√2 − 1).

f ′(x) = 2 cosx · (− sin x) − 2 cos x

= −2(cos x)(sin x + 1)

f ′′(x) = −2(cos2 x − sin2 x + cos x − sin x)

= 4

(

sin x − 1

2

)

(sin x + 1)

The sign of the first derivative depends on the sign of the cosine, which is negativein the 2nd and 3rd quadrants; the sign of the second derivative depends on thesign of sin x − 1

2. We tabulate the signs of the factors and the derivatives:

Interval cos x sin x + 1 f ′(x) f0 end point

0 < x < π2

+ + − decreasing on (0, π2)

π2

0 + 0 critical pointπ2

< x < 3π2

− + + increasing on (π2, 3π

2)

3π2

0 0 0 critical point3π2

< x < 2π + + − decreasing on (3π2

, 2π)2π end point


We haven’t classified the critical points yet: at the critical point π2

the functionchanges from decreasing to increasing, so the point is a local minimum; analogously,3π2

is a local maximum. The instructions asked for the extremal values of thefunction, so we need to compute f

(

π2

)

= 0 − 2 = −1 and f(

3π2

)

= 0 − 2(−1) = 2.

2

x-1

1

60

4

-2

20-2

Figure 15: Graph of the Function cos2 x − 2 sin x

Interval sin x − 12

sin x + 1 f ′′(x) f0 < x < π

6− + − concave downwards

π6

0 + 0 inflection pointπ6

< x < 5π6

+ + + concave upwards5π6

0 + 0 inflection point5π6

< x < 3π2

− + − concave downwards3π2

− 0 0 NOT inflection point3π2

< x < 2π − + − concave downards

The coordinates of the inflection points are(

π6,−1

4

)

and(

5π6

,−14

)

. There is noinflection point at x = 3π

2, even though the second derivative vanishes, because the

concavity does not change at that point. A sketch is given in Figure 15, on page2188 of these notes.

[1, Exercise 4.3.20, p. 305] (This problem was not discussed at all at the lecture.)You are asked to

1. “Find the intervals on which f is increasing or decreasing.

2. “Find the local maximum and minimum values of f , (if any).

3. “Find the intervals of concavity and the inflection points (if any).”


Here f(x) = x ln x.

Solution: First observe that the domain of f is (0,∞).

1. f ′(x) = 1 · ln x + x · 1

x= ln x + 1. f ′(x) > 0 ⇔ ln x > −1 ⇔ x >

1

e. Thus

the function is increasing over the interval(

1e,∞)

, and decreasing over theinterval

(

0, 1e

)

.

2. f ′′(x) =1

x. The second derivative exists at all points of the domain, and is

always positive: the graph is concave upwards throughout its domain. Thereare no inflection points.

While we have answered the questions that were asked, we still do not have enoughinformation to graph the function. Investigation of the graph of this function willbe continued in [1, Exercise 4.5.43, p. 323] (cf. Figure 16, [3, p. 165].) The graphcan be shown to be tangent to the y-axis at (x, y) = (0, 0).

0.4

10

-0.4

0.50-0.5

x

2

1.2

1.5

0.8

Figure 16: Graph of the Function x ln x


D.20 Supplementary Notes for the Lecture of Monday, Novem-ber 13th, 2006

Release Date: Monday, November 13th, 2006, (subject to correction)

D.20.1 §4.4 Indeterminate Forms and L’Hospital’s Rule

In [1, §2.3, §2.5] we discussed several limit laws that could be used to evaluate a combi-nation of functions when all of the components had limits, subject to certain restrictions.The first restriction we met, in [1, §2.3], was in the “Quotient Law”:

limx→a

f(x)

g(x)=

limx→a

f(x)

limx→a

g(x)if lim

x→ag(x) 6= 0 ,

in addition to the requirement that the two limits mentioned exist. Subsequently westudied the concept of derivative, whose definition involved just such a limit — wherethe limit of the denominator (of the ratio giving the slope of the secant to the graph) iszero. Then, in [1, §2.5], we generalized our definitions to permit limits to have “values”of ±∞. This raises the question of whether the “limit laws” hold for such limits also.In [1, Exercise 2.4.44, p. 124] some such laws are proved to hold. The generalizations toinfinite limits can be seen to hold so long as we can avoid certain “forbidden” operations:

like ∞ · 0, ∞−∞,∞∞ . I have mentioned, on page 2069 of these notes, the possibility

of “extending” the real number system to include two new objects: +∞ and −∞. Inpractice this is what we do informally, but it has the downside that we have to avoidcertain types of operations — like those combinations just mentioned — to which wecannot give a meaning. The issue is not that we don’t know the correct value to attach tothese combinations; it is that we can show that there is no number — either real or ±∞that can serve as a value, without creating inconsistencies in the rest of our algebraicsystem. If we wish to work with ±∞, then we must avoid these combinations. Thatmeans that the limit laws do not hold for such combinations; but the correspondinglimits may still exist, even though we can’t find them using the limit laws. This isthe subject of the present section — to evaluate the limits of certain combinations offunctions where the limit laws fail to hold. We call these combinations “indeterminateforms”. The first case we consider is a ratio of two functions whose limits are either both0 or both one of ∞ or −∞. The name of the theorem is, in practice, given either thearchaic French spelling, L’Hospital’s Rule, or the more modern spelling, where thesilent “s” is replaced by a circumflex on the preceding vowel: L’Hopital’s Rule.

Theorem D.42 (L’Hospital’s Rule) Suppose that

1. f and g are differentiable;


2. g′(x) 6= 0 near a, except possibly at a; and that one of the following pairs ofconditions holds:

(a) limx→a

f(x) = 0 and limx→a

g(x) = 0

(b) limx→a

f(x) = ∞ and limx→a

g(x) = ±∞

(c) limx→a

f(x) = −∞ and limx→a

g(x) = ±∞

(d) limx→a

f ′(x)

g′(x)exists or is ∞ or −∞.

Then

limx→a

f(x)

g(x)= lim

x→a

f ′(x)

g′(x).

An analogous result holds for limits from the right and limits from the left.85

Instructors in Calculus courses often discourage their students from using L’Hospital’sRule because it renders the computation of limits so easy: many of the problems thatyou will see in this section can be solved without the Rule, and you might wish to see ifyou can accomplish that. But a technique that is so powerful is too important to ignorefor macho reasons, so you are expected to become competent in the use of L’Hospital’sRule; sometimes it could be the only method you will have to determine a limit; and,even when other methods are possible, L’Hospital’s Rule will usually be faster. However,there are skills that you need to acquire to use the Rule effectively, and these will comefrom working many examples.

The theorem tells us that the last equation holds only if it known that the secondlimit exists . Sometimes we may need to apply the theorem several times in succession.When we do that, the equations we write are not known to be valid until the limit ofthe very last ratio in the sequence has been shown to exist; only then all interveningequations are justified.

85

Example D.43 Let’s apply L’Hospital’s Rule first to the case of the derivative of a continuous function

f at x = a. We cannot use the Quotient Law on the ratio f(x)−f(a)x−a

because both top and bottom havelimit equal to 0. If f is differentiable the conditions of L’Hospital’s Rule are satisfied, and the limit is

limx→a

f ′(x)1 = lim

x→af ′(x) which will equal f ′(a), provided we know that f ′ is continuous at x = a. This

verifies the Rule in a special case, shows that L’Hospital’s Rule is not completely efficient, as it requiresthe continuity of f ′, and is of no use otherwise, as it does not help us evaluate any limit that we couldn’tevaluate otherwise.


Equal signs in applications of l’Hospital’s Rule L’Hospital’s Rule, where applica-ble, replaces a given limit of certain types of ratios by the limit of another ratio, hopefullyeasier to evaluate. If the second limit cannot be evaluated, then the entire application isworthless. For this reason the authors of the solution manual for your textbook [3] have

introduced the novel symbolH=. The message conveyed by the H is that the equality

will not be justified until the sequence of =’s andH=’s ends in the limit of a ratio which

can be evaluated. When that last limit has been evaluated, all interveningH=’s can be

replaced by =’s. This is actually the spirit of some of the other Limit Laws that we havelearned earlier, but we haven’t used a modified equal sign in those cases. For example,the Produce Law replaces the problem of evaluating the limit of a product by evaluatingthe limits of its factors; but, if the factors do not all have a limit, or if the product ofthe limits is undefined, then the Law fails.

Example D.44 [1, Exercise 4.4.22, p. 313] Evaluate limx→0

ex − 1 − x − x2

2

x3.

Solution: First look at limx→0

ex − 1

x, and then at [1, Exercise 4.4.21, p. 313], which is solved

in the Student Solutions Manual, and also on one of the CD-Roms that comes with thetextbook. These are a progression of more difficult versions of the same problem. In thefirst case we may apply L’Hospital’s Rule since lim

x→0(ex−1) = 1−1 = 0, by the continuity

of the exponential function, and the fact that e0 = 1. Then we have

limx→0

ex − 1

x= lim

x→0

ex − 0

1= e0 = 1 .

We could also have proved this without L’Hospital’s Rule, by observing that

limx→0

ex − 1

x= lim

x→0

ex − e0

x − 0=

d

dxex

∣

∣

∣

∣

x=0

= ex|x=0 = 1 .

Proceeding to the present example,

limx→0

ex − 1 − x − x2

2

x3= lim

x→0

ex − 0 − 1 − 2x2

3x2

= limx→0

ex − 0 − 22

6x

= limx→0

ex − 0

6

whose limit we know by the continuity of ex to be 16. The intermediate = signs are not

justified until this last step, where we prove that the last limit exists, hence the precedingone, hence the one before that, etc.


Of course, just because a function whose limit is sought is presented as a quotientdoes not guarantee that L’Hospital’s Rule is applicable to the problem. Don’t forgetthat the numerator and denominator need to either both have limits equal to 0, or bothhave limits of either +∞ or −∞. So, for example, you must not use L’Hospital’s Ruleto solve the following problem:

Example D.45 [1, Exercise 4.4.29, p. 314] limx→0

x + sin x

x + cos x=?

Solution: While the limit of the numerator as x → 0 is, indeed, 0; the limit of thedenominator is 0 + 1 = 1. You may not apply L’Hospital’s Rule, and there is no needto attempt to do so, since the Quotient Rule applies here, and the limit is the ratioof the limits, i.e. 0

1= 0. (The “answer” that you might have obtained here, using the

inapplicable Rule, is irrelevant, since the operation cannot be justified by the theorem!)

Indeterminate Products L’Hospital’s Rule is available only for quotients of functionswith the properties described earlier: you do not have a counterpart to this theorem forproducts. So, if you are presented with a product of functions whose limit is required,you must either apply one of the other Rules you have available, or attempt to transformthe product into a quotient by moving a factor to its reciprocal in the denominator ofa fraction. This could require some experimentation: if you move the “wrong” factor,you may transform the function into a ratio where the ratio of the derivatives is “morecomplicated” than the original function; unless L’Hospital’s Rule leads to a ratio whoselimit can be found, all steps in the transition to that ratio become invalid.

Example D.46 [1, Example 4.4.50, p. 314] limx→∞

(xe1x − x) =?

Solution: The function may be factorized as xe1x − x = x

(

e1x − 1

)

. Now we can move

one of the factors as a reciprocal into the denominator of a fraction. There are twochoices, either

xe1x − x = x

(

e1x − 1

)

=e

1x − 1

1x

or

xe1x − x = x

(

e1x − 1

)

=x1

e1x −1

Of these the second does not appear to be amenable, and we shall try the first first.86

After ascertaining that limx→∞

(

e1x − 1

)

= 1 − 1 = 0, and limx→∞

1x

= 0, ensuring that

86But we could be wrong, and it might be that a very complicated-looking ratio might lend itself tosimplification after the differentiations are carried out.


L’Hospital’s Rule is applicable, we proceed:

limx→∞

(

xe1x − x

)

= limx→∞

e1x − 1

1x

= limx→∞

e1x ·(

− 1x2

)

− 1x2

= limx→∞

e1x

= elim

x→∞

1x (by continuity of ey at y = 0)

= e0 = 1 .

It might have been wiser to change the variable during this process, in order to simplifythe differentiation:

limx→∞

(

xe1x − x

)

= limx→∞

e1x − 1

1x

= limy→0+

ey − 1

y

(defining y = 1x

and letting it → 0+)

= limy→0+

ey

1= e0 = 1.

And we could have observed that the quotient was precisely the difference quotient whichdefines d

dyey evaluated at y = 0.

Indeterminate Differences The example just worked could also be interpreted asa difference of two functions, both of which become infinite. In such a situation theDifference Law is not applicable, but it may still be possible to transform the functioninto a quotient to which L’Hospital’s Rule is applicable.

Example D.47 [1, Exercise 4.4.48, p. 314]: limx→1

(

1

ln x− 1

x − 1

)

=?

Solution: In this case the transformation to a quotient appears to “complicate” thefunction, but we are still able to proceed.

limx→1

(

1

ln x− 1

x − 1

)

= limx→1

x − 1 − ln x

(x − 1) lnx

= limx→1

1 − 1x

1 lnx − (x − 1) · 1x


= limx→1

1x2

1x

+ 1x2

= limx→1

1

x + 1

=1

2(by continuity of 1

x+1at x = 1.)

In this last example we might have been tempted to eliminate the negative powers of x;the saving in effort would be insignificant:

limx→1

1 − 1x

1 ln x − (x − 1) · 1x

= limx→1

x − 1

x ln x + x − 1

= limx→1

1

1 ln x + x · 1x

+ 1 − 0

= limx→1

1

ln x + 2=

1

0 + 2=

1

2(by continuity of 1

ln xat x = 1).

Indeterminate Powers Since a function of the form f(x)g(x) with f(x) > 0 may be

interpreted as(

eln f(x))g(x)

= eg(x)·ln f(x), limits of functions of this type may be found byfinding the limit of the exponent, and then applying the continuity of the exponentialfunction to infer that the limit of the exponential is equal to the exponential of thelimit. Alternatively, one may use logarithmic differentiation and evaluate the limit ofthe logarithm of the given function. But, when the time comes to apply L’Hospital’sRule, be sure to check that the Rule is, indeed, applicable.

Example D.48 [1, Exercise 4.4.62, p. 314] limx→∞

(

(

2x − 3

2x + 5

)2x+1)

=?

Solution: To apply “logarithmic differentiation”, first assign a name to the function;then take its logarithm and find the limit of the logarithm:

y =

(

2x − 3

2x + 5

)2x+1

ln y = (2x + 1) ln

(

2x − 3

2x + 5

)

= (2x + 1) (ln(2x − 3) − ln(2x + 5))

=ln(2x − 3) − ln(2x + 5)

12x+1

⇒ limx→∞

ln y = limx→∞

ln(2x − 3) − ln(2x + 5)1

2x+1


But note that I have made a tactical error here: by expressing the logarithm of thequotient as the difference of the logarithms, I have created an indeterminate difference,and am unable to establish its limit, so I can’t show that l’Hospital’s Rule is applicable.I need to observe that

limx→∞

(

ln2x − 3

2x + 5

)

= ln

(

limx→∞

2x − 3

2x + 5

)

(by continuity of ln)

= ln

(

limx→∞

2 − 3x

2 + 5x

)

= ln 1 = 0 ;

and, obviously, limx→∞

1

2x + 1= 0. Now we can return to the earlier expression and apply

L’Hospital’s Rule:

limx→∞

ln y = limx→∞

ln(

2x−32x+5

)

12x+1

= limx→∞

ln(2x − 3) − ln(2x + 5)1

2x+1

= limx→∞

22x−3

− 22x+5

− 1(2x+1)2

· 2

= limx→∞

−8(2x + 1)2

(2x − 3)(2x + 5)

= limx→∞

−8(2 + 1x)2

(2 − 3x)(2 + 5

x)

= −8

Then

limx→∞

(

(

2x − 3

2x + 5

)2x+1)

= limx→∞

y

= limx→∞

eln y

= elim

x→∞ln y

(by continuity of the exponential)

= e−8


4.4 Exercises

[1, Exercise 4.4.6, p. 313] “Find the limit. Use l’Hospital’s Rule where appropriate.If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t

apply, explain why. limx→−2

x + 2

x2 + 3x + 2.”

Solution: This problem does not require l’Hospital’s Rule. If the denominator ofthe fraction is factorized, into (x+1)(x+2), then the fraction can be seen to equal

1

x + 1, whose limit may be found by the Quotient Law; or by using our result that

rational functions are continuous at points which are not discontinuities, since the

only discontinuity of1

x + 1is at x = −1, and we are taking the limit as x → −2.

Thus we see that the limit is 1−2+1

= −1.

Since both numerator and denominator approach 0 as x → −2, we may use

l’Hospital’s Rule. The ratio of the derivatives is1

2x + 3→ 1

−4 + 3= −1, as

before.

Mathematicians prefer not to use l’Hospital’s Rule where the problem has an easysolution without it. This may derive from the “macho” element of traditionalmathematics. Sometimes on examinations you may be asked to find a limit andexplicitly instructed not to use l’Hospital’s Rule.

[1, Exercise 4.4.12, p. 313] (This problem was not discussed in the lecture.) “Find

the limit. ... limt→0

e3t − 1

t”.

Solution: Since both numerator and denominator approach 0, this problem maybe solved by l’Hospital’s Rule. (More precisely, we may attempt to solve it usingl’Hospital’s Rule: the Rule does not always yield a solution — if the ratio of deriva-tives does not have a limit, or if it is not again amenable to another applicationof l’Hospital’s Rule then the Rule fails to provide any information.) The ratio of

derivatives is3e3t

1. By the continuity of the exponential, the limit may be obtained

by evaluating this last function at t = 0, which is 3e3t

1, to which the Quotient Law

may be applied, yielding a limit of 3e0 = 3. Of course, this problem could also beviewed as finding the value at t = 0 of d

dt(e3t), and l’Hospital’s Rule would not be

needed.

[1, Exercise 4.4.56, p. 314] (This problem was not discussed in the lecture.) “Find

the limit. ... limt→∞

xln 2

1+ln x ”.


Solution:

limt→∞

xln 2

1+ln x = limt→∞

(

eln x)

ln 21+ln x

= limt→∞

e((ln x)· ln 21+ln x)

= elimt→∞

(

(ln x) · ln 2

1 + ln x

)

We now investigate the limit in the exponent.

limt→∞

(

(ln x) · ln 2

1 + ln x

)

= (ln 2) · limt→∞

ln x

1 + ln x

= (ln 2) · limt→∞

11

lnx+ 1

= (ln 2) ·limt→∞

1

limt→∞

(

1ln x

+ 1)

= (ln 2) · 1

0 + 1= ln 2

which permits us to assert that

limt→∞

xln 2

1+ln x = eln 2 = 2 .

We didn’t have to use l’Hospital’s Rule, but needed the fact that limt→∞

ln t = ∞;

but we would have needed to use this fact to apply l’Hospital’s Rule anyhow. Thisfact follows from the observation that, as t → ∞, it passes through values whichare powers en, and ln (en) = n → ∞ as n → ∞.

Example D.49 I end the lecture with an example where a limit of an indeterminatefraction does exist, but where l’Hospital’s Rule breaks down and is unable to reveal that

limit. Consider limx→∞

x + sin x

x. Here lim

x→∞x + sin x = ∞ = lim

x→∞x , so we cannot use the

Quotient Law. But, if we apply L’Hospital’s Rule, we obtain

limx→∞

x + sin x

xH= lim

x→∞

1 + cos x

1.

Hitherto, in the examples of l’Hospital’s Rule that we have studied, the limit of thelast ratio could be calculated, and that would validate the differentiation operations,

permitting us to replaceH= by =. But, in this case, the limit of the numerator does not


exist, while the limit of the denominator is 1. So, in this case, the attempted applicationof l’Hospital’s Rule must be aborted, with no progress in evaluating the limit of theoriginal function. That does not imply that the original limit does not exist; in fact

limx→∞

x + sin x

x= lim

x→∞

(

1 +cos x

x

)

= limx→∞

1 + limx→∞

cos x

xby the Sum Law

= 1 + limx→∞

cos x

xsince the limit of a constant is that constant

= 1 + 0 .

(Since

−1 ≤ cos x ≤ 1 ⇒ −1

x≤ cos x

x≤ 1

xfor x > 0 .

As x → ∞, the extremes of the last pair of inequalities both approach 0, so the “Squeeze”

Theorem ensures that limx→∞

cos x

x= 0.)


D.21 Supplementary Notes for the Lecture of Wednesday, Novem-ber 15th, 2006

Release Date: Thursday, November 15th, 2006(subject to correction)

D.21.1 §4.5 Summary of Curve Sketching

Read the textbook! The author’s preamble to the section, [1, pp. 316-317] provides anexcellent introduction to the issues, and to the limitations of both plotting isolated pointson a curve and using calculators and computers to produce graphs.

Limitations of the method of plotting points. If you attempt to graph a curveonly by plotting a number of points — even a large number — you expose yourself tothe risk that either

• your selection of points is too sparse at critical places where the graph changesrapidly, and so you miss those important features; or

• your selection happens to show undue emphasis on features which are not repre-sentative of the global behavior of the function.

It is always important to obtain some information about specific points on a curve. Theauthor warns that such information alone is often misleading. A good policy is that youshould never base any serious conclusions on the behavior of a function at specific pointsunless you have some theoretical justification.87

Limitations of graphing calculators and computers. Graphing calculators andcomputers are limited by the technical sophistication of the software. In principle thesoftware should be able to do anything you can do, and more: but, at the moment itmay not be able to do that.

Here again, the issue is that the global behavior of a function may not be best renderedby selection of specific points in the domain, even if the points are close together.

87Sometimes we do have such justification! For example, when we tabulate the factors whose productis f ′, and then indicate where each of the factors is positive, we are justified in concluding that the signof the derivative in an interval can be determined by selecting any point in the interval. Or, when weuse a corollary to the Mean Value Theorem to prove that a function is constant over an interval, we arejustified in determining that constant by choosing any point we like in the interval.


Guidelines for Sketching a Curve The textbook recommends 8 steps to follow whensketching, and emphasizes that “not every item is relevant to every function”. Try touse this as a checklist when you solve problems from the chapter.

A. Domain: Remember that sometimes the domain is not the “natural domain” — thelargest set of values where the definition of the function makes sense; sometimesyou are explicitly instructed to consider the function over a restricted subset ofthat “natural” domain. But sometimes — particularly in problems stated verbally— the “instruction” to restrict the function is implicit in other information givenabout the problem. Thus it may happen that there are real numbers where thefunction is defined, but where the function in the question does not “make sense”.

B. Intercepts: Technically, this term refers to the distance between two designatedpoints on a line. In this context the line is usually one of the coordinate axes,and the points in question are the origin and the point where the curve crosses theaxis; we interpret distance as directed distance, so the y-intercept of the graph off is usually taken to be the y-coordinate88 of the point where the curve crossesthe y-axis, i.e., the value f(0); and each x-intercept is the x-coordinate89 of apoint where the curve crosses the x-axis, i.e., where f(x) = 0. If you are given anexplicit formula for f , then it is often easy to determine f(0); if the function isdefined implicitly, you may be faced with solving an equation to determine f(0).

I have often used the word “intercept” to describe the corresponding coordinates ofpoints of intersection of the graph with lines parallel to the coordinate axes. Thiskind of information is often helpful, particularly if the graph has symmetry aboutlines parallel to but distinct from the coordinate axes.

C. Symmetry: The textbook reminds you of the concepts of evenness and oddnessdiscussed in [1, §1.1, pp. 20, 21], and in these notes. Where a function is known tobe even or odd it suffices to consider its behavior on only half of its domain. Thefull graph of an even function can be obtained from the graph for non-negativepositive x by reflecting the graph in the y-axis; the full graph of an odd functioncan be obtained from the graph for non-negative x by rotating that portion of thegraph through an angle of π radians around the origin.

The textbook also discusses the property of periodicity under which a function mayecho over its entire domain the behavior over one portion of the domain. If f hasperiod p, it has the property that

f(x) = f(x + p) for all real numbers x. (195)

88the ordinate89the abscissa


Examples are

• sin and cos, which have period 2π

• tan and cot which have period π

• x − [x] (or x − ⌊x⌋), which have period 1.

The author defines period to be the shortest distance for repetition to take place.Sometimes we are more casual with this term, and may use it to mean any distancefor which (195) holds.

D. Asymptotes: The textbook reminds you of the two types of asymptotes we havestudied:

Horizontal Asymptotes: These are lines y = L with the property that eitherlim

x→∞f(x) = L or lim

x→−∞f(x) = L. It can happen that a graph has 2 distinct

horizontal asymptotes, just one asymptote to which it is asymptotic both asx → ∞ and x → −∞, or no horizontal asymptotes at all. You should be ableto give examples of these situations.

Vertical Asymptotes: These are lines y = a with the property that one of thefollowing 4 situations holds:

limx→a+

= +∞ limx→a−

= +∞lim

x→a+= −∞ lim

x→a−= −∞

Of course, if we know either that limx→a

= +∞ or limx→a

= −∞, we have more

than enough information to claim that y = a is a vertical asymptote.

The textbook also discusses the possibility of “Slant Asymptotes”. This topic isomitted from the syllabus of Math 140.

E. Intervals of Increase or Decrease: We have discussed methods for obtaining thisinformation and presenting it in tabular form in connection with [1, §4.3].

F. Local Maximum and Minimum Values: This topic has also been discussed inconnection with [1, §4.3]; remember that you have 2 tests available for classifyingcritical points, and it may happen that one of these is much more useful than theother in a particular application.

G. Concavity and Points of Inflection: This also has been discussed in connectionwith [1, §4.3]. Remember that it is not enough to find points where f ′′(x) = 0. Wehave seen instances where this happens, and yet the point is not an inflection point;


and instances where there is an inflection point and yet there is no 2nd derivativeavailable. For example, the continuous function x · |x| changes concavity at x = 0,but has no 2nd derivative there; so x = 0 is an inflection point.

H. Sketch the Curve: The textbook offers practical assistance in the use of the pre-ceding information to plot the graph. The author suggests that, where you havedetermined a point (x, f(x)) on the curve, you may sometimes wish to determinethe value of f ′(x), in order to know the slope at which the curve passes throughthe point.

Slant Asymptotes Students in MATH 140 are not expected to be familiar with theconcept of slant asymptotes . More generally, we could wish to explore situations wherecurves are asymptotic to other kinds of curves than lines; such topics will not be consid-ered in MATH 140.

4.5 Exercises I will work several problems to indicate the way in which I would ap-proach problems of this type. But there are so many essentially different kinds of func-tions, you are advised to work many problems in this section. This is time-consuming,but essential.

[1, Exercise 4.5.22, p. 323] Use the guidelines of this section to sketch the graph of

f(x) =

√

x

x − 5.

Solution: (cf. Figure 17, page 2204)


y

8

6

4

2

0

x

1086420-2-4

Figure 17: Graph of f(x) =

√

x

x − 5and its horizontal asymptote


A. Domain: f is the composition of more than one function. We can interpret itas resulting from the application of the square root function to the function

x 7→ x

x − 5. This last function is defined for all x except 5. But the square

root function is defined when its argument is non-negative; thus we have torestrict x according to the following sequence of inequalities:

x

x − 5≥ 0

⇔

x ≥ 0 and x − 5 > 0or

x ≤ 0 and x − 5 < 0

⇔ x > 5 or x ≤ 0 .

Thus the domain is made up of two non-intersecting infinite intervals: thefunction is not defined for any x in the interval (0, 5].

B. Intercepts: The y-intercept is f(0) = 0. And the only solution to the equationf(x) = 0 is the point x = 0. Thus both intercepts of f with the coordinateaxes are both 0.

C. Symmetry: We can check that f(−x) 6= f(x) (for any x in the domain), sothe function is not even; it is surely not odd, since an odd function wouldhave a positive value at some x paired with a negative value at −x, and thepresent function is never odd. It is also not periodic, although we have notdiscussed in this course the details of proving that.

D. Asymptotes: For values of x that are large in absolute value, it is convenient

to view the function as

√

1

1 − 5x

or as

√

1 +5

x − 5. Either of these functions

can be seen to approach√

1 = 1 as x → ±∞. Thus y = 1 is the onlyhorizontal asymptote (and the graph is asymptotic to it both as x → ∞and as x → −∞). Indeed, the function approaches the value 1 from belowas x → −∞, and approaches the value 1 from above as x → +∞. Thisobservation is made only to help in the sketching: a limit as x → ∞ or asx → −∞ need not be approached “monotonically”: it is possible for a curveto cross its horizontal asymptote, although that does not happen here.

The graph is also asymptotic to the vertical line x = 5, since limx→5+

√

xx−5

=

+∞.

E. Intervals of Increase and Decrease:

f ′(x) =1

2

√

x − 5

x· −5

(x − 5)2


=

− 5

2x12 (x − 5)

32

if x > 5

− 5

2(−x)12 (5 − x)

32

if x < 0

which is always negative. Thus the function is decreasing over its entire do-main, except for the end-point x = 0, where f ′ is not defined. Since thedomain consists of two separate disconnected intervals, this property of be-ing decreasing applies separately to each of the two intervals (−∞, 0) and(5, +∞).

F. Local Maximum and Minimum Values: There are no critical points; (wefollow the textbook’s terminology, under which an end-point is not a criticalpoint). By Fermat’s Theorem there is only one point where an extremumcould occur, and that is at x = 0, when f(0) = 0. This is evidently alocal minimum. Indeed, this is the global minimum of the function, since asquare root cannot be less than 0. There is no local maximum, and no globalmaximum either.

G. Concavity and Points of Inflection: Since

f ′(x) = −5

2· 1√

x(x − 5)3,

f ′′(x) = −5

2·(

−1

2

)

· 1 · (x − 5)3 + 3x(x − 5)2

(√

x(x − 5)3)3· ,

=5

4· (x − 5)2(4x − 5)(

√

x(x − 5)3)3

=5

4(4x − 5)

√

x − 5

x3=

5

4

(

4x − 5

x2

)

√

x(x − 5) .

We could set up a chart, analyzing the various factors. We would find thatthe second derivative is positive for x > 5 and negative for x < 0: the graphis concave downward for x < 0, and upward for x > 5. There are no pointswhere the concavity changes within the domain, so no inflection points.

H. Sketch the Graph: This will be done at the lecture.

[1, Exercise 4.5.50, p. 323] To sketch the curve y = ex − 3e−x − 4x (cf. Figure 18,page 2207).


y

4

2

0

-2

-4

x

3210-1-2

Figure 18: Graph of y = ex − 3e−x − 4x


Solution:

A. Domain: The exponential function ex is defined for all real x; e−x also hasthe whole of R as its domain; and 4x also has domain R: so the domain off(x) = ex − 3e−x − 4x is R.

B. Intercepts: The y-intercept is f(0) = 1− 3− 0 = −2. We will defer consider-ation of the x-intercepts until we have more information about f .

C. Symmetry: f(−x) = e−x − 3ex + 4x. The equality f(x) = f(−x) would thusbe equivalent to the equation ex−e−x

x= 2, or to sinh x = x. While this equation

may have some solutions — x = 0 is one such value — it is not always true; ifit were true, then the derivatives of the two sides of the equation would alsobe equal, and cosh x would be constant, 1, which it certainly is not; thus fis not an even function. Similarly, it would follow from f(x) = −f(−x), thatex is negative; from such a contradiction we know that f is not odd either.After we have investigated the limits as x → ∞ we will be able to concludethat the function is not periodic either.

D. Asymptotes: The graph has no vertical asymptotes, since, being continuouseverywhere, it has a finite limit at every point in its domain.

limx→∞

f(x) = limx→∞

ex

(

1 − 3

ex− 4

x

ex

)

We can show, by L’Hospital’s Rule, that limx→∞

xex = lim

x→∞1ex = 0; evidently

limx→∞

3ex = 0. Hence, by the Product Law,

limx→∞

f(x) = limx→∞

ex

(

1 − 3

ex− 4

x

ex

)

= limx→∞

ex · limx→∞

(

1 − 3

ex− 4

x

ex

)

= limx→∞

ex · limx→∞

(

limx→∞

1 − limx→∞

(

3

ex

)

− limx→∞

(

4x

ex

)

)

= limx→∞

ex · (1 − 0 − 0)

= limx→∞

ex · 1 = +∞

In a similar way we can show that

limx→−∞

f(x) = limx→−∞

e−x(

e2x − 3 − 4x

e−x

)

= limx→−∞

e−x · limx→−∞

(

e2x − 3 − 4x

e−x

)


= limx→−∞

e−x ·(

limx→∞

e−2x − 3 − limx→−∞

(

4x

e−x

)

)

= limx→−∞

e−x · (0 − 3 − 0) = −∞

Thus there are no horizontal asymptotes to the graph of f .

E. Intervals of increase or decrease:

f(x) = ex + 3e−x − 4 =e2x − 4ex + 3

ex=

(ex − 1)(ex − 3)

ex.

The critical points are x = ln 1 = 0 and x = ln 3. We tabulate the signs ofthe derivative:

Interval ex − 1 ex − 3 ex f ′ fx < 0 − − + + increasingx = 0 0 −2 1 0 critical point

0 < x < ln 3 + − + − decreasingx = ln 3 + 0 + 0 critical pointln 3 < x + + + + increasing

By the First Derivative Test, both critical points are local extrema: x = 0 isa local maximum, and x = ln 3 is a local minimum.

We can now return to the question of the x-intercepts. There can be no zeroa of f such that a < 0, since then, applying the Mean Value Theorem tof on the interval a ≤ x ≤ 0, we would find that f(0)−f(a)

0−a= f ′(c) > 0, so

−2−00−a

> 0, which would imply that a > 0, contradicting the hypothesis. Sincef(0) = −2, and since f ′ < 0 for 0 < x < ln 3, the function values in thisinterval will be negative. f(ln 3) = 2−4 ln 3 = 2− ln 81 < 0. Finally, considerf on the interval ln 3 < x: here the function is increasing; by a familiar useof the Mean Value Theorem, the graph cannot cross the x-axis more thanonce. Since f(3) = e3 − 3

e3 − 12 > (2.7)3 − 3(2.7)3

− 12 > (2.7)3 − 12 > 0, weknow by the Intermediate Value Theorem that there is a point x between ln 3and 3 where f(x) = 0 — hence it is a unique point, and this is the secondx-intercept.

G. Concavity and Points of Inflection: Differentiating, we find that f ′′(x) =

ex − 3e−x =(ex+

√3)(ex−

√3)

ex . The factor ex +√

3 is always positive, as is the

factor ex. The factor ex −√

3 is negative for x < ln 32

and positive for x > ln 32

.Thus, to the left, the graph is concave downwards; and, to the right of ln 3

2,

the graph is concave downwards. Thus x = ln 32

is an inflection point.

Example D.50 One student asked that I consider the graph of the function f(x) =sin x

x.


1

0.6

-0.2

0.8

0.4

20 255 3015

x

0.2

010

Figure 19: Graph of f(x) =sin x

xfor x > 0


A. Domain: This function has as its domain R − {0}, since the fraction is not definedwhen the denominator is 0. However, we know that limx → 0f(x) = 1

B. Intercepts: There is no y-intercept, since 0 is not in the domain. The graph willcross the x-axis whenever sin x = 0, i.e., precisely at the integer multiples of π —infinitely often.

C. Symmetry: This function is even:

f(−x) =sin(−x)

−x=

− sin x

−x=

sin x

x= f(x)

for all x. It is not periodic, since the maximum magnitude of f(x) gets smaller asx grows.

D. Asymptotes: The curve has no vertical asymptotes, since there is no place wherea one-sided limit is infinite: even at x = 0 the limit is finite. As x → ∞,

−1

x≤ sin x

x≤ 1

x

and the extremes both approach 0; hence, by the Squeeze Theorem, limx→∞

= 0, and

y = 0 is a horizontal asymptote. The same limit holds as x → −∞, since thefunction is even, and the graph to the left of x = 0 is the mirror image of the graphto the right.

E. Intervals of Increase and Decrease: At this point the author’s proposed rubricsfor investigation become difficult for this function. We find that the derivative is

f ′(x) =x cos x − sin x

x2,

so the tangent is horizontal when tanx = x, i.e., where the curve y = x crosses thegraph of the tangent function. It is beyond this course to be able to extract furtherinformation about precisely where this happens, except to observe that there isgoing to be one crossing in every interval of length π. So there are going to beinfinitely many critical points; they are located “near” the odd multiples of π

2.

When x is equal to an odd multiple of π2, the graph touches one of the curves y = ± 1

x; it

is, in that sense, asymptotic to that pair of curves; more precisely, the distance betweeny = f(x) and y = 1

xapproaches 0 as x → ±∞ and also the distance between the given

curve and y = − 1x. So the curve is oscillating between those two curves, all the while

approaching the y-axis. A sketch is given in Figure 19


D.21.2 §4.6 Graphing with Calculus and Calculators

(This portion of the notes covers material for which there was not time for a classroomdiscussion.)

Because we are asking student to avoid the use of calculators in MATH 140/141, thissection is not examination material. You are, however, urged to peruse it, since someof the skills required in solving the problems is similar to what you will be expected toknow.

4.6 Exercises You might wish to try some of the problems to see how far you can getwithout having to resort to the use of a calculator.

[1, Exercise 4.5.26, p. 331] : “Describe how the graph of f varies as (the parameter)c varies...You should investigate how maximum and minimum points and inflectionpoints move when c changes. You should also identify any transitional values of cat which the basic shape of the curve changes. f(x) = x3 + cx.”

Solution:

A. Domain: The domain of f is the entire real line.

B. Intercepts: The y-intercept is f(0) = 0. The x-intercepts are the solutions tox3 + cx = 0: when c > 0 the only x-intercept is 0; when c ≤ 0 the x-interceptsare 0 and ±

√−c.

C. Symmetry: f(−x) = (−x)3 + c(−x) = −f(x). Thus these functions arealways odd . They are never even, and never periodic.

D. Asymptotes: As limx→±∞

= ±∞, there are no horizontal asymptotes. There are

also no vertical asymptotes, as the function is continuous on the whole realline.

E. Intervals of Increase or Decrease: f ′(x) = 3x2 + c. When c > 0, f ′ > 0for all x, and the function is always increasing. When c = 0, the functionis increasing for x > 0, and decreasing for x < 0. And, when c < 0, thefunction is increasing for x < −

√

− c3, decreasing for −

√

− c3

< x <√

− c3,

and increasing again for x >√

− c3. The points x = ±

√

− c3

are then criticalpoints.

F. Local Maximum and Minimum Values: For c > 0 there are no criticalpoints, and no local extrema. For c = 0 there is one critical point, x = 0;but this point is not a local extremum since f < 0 to its left and f > 0 toits right; so here also there are no local extrema. But, for c < 0, the criticalpoint x = −

√

− c3

is a local maximum, and the critical point x =√

− c3

is


a local minimum. These properties may be shown by either the First or theSecond Derivative Tests.

G. Concavity and Points of Inflection: f ′′(x) = 6x for all c: The graph isalways concave upward for x > 0, and concave downward for x < 0. Thusx = 0 is always an inflection point!

H. Sketch the Curve: This may be done at the lecture.


D.22 Supplementary Notes for the Lecture of Monday, Novem-ber 20th, 2006

Release Date: Monday, November 20th, 2006subject to correction

Before presenting the notes for today’s lecture, I am including, from my files, sketchesof solutions to problems on the December final examinations in this course in 2003 andin 2004. These will not be discussed in the lectures. However, I plan to discuss thesolutions to the December final examination for 2005 once I have finished covering thesyllabus.

D.22.1 Sketch of Solutions to Problems on the December, 2003 Final Ex-amination

We do not have files of solutions to old examinations, as I try to discourage students fromstudying from old exams. However, the following solutions were prepared for discussionin my classes in November, 2004, and I am making them available to you. I plan todiscuss a different examination with this year’s class, but you are invited to ask me orthe TA’s about any of the following examination questions. The sketches I provide inthese notes do not constitute “official” solutions of the examination questions, but aredesigned to explain the questions to current student.

Students were advised that there were two kinds of problems on this examination,each clearly marked as to its type:

• “Some of the questions on this paper require that you SHOW ALL YOUR WORK!

Their solutions are to be written in the space provided on the page where thequestion is printed. When that space is exhausted, you may write on the facingpage. Any solution may be continued on the last pages, or the back cover of thebooklet, but you must indicate any continuation clearly on the page where thequestion is printed!

• “Some of the questions on this paper require only BRIEF SOLUTIONS ; for theseyou are expected to write the correct answer in the box provided; you are not askedto show your work, and you should not expect partial marks for solutions that arenot completely correct.

“You are expected to simplify your answers wherever possible.”

1. BRIEF SOLUTIONS

[2 MARKS EACH] Give the limit in each of the following cases. If the limit doesnot exist, or is +∞ or −∞, write “DOES NOT EXIST”, +∞, or −∞ respectively.


(a) limx→3−

15 − 2x − x2

x − 3=?

Solution: The intention was that students would observe that the quadraticnumerator factors into −(x−3)(x+5), so that the function is equal to −(x+5)at all points except x = 3, where it is undefined. For the purpose of findingthe one-sided limit as x → 3−, where the value of the function at x = 3 isirrelevant, we can use the textbook result that polynomials are continuouseverywhere; hence the limit of −x − 5 is obtained by replacing x + 5 by thatfunction’s value at x = 3, i.e., by −3 − 5 = −8.

Of course, since students did not have to show their work, they could havefound this limit also by using l’Hospital’s Rule:

limx→3−

15 − 2x − x2

x − 3= lim

x→3−

−2 − 2x

1= −2 − 6 = −8

(b) limx→−∞

7x5 + 2x3 − x2 + 11

x5 − 3x4 + 2=?

Solution: For the limit of a quotient of polynomials as x becomes infinitepositively or negatively, I recommend dividing out the leading power of x:

limx→−∞

7x5 + 2x3 − x2 + 11

x5 − 3x4 + 2= lim

x→−∞

x5(

7 + 2x2 − 1

x3 + 11x5

)

x5(

1 − 3x

+ 2x5

)

= limx→−∞

7 + 2x2 − 1

x3 + 11x5

1 − 3x

+ 2x5

=7

1

Here also, l’Hospital’s Rule could have been used (5 times).

(c) limx→0

tan 5x

sin 4x=?

Solution: We try to transform the function into a product of quotients of theform sin x

x:

limx→0

tan 5x

sin 4x= lim

x→0

5x sin 5x

5x· 1

cos 5x· 4x

4x sin 4x

= limx→0

sin 5x

5x· 1

cos 5x· 4x

sin 4x· 5

4

= limx→0

sin 5x

5x· lim

x→0

1

cos 5x· lim

x→0

4x

sin 4x· lim

x→0

5

4

= 1 · 1

1· 1 · 5

4=

5

4

This problem could have been solved using l’Hospital’s Rule.


(d) limx→−5−

30 + 6x

|5 + x| =?

Solution: This is a limit as x → 5−; to the left side of 5, |5 + x| = −(5 + x).Accordingly

limx→−5−

30 + 6x

|5 + x| = limx→5−

−6 = −6 .

(e) limx→∞

(√

x2 + x −√

x2 − 4x) =?

Solution: This function is a difference of two functions, both of which becomeinfinite as x → ∞. We need to transform it into a function which does notinvolve an undefined combination of functions which are becoming infinite.The “standard” way of attacking this problem is to rationalize the surds, bymultiplying by

√x2 + x −

√x2 − 4x and dividing by the same quantity. We

obtain(x2 + x) − (x2 − 4x)√

x2 + x +√

x2 − 4x=

5x√x2 + x +

√x2 − 4x

At this point we could, perhaps, use l’Hospital’s Rule. But a simpler attackis to divide numerator and denominator by x; since x is positive, this isequivalent to dividing the surds by x2 under the root signs. We obtain

5x√x2 + x +

√x2 − 4x

=5

√

1 + 1x

+√

1 − 4x

which approaches 52

as x → ∞.

(f) limx→−∞

x2ex =?

Solution: While some of the preceding problems could be solved by l’Hospital’sRule, but also by other — preferred — methods, this problem, for most stu-dents, would need l’Hospital’s Rule.

limx→−∞

x2ex = limx→−∞

x2

e−x= lim

x→−∞

2x

−e−x= lim

x→−∞

2

e−x= 0

2. BRIEF SOLUTIONS

[2 MARKS EACH] Determine each of the following derivatives.

(a)d

dtesin t2 =?

Solution: This function is the composition of three functions: first t 7→ t2; thenu 7→ sin u, finally v 7→ ev. To differentiate we begin with the last function:


the derivative of ev is ev; here v = sin t2. We multiply by the derivative withrespect to u of sinu; here u = t2. Finally we multiply by the derivative of t2.The product is

esin t2 ·(

cos t2)

· 2t .

(b)d

dxln(e4x) =?

Solution: We expect students to observe that, as the exponential and loga-rithm functions are mutual inverses, this function is equal to 4x, so its deriva-tive is 4. Students could fail to make this observation and still solve theproblem correctly, using the Chain Rule.

(c)d

dx|x4| =?

Solution: Since the exponent is even, |x4| = x4, so the derivative is simply4x3. Students who attempted to solve the problem from first principles wouldlikely fail to observe that it is differentiable at x = 0, but could still determinethe correct derivative elsewhere.

(d)d

dxarcsin

√x =?

Solution: This is a straight-forward Chain Rule problem:

d

dxarcsin

√x =

1√

1 − |x|· 1

2√

x=

1

2√

x − x2

and is defined only for 0 < x < 1.

(e) If y(x) satisfies the equation(y(x))5 + x (y(x))2 + 9x4 = 1 , then y′(0) =?

Solution: It was intended that this problem be solved by implicit differentia-tion with respect to x:

5y4y′ + 1y2 + x · 2y · y′ + 36x3 = 0

which can be solved for y′ in terms of x and y.

(f) If g(x) =(

x6)

x2 , g′(0) =?

Solution: We might be tempted to solve this problem by “logarithmic differ-entiation”. In the latter we find the logarithm of g(x):

ln g(x) =x

2· ln(x6) =

x

2· 6 lnx = 3x ln x

whose derivative, g′(x)g(x)

is 3(ln x + 1). Hence the derivative of g(x) is

3x3x · (ln x + 1) .


However, this derivation applies only for x > 0, so it is not relevant for theproblem given. We do see, though, that as x → 0+, this function approaches−∞, suggesting that there is no derivative anyhow, as the slope is approachingthe vertical.

So how can we attack this problem? We will have to appeal to the definitionof the derivative, i.e.,

limx→0

x3x − 1

x.

In order to compute this limit, we will have to determine the two one-sidedlimits separately. Since

limx→0+

x3x − 1

x= lim

x→0+

e3x ln x − 1

x

and we can show by l’Hospital’s Rule that x ln x → 0 as x → 0+, the nu-merator approaches 0, as does the denominator, we may apply l’Hospital’sRule:

limx→0+

e3x ln x − 1

x= lim

x→0+3(ln x + 1)x3x .

Thus the earlier formula does apply in the limit, which is found to be −∞.And, as x → 0−, we have

limx→0−

x3x − 1

x= lim

x→0−

−e3x ln(−x) − 1

x.

An analysis similar to the case from the right shows that the limit here is +∞.Our definition of derivative does not permit infinite values: this function is notdifferentiable at 0. In fact, the graph has a cusp at the point (x, y) = (0, 1).

3. BRIEF SOLUTIONS

[3 MARKS EACH]

(a) Give equations for all of the vertical asymptotes of the graph of

f(x) =x

x2 + x − 2.

If there is none, write “NONE”.

Solution: The denominator factorizes into (x + 2)(x − 1) which is zero atx = −2 and x = 1. The numerator remains non-zero at these points, so thequotient is undefined and approaches ±∞ as x → −2 and x → 1 from eitherside. Hence x = −2 and x = 1 are both asymptotes of this graph.


(b) Determine all values of the constant c that will make the function

f(x) =

{

−4x + c when x < 0(x + c)2 when x ≥ 0

continuous from the right at x = 0. If there is none, write “NONE”.

Solution: Suppose first that the question had read “Determine all values of theconstant c that will make the function continuous at x = 0.” Then we couldproceed as follows. The functions that are being pieced together are bothpolynomials, and have one-sided limits at all values. As x → 0−, f(x) →−4(0) + c = c. As x → 0+, f(x) → (0 + c)2 = c2. For continuity these valuesmust be equal: c = c2 or c(c − 1) = 0, so c = 0 or 1. Both of these valuesmake the function continuous (not just continuous from one side).

But the problem asked that the function be “continuous from the right atx = 0”. Since the second line of the definition of the function gives its valueas (x + c)2 whenever x ≥ 0, this function is continuous from the right for allvalues of c.

(c) Determine all horizontal asymptotes to the following curve; if there is none,write “NONE”.

y = arctan(x2)

Solution: As x → ±∞, x2 → ∞, and its arctangent→ π2. Thus there is just

one horizontal asymptote, y = π2.

(d) If tanh x =3

5, sinh x =?

Solution: This problem can be approached in a variety of ways. If a studentremembered the identity, tanh2 x + sech2x = 1, he could then argue thatsech2x = 16

25. Then, observing that the hyperbolic secant is always positive,

he could conclude the correct value.

But we don’t expect you to memorize this identity. So, instead, solve theequation

ex − e−x

ex + e−x=

3

5

to obtain ex = ±2, where the + sign must be taken, because real exponentialsare never negative, so

sinhx =ex − 1

ex

2=

2 − 12

2=

3

4.

4. BRIEF SOLUTIONS

[4 MARKS EACH] The examiners are aware that you do not have a calculator.


(a) Determine all points on the curve

y3 − x2 = 4

where the tangent is horizontal. If there is none, write “NONE”.

Solution: By implicit differentiation with respect to x we see that 3y2·y′−2x =0. There is no point on the curve where y = 0, as this would entail that x2

be negative. Hence the only way in which y′ = 0 is that x = 0. The point ofcontact of this tangent with the curve is (0, 3

√4).

(b) On the interval −2 < x < 0, determine the values of x at which the functionf has a local (=relative) minimum, if it is known that

f ′(x) = (2x + 1)(x + 1)2(x + 3) .

If there is no local minimum, write “NONE”.

Solution: The derivative vanishes at x = −12, x = −1, and x = −3. Of these

3 points, the last is outside the domain of the function. The other 2 yieldtwo critical points. To determine which — if either — of them yields a localminimum, we must use either the First or the Second Derivative Test. Becauseof the complication of the function, it is easiest to use the First DerivativeTest:

x 2x + 1 x + 1 (x + 1)2 x + 3 f ′ f−2 ≤ x < −1 − − + + + decreasing

x = −1 critical point−1 < x < −1

2− + + + − decreasing

x = −12

critical point−1

2< x ≤ 0 + + + + + increasing

At x = −1 the function does not change from increasing/decreasing to de-creasing/increasing, so this is NOT a local maximum; at x = −1

2it changes

from decreasing to increasing, so the function has a local minimum.

(c) Find f(2) if it is known that

f ′(x) =x2 − x + 1

x, and

f(1) = 5 .

Solution: The general antiderivative of f is

F (x) =

∫(

x − 1 +1

x

)

dx =x2

2− x + ln x + C.


Imposing the condition that f(1) = 5 yields 5 = 12− 1 + C, so C = 11

2. It

follows that

f(2) =22

2− 2 + ln 2 + C = 2 − 2 + ln 2 +

11

2= ln 2 +

11

2.

5. SHOW ALL YOUR WORK!

[3 MARKS EACH]

For each of the following descriptions of a function or functions, either

• give an example of functions with the properties stated, or

• name or state a theorem, law, or rule which can be used to show that no suchfunction or functions exist.

(a) f(3) = 0, f(1) = −4, f ′(x) ≤ 1 for all x.

Solution: By the Mean Value Theorem there exists c such that 1 < c < 3 andf ′(c) = 0−(−4)

3−1= 2 > 1, which contradicts the hypothesis that the value of the

derivative can never exceed 1. From this contradiction we conclude that nosuch function f can exist.

(b) f is continuous on [−4, 4], f(−3) = −1, f(3) = 2, f(x) 6= 0 for all x.

Solution: A continuous function must, by the Intermediate Value Theorem,assume all values between any two values it assumes. Thus there must be apoint where f takes on the value 0 that lies between the values of −1 and2 that is is known to assume. From this contradiction we conclude that nofunction f of this type exists.

(c) limx→4

f(x) = 3, f(

limx→4

x)

= 2 .

Solution: If f were continuous at 4, then, since limx→4

x = 4, the two hypotheses

would be contradictory. We conclude that f cannot be continuous at 4. Anexample of a function f of this type is

f(x) =

{

3 when x 6= 42 when x = 4

(d) f ′(1) = 5 and f is not continuous at x = 1 .

Solution: We know that a function which has a derivative at x = 1 must becontinuous at x = 1. Thus the description is self-contradictory, and therecannot exist any such function f .



[12 MARKS] A balloon leaves the ground 100 metres from an observer, and risesvertically at the rate of 40 metres per minute. Determine the rate at which theangle of inclination of the observer’s line of sight (the angle between the line ofsight and the horizontal) is increasing at the instant when the balloon is exactly100 metres above the ground?

Solution: It appears that the examiners intended that the observer’s eye be atground level. Let’s place that eye at the origin, and the balloon on the x-axisat position (100, 0) at time t = 0. The position of the balloon at time t willbe (100, 40t). The angle subtended at the observer’s eye at time t is, therefore,

arctan 40t100

, whose derivative is25

1+( 2t5 )

2 . The balloon will be at height 100 when

t = 52, at which time the value of the derivative is 1

5radian/minute.


Let f(x) = x√

x + 3 .

(a) [2 MARKS] State the domain of f .

(b) [4 MARKS] Find the intervals of increase and the intervals of decrease of f .

Solution:√

x + 3 is defined precisely when x ≥ −3; the product with x doesnot restrict the domain. Thus the domain of f is x ≥ −3.

(c) [4 MARKS] Find the absolute (global) maximum and minimum values of f— if any — and the points where they are attained.

Solution: Differentiation by the product rule, followed by algebraic simplifi-cation, yields

f ′(x) =3

2· x + 2√

x + 3.

This function is 0 when x = −2, and is not defined when x = −3. For−3 < x < −2 f ′ < 0 so f is decreasing; for −2 < x f ′ > 0 so f is increasing.It follows that x = −2 is a global minimum of f . The global minimum valueof f is, therefore f(−2) = −2.

(d) [3 MARKS] Determine the intervals of concavity upwards and the intervals ofconcavity downwards, and the inflection points, if any.

Solution: We find after another differentiation and simplification that

f ′′(x) =3

4· x + 4

(√

x + 3)3.

This function is positive throughout the domain of f , so the graph is alwaysconcave upward, and there are not inflection points.


(e) [3 MARKS] Sketch the graph of y = f(x).


Let t represents time measured in seconds, and let C be a constant that is tobe determined. A particle moves so that its position on the x axis at time t isx(t) = t3 − 12t2 + 36t + C.

(a) [3 MARKS] Determine the acceleration of the particle as a function of time.

Solution:

v(t) = x′(t) = 3t2 − 24t + 36

a(t) = v′(t) = x′′ = 6t − 24

(b) [6 MARKS] The speed of the particle is the absolute value of its velocity.Determine the time intervals when the speed is increasing.

Solution: v(t) = 3(t − 2)(t − 6), while a(t) = 6(t − 4). The various signs willchange only at values t = 2, 4, 6. Let us tabulate the behavior of the velocityand speed first:

t t − 2 t − 6 v |v| d|v|dt

t < 2 − − + v 6(t − 4)2 < t < 6 − + − −v −6(t − 4)

t > 6 + + + v 6(t − 4)

We see that the derivative of the speed will be changing sign at t = 4. Wehave

t d|v|dt

t < 2 −2 < t < 4 +4 < t < 6 −

6 < t +

Thus the speed is increasing in the intervals 2 < t < 4 and 6 < t.

(c) [3 MARKS] Determine the value of C if it is known that the particle is at theorigin the whenever the acceleration is 0.

Solution: The acceleration has already been determined to equal 6(t−4), andis 0 precisely at time t = 4. We are thus being told that

0 = x(4) = 43 − 12(42) + 36(4) + C = 16 + C

so C = −16.


D.22.2 Sketch of Solutions to Problems on one of several versions of theDecember, 2004 Final Examination

Students were advised that there were two kinds of problems on this examination, eachclearly marked as to its type:





1. BRIEF SOLUTIONS

[3 MARKS EACH] Give the numeric value of each of the following limits if it exists;if the limit is +∞ or −∞, write +∞ or −∞ respectively. In all other cases write”NO FINITE OR INFINITE LIMIT”.

(a) limy→−∞

|y|y

=

ANSWER ONLY

Solution: For negative y, |y| = −y. It follows that

limy→−∞

|y|y

= limy→−∞

−y

y= lim

y→−∞(−1) = −1 ,

the limit of a constant function.

(b) limu→−∞

sin u

u=

ANSWER ONLY


Solution: Don’t confuse this question with the theorem that limu→0

sin uu

= 1. As

u → ∞ the numerator here oscillates between −1 and +1, but the denom-inator is approaching −∞, so the quotient approaches 0. Here is a formalproof:

Beginning with the inequalities

−1 ≤ sin u ≤ 1

we divide all members by u < 0, reversing the inequalities:

−1

u≥ sin u

u≥ 1

u

and allow u → ∞. The extremes approach 0; hence, by the Squeeze Theorem,

limu→−∞

sin u

u= 0.

(c) limn→∞

(

1 +1

n

)2n

=

ANSWER ONLY

Students should know the result [1, Theorem 6, p. 248] that limn→∞

(

1 +1

n

)n

=

e; the present function is the square of that function, so the limit is e2.

(d) limx→0

(

1

x− 1

sin x

)

=

ANSWER ONLY

Solution: This was [1, Exercise 45, p. 314]. It was a bad choice for a questionwhich requires no solutions, since we saw that some students solved it usingfallacious logic, and we still gave them the marks because I did not wish topenalize them for proofs that were not required, even if they were very, verywrong.

What was wrong was that some students reasoned that the two subtractedterms both approached +∞, so the difference was 0. This is nonsense, but itdoes happen to yield the correct answer in this case. Here is a correct solution:

limx→0

(

1

x− 1

sin x

)

= limx→0

(

sin x − x

x sin x

)


where numerator and denominator both approach 0. We apply l’Hospital’sRule:

limx→0

(

sin x − x

x sin x

)

= limx→0

cos x − 1

sin x + x cos x

= limx→0

sin x

cos x + cos x − x sin x

after a second application of l’Hospital’s Rule. In this last fraction the de-nominator approaches 2, but the numerator approaches 0. We may use the

Quotient Law, and conclude that the limit of the original function is0

2= 0.

(e) limt→∞

(

ln(3t2) − ln(t2 + 7))

=

ANSWER ONLY

This problem resembles [1, Exercise 63, p. 314].

limt→∞

(

ln(3t2) − ln(t2 + 7))

= limt→∞

ln3t2

t2 + 7

= limt→∞

ln3

1 + 7t2

= ln

(

limt→∞

3

1 + 7t2

)

= ln 3 .

Note that, contrary to the inference you might make from the location of theproblem in the textbook, you do not need to use l’Hospital’s Rule here.

(f) limx→8

√x + 8 −

√2x

x2 − 8x=

ANSWER ONLY

Solution:

limx→8

(√x + 8 −

√2x

x2 − 8x

)

= limx→8

√x + 8 −

√2x

x2 − 8x·√

x + 8 +√

2x√x + 8 +

√2x

= limx→8

(

(x + 8) − 2x

(x2 − 8x)(√

x + 8 +√

2x)

)


= limx→8

−1

x(√

x + 8 +√

2x).

The denominator approaches 8× (4+4) = 64, while the numerator → −1, sothe function approaches − 1

64as x → 8.

2. BRIEF SOLUTIONS

[3 MARKS EACH] Determine each of the following derivatives, and simplify youranswers as much as possible.

(a)d

dx

(

x2 + 3x

x

)

=

ANSWER ONLY

Solution:d

dx

(

x2 + 3x

x

)

=d

dx(x + 3) = 1 + 0 = 1.

(b)d

dt

(

t−3t)

=

ANSWER ONLY

d

dt

(

t−3t)

=d

dt

(

eln t)−3t

=d

dte−3t ln t

= e−3t ln t · d

dt(−3t ln t)

= e−3t ln t · (−3)

(

1

t· (t) + ln t · 1

)

= e−3t ln t · (−3) · (1 + ln t)

= t−3t · (−3) · (1 + ln t) .

(c)d

ds

(

tan(e2s) − e2 tan s)

=


ANSWER ONLY

Solution:

d

ds

(


= sec2(e2s) · e2s · 2 − e2 tan s · 2 sec2 s .

(d)d

dy

(

cosh2(3y))

=

ANSWER ONLY

Solution:d

dy

(

cosh2(3y))

= 2 cosh(3y) · sinh(3y) · 3 .

(This answer could be further simplified, as follows, but I didn’t expect stu-dents to know the identity required for hyperbolic functions:

6 cosh(3y) · sinh(3y) = 3 sinh(6y) .)

(e)d

dX

(

cos2 X − cos(X2))

=

ANSWER ONLY

d

dX

(

cos2 X − cos(X2))

= 2 cos X · (− sin X) +(

sin X2)

· 2X

3. BRIEF SOLUTIONS

(a) [5 MARKS] Find an equation for a line through the point (−2, 0) which istangent to the curve y = x2 and is not horizontal.


ANSWER ONLY

Solution: On the curve y = x2, the slope of the tangent at the point (x0, x20)

is 2x0. An equation for the tangent line is, thus,

y − x20 = 2x0(x − x0)

or y = 2x0x−x20. For this tangent line to pass through (−2, 0), the coordinates

of that point must satisfy the equation of the tangent, i.e., 0 = 2x0(−2)− x20,

implying that x0 = 0 or x0 = −4. The first of these alternatives gives ahorizontal tangent; hence x0 = −4, and the tangent line sought is

8x + y + 16 = 0 .

(Some students grabbed the only constant they could find in the problem —−2 — and simply found the equation of the tangent line at the point on thecurve with x = −2. This tangent line does not pass through the point (−2, 0),and is not the line that the question is looking for.)

(b) [5 MARKS] Determine values of the constants a and b that will make thefollowing function continuous at x = −6.

f(x) =

√x + 31 if x < −6a + b if x = −6

a(x + 5) if x > −6

ANSWER ONLY

Solution: This function is pieced together from a root function, a constant,and a linear function. The functions a + b (a constant), and a(x + 5) arecontinuous everywhere, and

√x + 31 is continuous on (−31,∞); but there

could be discontinuities where the piecing together occurs, i.e., at x = −6.The function is defined at that point. Thus all we need to check is that thelimit should exist at x = −6, i.e., that the limits from left and right separatelyshould exist and they should be equal; and that the common value of thesetwo one-sided limits should be the function value.


The limit from the left will be limx→−6−1

√x + 31 = 5; the limit from the right will

be limx→−6+

a(x + 5) = −a. Thus the equality a = −5 is required for continuity,

and will ensure that the 2-sided limit of the function exists at x = −6. Wemust now impose the condition that this limit, 5, be equal to the functionvalue, a + b, i.e., to −5 + b. It follows that b = 10.

(c) [5 MARKS] Determine values of the constants k and ℓ that will make thefollowing function differentiable at x = 1.

g(x) =

{

kx2 + ℓ if x ≤ 16x − 4 if x > 1

ANSWER ONLY

Solution: The functions being pieced together here are polynomials, whichwe know to be differentiable everywhere. As in the preceding problem, theonly point where there could be a question is where they are pieced together,at x = 1. At that point we know that differentiability implies continuity.We will first make the function continuous there. The limit from the left islim

x→1−1(kx2 + ℓ) = k + ℓ. The limit from the right is lim

x→1+= 6− 4 = 2; thus we

havek + ℓ = 2 .

The difference quotient from the right is (6x−4)−(k+ℓ)x−1

= 6x−6x−1

= 6 . Thus, if thefunction is differentiable at x = 1, the value of the derivative will be 6. Wenow look at the difference quotient from the left:

kx2 + ℓ − (k + ℓ)

x − 1= k(x + 1) .

As x → 1, this quotient approaches 2k, which must be equal to 6; hence k = 3and ℓ = −1.


(a) [6 MARKS] Coffee is draining from a conical filter of depth 10 cm and diameter10 cm (at the top) into a cylindrical coffee pot of diameter 12 cm, at the rateof 100 cm3/min. Determine how fast, in cm/min, the level of coffee in the potis rising when the coffee in the filter is 3 cm deep?


Solution: This was the easier part of a 2-part problem that had constituted awritten assignment for the students. This part of the problem did not dependon the shape of the filter. If the coffee is h cm deep in the cylinder, the volumeV is π62h, so the rate of change of volume with time t is

100 =dV

dt= 36π

dh

dt.

The rate of increase of volume is, therefore the constant,100

36π.

(b) [6 MARKS] You are given that y = y(t) is a function of t satisfying t3y+ty3 =2. Assuming that y(1) = 1, determine the values of y′(1) and y′′(1).

Solution: Differentiating the equation for y implicitly with respect to t yields

3t2y + t3 · dy

dt+ y3 + t · 3y2 · dy

dt= 0

or

t(t2 + 3y2)dy

dt+ y(3t2 + y2) = 0 .

Setting t = 1 in the original equation yields y3+y−2 = 0; y3+y−2 factorizesinto (y−1)(y2 +2y +2) = 0, of which y = 1 is the only solution. Substituting

t = 1, y = 1 in the equation for the derivative yieldsdy

dt= −1. Finally we

must differentiate, for example, the equation

y′ = −y(3t2 + y2)

t(t2 + 3y2).

We obtainy′′

= −y′(3t2 + y2)(6ty + 2y2y′)(t3 + 3ty2) − y(3t2 + y2)(3t2 + 3y2 + 6tyy′)

t2(t2 + 3y2)2

which vanishes when (t, y, y′) = (1, 1,−1).


(a) [4 MARKS] Prove that the function x3 + 9x2 + 33x assumes the value −8 atleast once.

Solution: The function takes on the value 0 > −8 when x = 0. It takes on thevalue −1+9−33 = −25 < −8 at x = −1. Since the function is continuous, wemay apply the Intermediate Value Theorem to infer that the function takeson the value −8 somewhere in the interval −1 < x < 0.


(b) [8 MARKS] Using the Mean Value Theorem or Rolle’s Theorem — no othermethods will be accepted — prove carefully that x3 + 9x2 + 33x takes on thevalue −8 at most once.

Solution: Suppose the function took on the value −8 twice. Since the func-tion is differentiable everywhere, it satisfies the conditions of the Mean ValueTheorem. There would have to be a point x = c in the interval −1 < x < 0at which f ′(c) = 0. But this is contradicted by

f ′(x) = 3(x2 + 6x + 11) = 3(x + 3)2 + 6 ≥ 6 > 0

for all x. We conclude that our assumption of the existence of 2 roots is false— the function attains the value −8 only once.


[10 MARKS] Using the calculus carefully, determine how to express 8 as the sumof 2 nonnegative real numbers such that the sum of the square of the first and thecube of the second is as small as possible.

Solution: Denote the second number by b, so the first will be 8− b. Then we wishto minimize the sum f(b) = (8 − b)2 + b3. We find that

f ′(b) = 2(8 − b)(−1) + 3b2 = (3b + 8)(b − 2)

so the only critical point in the domain 0 ≤ b ≤ 8 is b = 2, at which f ′′(b) = 6b + 2evaluates to 14 > 0, so b = 2 is a local minimum, where the function value isf(b) = 62 +23 = 44. At the end points of the domain we have f(0) = 0+512 > 44and f(8) = 64 + 0 > 44. Hence the sum is minimized globally when the secondsummand is 2, and the first is 8 − 2 = 6.


Let f(x) =

(

x

x + 3

)2

.


Solution: The fraction is meaningful only where the denominator is non-zero.Thus the domain of the fraction, and hence of its square f , is R − {−3}.


Solution:

f ′(x) =

(

x

x + 3

)

· d

dt

(

1 − 3

x − 3

)

=6x

(x + 3)3



which vanishes only at x = 0. Because the degree of the factor x − 0 is 1,which is odd, the function changes sign at 0. It is decreasing for −3 < x < 0and increasing for 0 < x. In the interval x < −3, the degree of the negativefactor x + 3 is odd, and the degree of the negative factor x is also odd, so thederivative is positive for x < −3.

(c) [4 MARKS] Determine the intervals of concavity upwards and the intervals ofconcavity downwards, and the inflection points, if any.

Solution: Another differentiation yields

f ′′(x) = −12x − 3

2

(x + 3)4.

The odd factor x − 32

changes sign at x = 32, so x = 3

2is an inflection point:

the graph is concave upward in both the intervals (−∞,−3) and(

−3,−32

)

.For x > 3

2the graph is concave downwards.

(d) [3 MARKS] Sketch the graph of y = f(x), showing — clearly labelled — allhorizontal and all vertical asymptotes.

Solution: There is a vertical asymptote at x = −3, because the functionapproaches ∞ from both sides of x = −3. As x → ±∞, f(x) → 1, so y = 1is a horizontal asymptote: the graph is above the horizontal asymptote forx,−3, and below it in the first quadrant. It crosses that asymptote from aboveto below at x = −3

2.


[6 MARKS] Consider the function f(x) = 5x + 9 near x = −1. Is it, or is it nottrue that f is continuous at x = −1? If the statement is true, prove it carefully,using the ǫ-δ definition. If it is false, prove that carefully.

Solution: This topic has been deleted from this year’s syllabus, and I will notprovide a solution here.

D.22.3 §4.5 Summary of Curve Sketching (conclusion)

I include one more long example.


[1, Exercise 4.5.32, p. 323] To sketch the graph of f(x) = sin x − tanx.

Solution: (cf. Figure 20, page 2234 in these notes)


x

420-2-4-6y

4

2

0

-2

-4

Figure 20: Portion of the graph of y = sin x − tanx, showing its vertical asymptotes

A. Domain: The sine function is defined for all real numbers; but the tangentfunction is not defined at odd multiples of π

2; hence the domain of f is R −

{(2n + 1)π2|n = any integer}.

B. Intercepts: The y-intercept is f(0) = 0 − 0 = 0. The x-intercepts will bethose x for which sin x = tan x, equivalently, for which sin x(1 − sec x) = 0,equivalently, for which either sin x = 0 or cosx = 1. sin x = 0 when x is aninteger multiple of π, while cos x = 1 when x is an integer multiple of 2π: thusthe second factor vanishes only when the first vanishes, and it suffices to lookonly at the first. To summarize, the x-intercepts are the integer multiples ofπ.

C. Symmetry: This function is odd , since f(−x) = sin(−x)−tan(−x) = − sin x+tanx = −f(x). It is also periodic, since the sine function has period 2π andthe tangent function has period π: f has period 2π. Thus it suffices to study


the function over any interval of length 2π — the local extrema, inflectionpoints, etc, will be repeated periodically over the whole domain.

D. Asymptotes: We know that

limx→−(π

2 )+

f(x) = 0 − (−∞) = ∞

limx→(π

2 )−

f(x) = 0 − (∞) = −∞

limx→(π

2 )+

f(x) = 0 − (−∞) = ∞

limx→( 3π

2 )−

f(x) = 0 − (∞) = −∞

and the pattern will repeat over the whole domain. There are thus verticalasymptotes as all odd multiples of π

2; in each case the asymptote occurs for

two reasons, because a single infinite one-sided limit is sufficient for a verticalasymptote.

A function with a positive period cannot have a horizontal asymptote, sincethe maximum distance of the graph from any horizontal line will be oscillatingas we pass to ±∞.

E. Intervals of increase or decrease:

f ′(x) = cos x − sec2 x =cos3 x − 1

cos2 x=

(cos x − 1)(cos2 x + cos x + 1)

cos2 x.

The numerator is a product of two functions. Of these, the second function isa quadratic function which cannot be zero, since its discriminant is 12−4 < 0.Thus the only way in which f ′(x) = 0 is that cosx = 0, which, as mentionedearlier, occurs when x = 2nπ, where n is any integer. These are the onlycritical numbers, and they are infinitely many. The odd multiples of π

2are,

indeed, points where the derivative is not defined; but, as they are not in thedomain of the function, they are not critical numbers. More generally, weneed to determine where the derivative is positive and negative. It is neverpositive, since the factor cos2 x + cos x + 1 always has the sign of the squareterm of the quadratic, here +; and the denominator is a square, so it also ispositive; the factor cos x− 1 cannot be positive, but it can be 0 — when x isan even integer multiple of π. Thus the function is decreasing on the intervals(

−π2, 0)

,(

0, +π2

)

,(

π2, +3π

2

)

, and then the pattern repeats itself periodically:two intervals of length π

2, and one interval of length π.

F. Local Maximum and Minimum Values: We found the critical numbers inthe preceding paragraph. As the function is decreasing on both sides of eachof these points, the function has no local extrema.


G. Concavity and Points of Inflection: Differentiating, we find that

f ′′(x) = − sin x − 2 sec2 x · tan x = − sin x(

1 + 2 sec3 x)

.

Here the factor 1 + 2 sec3 x cannot be 0, but will alternate between positiveand negative as x passes over the discontinuities at the odd multiples of π

2.

The factor sin x can be 0, and will change sign at the mid point of each of theintervals over which f is defined. Thus the sign of f ′′ will change at each ofthose midpoints, so each of those mid points — the integer multiples of π —will be an inflection point of the graph.

All of the foregoing could be suggesting that the function has branches that alllook the same. But, while the function is periodic with period 2π, there will be2 branches in each interval of length 2π, and they will not have exactly the sameshape. The difference is that the curve will be tangent to the x-axis at points 0,±2π, ±4π, . . ., but not at the intervening x-asymptotes. The tangency occurs atthe points separating the two shorter intervals of decrease.

D.22.4 §4.7 Optimization Problems

Here is a variation of the steps the textbook suggests for solving optimization problems.

Spend some time understanding the problem: What quantity is to be optimized? Whatis the “independent” variable? In Calculus 1 we can consider only problems with asingle independent variable; if your formulation of the problem appears to involvemore than one variable, there must be a relationship between those variables, andyou will have to determine that before you can complete your solution. (In Calculus3 you will consider functions of several independent variables.)

The textbook suggests that you sketch the geometry, if the problem has been statedin geometric language. But remember that your sketch is not a formal part of amathematical proof: it will be very helpful to you, but you must not attempt tobase any of your reasoning on the sketch.

Introduce notation for the independent variable(s) and the function (=“dependent vari-able”), and express the dependencies that hold between them. You may subse-quently find that your formulation of the problem is not an easy one to solve, andyou may wish to return to this step to express the dependent quantity in a differentway. Don’t be stubborn to continue with a formulation that proves to be difficultfor you.


In your formulation be careful to describe precisely the domain of the function that youare optimizing. Remember that the domain may not be the entire natural domainof the function, as there may be constraints in the problem that restrict certainquantities.

It is usually the global extrema that are of interest in optimization problems. Thismeans that you will have to consider the end points of closed intervals. You mayalso have to consider functions whose domains are not closed intervals.

4.7 Exercises

[1, Exercise 4.7.10, p. 337] “A box with a square base and open top must have avolume of 32,000 cm3. Find the dimensions of the box that minimize the amountof material used.”

Solution: The box has three spatial dimensions, but two are the same. Is this aproblem with 2 independent variables? No. We can call the side of the base x,and the height h; but these two variables are related, since x2h = 32000. We canuse this last equation to express either of x and h in terms of the other. Let’s waitbefore we decide which of them to take as our independent variable.

The area of the base is x2, and the area of the sides is 4x · h, so the quantity wewish to minimize is x2 + 4xh. If we were to take h as the independent variable,this area expression would involve fractional powers of h, which would be slightlyunpleasant to work with; so I decide to express the area in terms of x, and toeliminate h, even though the alternative approach could have been used, with abit more difficulty. Denote the area by

A(x) = x2 + 4xh = x2 + 4x · 32000

x2= x2 +

128000

x.

The domain of the function is 0 < x < ∞. We determine the derivatives of A(x):

A′(x) = 2x − 128000

x2

A′′(x) = 2 +256000

x3

The only critical point of A is at x = 40. Since A′′(40) > 0, this is a localminimum. Alternatively, we could have used the First Derivative Test. Indeed,the First Derivative Test is better since the fact that A′ is decreasing for x < 40 andincreasing for x > 40 shows that this local minimum is, in fact, a global minimum(cf. [1, First Derivative Test for Absolute Extreme Values, p. 334]). Finally wehave to answer the question as stated: the dimensions of the box using the smallestamount of material are


side of base = 40 cm.height = 32000

1600cm. = 20 cm.

(Note that the same method could be used for variants of this problem. For ex-ample, the box could have a top, or the cost per unit area of the base could bedifferent from that cost for the sides, as in [1, Exercise 4.7.12, p. 337].)

[1, Exercise 4.7.22, p. 337] (This problem was discussed only very briefly at the lec-ture.) “Find the dimensions of the rectangle of largest area that has its base onthe x-axis and its other 2 vertices above the x-axis, and lying on the parabolay = 8 − x2.”

Solution: In this problem the author has prescribed a coordinate system, so youdon’t have to decide where to place it. He has not named the points, so you maywish to do so; while this naming could be shown on your figure, a formal proofshould have the points named in the wording of the proof. For example, I willassert that the vertices of the base are A(a, 0) and B(b, 0), and, without limitingthe generality, I will decide to let A be the leftmost of the two, so that a < b.The vertices above them will then be C(a, 8 − a2) and D(b, 8 − b2). As this is tobe a rectangle, the side CD must be parallel to AB, which is on the x-axis; thus8 − a2 = 8 − b2, so b = ±a; since B is distinct from A, I will have to have b = −a.Since I decided that B should be to the right of A, I will now specify that a < 0:the vertices of the rectangle are now A(a, 0), B(−a, 0), C(a, 8−a2), D(−a, 8−a2);remember, a < 0. The base of this rectangle has length |2a| and the height is8 − a2, so the area — the function we wish to maximize — is

f(a) = (8 − a2)|2a| .

And what is the domain of f? We are told that C and D must lie above the x-axis,so 8− a2 ≥ 0 and −

√8 ≤ a ≤

√8. Earlier we restricted a to be non-positive; thus

the domain is −√

8 ≤ a ≤ 0. We can simplify the formula we gave for f(a):

f(a) = 2a3 − 16a .

Hence f ′(a) = 6a2−16, for −√

8 < a < 0. The critical points of this function would

be a = ±√

83

if a could take any value. But our restricting a to be negative leaves

only one of these points in the domain: a = −√

83. To determine the extrema of

f we evaluate the function at the critical point and at the end point of the closed

interval that is the domain, and find that f = 0 at the end points, and f = 643

√

23

at the critical point. We conclude that the dimensions of the largest rectangle are:

width is 4√

23, and height is 16

3.


In this solution I did not follow one of the pieces of advice I gave you: I persistedwith an awkward formulation of the problem (with my variable a restricted to benegative) rather than going back to the drawing board and reformulating it in amore intuitive way. While I did obtain the answer, it was not the simplest way ofattacking the problem.

[1, Exercise 4.7.54, p. 339] “A steel pipe is being carried down a hallway 9 feet wide.At the end of the hall there is a right-angled turn into a narrower hallway 6 feetwide. What is the length of the longest pipe that can be carried horizontally aroundthe corner?”

Solution: The textbook provides a figure. The figure shows an angle named θbetween the pipe and the outer wall of the 9-foot hallway, where it is touching theinner corner of the hallway and tight against the two outer walls. In this way theauthor has helped you by eliminating the doubts as to which independent variableto use.

Let’s call the maximum length of the pipe f(θ); the permitted values for θ are0 < θ < π

2: note that θ cannot equal either 0 or π

2. Call the length of pipe that can

fit tightly at angle θ f(θ); expressing it in terms of the widths of the two hallwayswe find it is a sum,

f(θ) = 9 csc θ + 6 sec θ .

Its derivative, after simplification, is

f ′(θ) =6 sin3 θ − 9 cos3 θ

sin2 θ · cos2 θ=

(cos θ) · (6 tan3 θ − 9)

sin3 θ.

We need to determine the global minimum for f , in order to determine the longestpipe that can turn the corner horizontally. The function has just one critical

point in the interval, when tan θ = 3

√

96, corresponding to θ = arctan 3

√

32. For

tan θ < 3

√

96

f ′ < 0, and for tan θ > 3

√

96

f ′ > 0: thus θ = arctan 3

√

32

is a

local minimum of f . Since these inequalities on the sides of the critical numberoccur throughout the domain, we can conclude that the point is, in fact, a globalminimum (cf. “First Derivative Test for Absolute Extreme Values” [1, p. 334]).

When tan θ = 3

√

32,

f(θ) = 9 csc θ + 6 sec θ

= (6 + 9 cot θ)√

1 + tan2 θ

=

(

6 + 93

√

2

3

)

√

1 +

(

3

2

)23


and this is the longest length of pipe that can turn the corner.

Example D.51 (This problem was not discussed in the lecture.) Consider the followingproblem from the Review Exercises: [1, Review Exercise 52 p. 363] “Find the point onthe hyperbola xy = 8 that is closest to the point (3, 0).”Solution: As independent variable take the abscissa of a point

(

x, 8x

)

on the hyper-bola; thus the domain will be the positive and negative coordinate axes, with the originremoved. As a first attack on the problem we can try to minimize the function

f(x) =

√

(x − 3)2 +

(

8

x− 0

)2

.

This will work, but will produce a computationally difficult problem. Instead we willminimize the square of the distance:

g(x) = f(x)2 = (x − 3)2 +

(

8

x− 0

)2

, (196)

since the square and square root functions are both increasing functions, so the minimumdistance occurs precisely where the minimum square distance occurs.

We compute the derivatives, and find that

g′(x) = 2(x − 3) − 128x−3

g′′(x) = 2 + 384x−4 > 0

which tells us, by the Second Derivative Test, that any local extremum is a local mini-mum. Examining the first derivative more carefully, we find that

g′(x) =2(x4 − 3x3 − 64)

x3=

2(x − 4)(x3 + x2 + 4x + 16)

x3. (197)

Let’s examine the domain of g more carefully. The graph has two branches — one inthe first quadrant, and one in the third quadrant. The distance of (3, 0) from the 3rdquadrant is surely at least 3, while its distance to the branch of the graph in the firstquadrant is no more than the vertical distance, 8

3. Thus the closest point to the given

point will be on the branch in the first quadrant, and we need not even consider the otherbranch. This observation has been made be examining the geometry. Another argumentwould simply examine the sum in (196), and observe that the second summand is asquare, so it is positive; if x < 0, the first sum cannot be less than 32. We see fromfactorization (197) that one critical number is x = 4. There cannot be any more criticalnumbers on the positive axis (0,∞), since the polynomial factor x3 +x2 +4x+16 cannot


be zero for positive x. But there could be critical numbers on the negative coordinateaxis; in fact, there is one at approximately x = −1.344389236. By arguing that theclosest point we seek could not be on the negative branch of the graph, we avoid havingto deal with this second critical point.

As for the critical number x = 4, the derivative is negative to the left of it, andpositive to the right of it, so, by the First Derivative Test, x = 4 is a local minimum.Indeed, x = 4 is a global minimum [1, First Derivative Test for Absolute Extreme Values,p. 334]. The y-coordinate is 8

4= 2, so the closest point is (4, 2).

The point that we have ignored on the negative branch of the graph is of no interestin this problem as stated; but a small change in the problem would make it relevant. Wecould simply ask for the closest point on the negative branch of the graph to the point(3, 0); or, equivalently, for the closest point on the given curve to the point (−3, 0).

D.22.5 §4.8 Applications to Business and Economics – OMIT

This section is not part of the syllabus of MATH 140.

4.8 Exercises

D.22.6 §4.9 Newton’s Method – OMIT

This section is not part of the syllabus of MATH 140.


D.23 Supplementary Notes for the Lecture of November 22nd,2006

Release Date: Wednesday, November 22nd, 2006subject to correction

The First Derivative Test for Absolute Extreme Values. In the discussion onpage 2237 of the solution of [1, Exercise 4.7.10, p. 337]. I drew a conclusion from thefacts that there was just one critical point, and that the derivative was always of onesign on either side of the point, and that those signs were different. We were able to inferthat the critical point was a global extremum. Hitherto most of the problems in whichwe considered global extrema were defined on a finite, closed interval. We had used thistype of reasoning previously in connection with the First Derivative Test, which is a testto a local extremum. The textbook calls this reasoning the First Derivative Test forAbsolute Extreme Values [1, p. 334].

D.23.1 §4.10 Antiderivatives

Definition D.8 A function F is an antiderivative of a function f on an interval I ifF ′ = f on that interval.

The term antiderivative is a “calculus book term” rather than a “mathematician’s term”.Mathematicians are more likely to use the traditional term indefinite integral . Textbookssometimes differentiate between the terms antiderivative and indefinite integral , butthese are likely to be rationalizations to justify the newer term. In practice the termsare interchangeable. The traditional notation for an antiderivative of a function f(x) is

∫

f(x) dx .

We can reformulate Corollary D.35 to the Mean Value Theorem as

Corollary D.52 [to the Mean Value Theorem] If F is one antiderivative of f on aninterval I, then the most general antiderivative of f on I is

F (x) + C ,

where C is an arbitrary constant.

Where we wish to denote the family of all antiderivatives of f(x), we usually write

∫

f(x) dx + C ,


Function One antiderivativef(x) F (x)g(x) G(x)

f(x) + g(x) F (x) + G(x)

xn (n 6= −1)xn+1

n + 11

xln |x|

ex ex

cos x sin xsin x − cos xsec2 x tan x

sec x tan x sec xcsc x cot x − csc x

1

1 + x2arctan x

1

1 + x2−arccot x

1√1 − x2

arcsin x

1√1 − x2

− arccos x

Table 12: Some Antiderivatives

where C or some other symbol is understood to represent a constant of integration thatranges over all real numbers, and will possibly be determined by additional informationthat is yet to be supplied about the particular antiderivative that is sought. C is not theonly letter that is used to represent this constant.

Tables of Antiderivatives Much of the information we have been collecting aboutderivatives can be recast into information about antiderivatives. For example, if we knowthat F and G are antiderivatives of f and g, then we know the antiderivatives shown inTable D.52 on page 2242. There are several observations to be made from this table:

1. There are several functions having two or more apparently different antiderivatives.This is, in one sense, no surprise, as we saw above that a function with at leastone antiderivative has infinitely many. What may be surprising is that we areseeing functions that we may not have observed earlier to be essentially the same— differing only by an additive constant.


2. While there is an antiderivative given for the sum of two functions with knownantiderivatives, there is none given for their product. There will be situationswhen we will know the antiderivatives of two functions, but we will not have oneavailable for their product! This is not because an antiderivative “does not exist”,but because we may not have constructed the elements out of which to describean antiderivative. We will return to this topic in MATH 141. One example will bethe functions

xex2

and1

x

of which we know such antiderivatives as

ex2

2and ln |x|

respectively, but where we will not have available an antiderivative of ex2

.

3. A substantial portion of our time in MATH 141 will be spend in investigating pro-cedures for determining antiderivatives for functions. While the tedious determina-tion of antiderivatives is more and more being done by computers, the proceduresfollowed in those searches are very much the business of mathematicians and theother users of mathematics.

The Geometry of Antiderivatives Read the textbook to see how a “derivative field”enables one to sketch the graphs of antiderivatives. This subsection is intended to preparereaders for parts of Chapter 9, which chapter is omitted from the syllabus of MATH 141(because students at McGill meet the subject matter — “ordinary” differential equations— in other courses, e.g., MATH 315). You should, however, understand that the graphsof antiderivatives of any function form a family of parallel curves, separated by a constantvertical distance.

Rectilinear Motion The theory in this section is a restatement of the material in [1,§3.3, pp. 199-200]. To solve some of the problems you will need to know the accelerationdue to gravity, usually taken to be approximately 32 feet/sec−2. I will illustrate withsome problems.

4.10 Exercises

[1, Exercise 10, p. 358] Find the most general antiderivative of the function

g(x) =5 − 4x3 + 2x6

x6


and check your answer by differentiation.

Solution: The “trick” here is not to attempt to work with the given ratio, butto transform the function into a sum: you know the antiderivatives of a sum offunctions with known antiderivatives, but you know little about the antiderivativesof a quotient. Since

5 − 4x3 + 2x6

x6=

5

x6− 4

x3+

2

1= 5x−6 − 4x−3 + 2x0 ,

one antiderivative is −x−5 + 2x−2 + 2x1 , and the most general antiderivative is

−x−5 + 2x−2 + 2x1 + C

where C is any real constant. We check by differentiation:

d

dx

(

−x−5 + 2x−2 + 2x1 + C)

= −(−5)x−5−1 + 2(−2)x−2−1 + 2(1)x1−1 + 0

which can be seen to equal the given function after the summands are taken to acommon denominator.

Note, however, that the given functions are not defined at x = 0, and can surelynot be continuous there. Thus there could be different constants of integration thatapply for the two intervals, (0,∞) and (−∞, 0) where the functions are defined.

[1, Exercise 14, p. 358] Find the most general antiderivative of the function

h(θ) =sin θ

cos2 θ

and check your answer by differentiation.

Solution: In MATH 141 we will see a general way of attacking problems of thistype. For now the solution depends upon observing, for example, that

sin θ = − d

dθcos θ .

Thus

sin θ

cos2 θ= −

d

dθcos θ

cos2 θ=

d

dθ

1

cos θ= sec θ

so one antiderivative is sec θ. This could also have been noticed if it had beenobserved that

h(θ) =sin θ

cos2 θ=

sin θ

cos θ · 1

sec θ

= tan θ · sec θ ,


which, it may be recalled, is the derivative of sec θ. The most general antiderivativeis sec θ + C.

As in the previous problem solved, the functions are undefined at certain real num-bers: the constants of integration could be different on the different intervals wherethe functions are defined. In this case the singularities occur at the odd integermultiples of π

2, so there are infinitely many intervals of length π for which a con-

stant of integration needs to be specified, and these could be combined arbitrarily,depending on the other information known about the function.

[1, Exercise 26, p. 358] Find f , if it is known that f ′(x) = 8x3+12x+3, and f(1) = 6.

Solution: We have seen that we can find the antiderivative of a sum of powers bytaking the appropriate constant multiples of powers of x. Since d

dxx4 = 4x3, an

antiderivative of 8x3 is 2x4; analogously we find antiderivatives of the other twosummands; hence the most general antiderivative of f ′(x) is

f(x) = 2x4 + 6x2 + 3x + C ,

were C is a constant to be determined. Imposing the condition that f(1) = 6 yields

6 = f(1) = 2(1)4 + 6(1)2 + 3(1) + C = 11 + C

from which we conclude that C = −5, and that

f(x) = 2x4 + 6x2 + 3x − 5 .

[1, Exercise 36, p. 358] Find f , if it is known that f ′′(t) = 3√t, f(4) = 20, f ′(4) = 7.

Solution: The most general antiderivative of f ′′ is of the form

f ′(t) = 3 (k) t−12+1 + C

where C is a constant of integration, and k is a constant that needs to be determinedby differentiation. Doing that, we find that

f ′′(t) = 3(k)

(

1

2

)

t−12 + 0

so 3 = 3(k)

(

1

2

)

, k = 2 and

f ′(t) = 6t12 + C ; (198)


a second such operation gives

f(t) = 6

(

2

3

)

t12+1 + Ct + E (199)

where E is a second constant of integration. The constant C may be determinedin equation (198) by setting t = 4 there:

7 = f ′(4) = 6(

412

)

+ C = 12 + C ,

implying that C = 7 − 12 = −5. Substituting x = 4 in equation (199) yields

20 = f(4) = 4(

432

)

+ (−5)(4) + E ,

implying that E = 20 − 4(8) − (−20) = 8, so

f(t) = 4t32 − 5t + 8 .

[1, Exercise 4.10.42, p. 359] “Find f if it is known that f ′′′(x) = sin x, f(0) =f ′(0) = f ′′(0) = 1.”

Solution: What we will be finding is an antiderivative of an antiderivative of anantiderivative of f , and we will impose the preceding 3 “initial” values on thefunction to determine the values of the “constants of integration”. We begin byinterpreting the equation f ′′′(x) = sin x as

(f ′′)′(x) = sin x .

One antiderivative of sin x is − cos x, so the most general antiderivative is − cos x+C, where C is a constant to be determined. Thus we know that

f ′′(x) = − cos x + C

and hence1 = f ′′(0) = − cos 0 + C = −1 + C ,

from which we conclude that C = 2, so

f ′′(x) = − cos x + 2 .

We interpret this last equation as

(f ′)′(x) = − cos x + 2 .


An antiderivative of − cos x is − sin x, and an antiderivative of 2 is 2x, so we inferthat

f ′(x) = − sin x + 2x + K ,

where K is another constant of integration; use another letter to name that constantso that it is not confused with the earlier constant. Imposing another of the initialvalue conditions, we have

1 = f ′(0) = − sin 0 + 2 · 0 + K = K ,

which determines this constant of integration:

f ′(x) = − sin x + 2x + 1 .

Finally, an antiderivative of − sin x + 2x + 1 is cos x + x2 + x, so

f(x) = cos x + x2 + x + L ,

where L is a third constant of integration. The third initial condition gives

1 = f(0) = cos 0 + 02 + 0 + L

so L = 0, and f(x) = cos x + x2 + x.

(In some problems the information given will be such that the evaluation of theconstants may have to wait until the general antiderivative has been determined,and then a set of equations will be determined, to be solved for the constants.)

[1, Exercise 4.10.44, p. 359] “Find a function f such that f ′(x) = x3 and the linex + y = 0 is tangent to the graph of f .”

Solution: The candidates are all functions f(x) =x4

4+C, where C is some constant,

to be determined. A general point on this curve is

(

a,a4

4+ C

)

, at which the slope

of the tangent is f ′(a) = a3. (I am using the letter a to denote the abscissa of thegeneral point in order that I can comfortably talk about the equation of the tangent,and have the letter x available for its traditional use.) An equation for the tangentat this general point is

y −(

a4

4+ C

)

= a3(x − a) .

For this to be the same line as x + y = 0, they must have the same slope, so

−1 = a3 ,


implying that a = −1, i.e., that the point of tangency is

(

−1,1

4+ C

)

. Thus the

tangent line is

y −(

1

4+ C

)

= −(x + 1)

or x + y = C − 3

4.

We have imposed the condition that the tangent line have the desired slope, but thegiven line passes through the origin; imposing that condition (that (x, y) = (0, 0)satisfy the equation) here gives

0 + 0 = C − 3

4,

from which we conclude that C = 34, so the desired function is

f(x) =x4 + 3

4.

(This problem could be solved more elegantly by a more algebraic treatment; I haveavoided that route since not all students in the course are taking or have taken acourse in linear algebra.)

[1, Exercise 64, p. 360] (not discussed in the lecture) A particle is moving with thefollowing data; find the position of the particle:

a(t) = 10 + 3t − 3t2

s(0) = 0

s(2) = 10

Solution: We know that position at time t is given by the displacement functions(t); that velocity is v(t) = s′(t), and that acceleration is a(t) = v′(t) = s′′(t).Note that this is the first problem I have considered where the data are not theinitial values of the function and its derivatives, but are “boundary” values of thefunction at different times.

The most general antiderivative of a(t) = s′′(t) is

v(t) = s′(t) = 10t +3

2t2 − t3 + C1 ,


where C1 is a constant of integration. A second operation of antidifferentiationgives

s(t) = 5t2 +3

2· 1

3t3 − 1

4t4 + C1t + C2 , (200)

where C2 is a second constant of integration. Imposing the boundary data gives

0 = s(0) = 5 · 0 +1

2· 0 − 1

4· 0 + C1 · 0 + C2

10 = s(2) = 5 · 4 +1

2· 8 − 1

4· 16 + C1 · 2 + C2

implying that C1 = −5 and C2 = 0, so

s(t) = 5t2 +1

2t3 − 1

4t4 − 5t + 0 ,

[1, Exercise 4.10.68, p. 360] (not discussed in the lecture) “Two balls are thrownupward from the edge of a cliff (432 feet above the ground). The first is thrownwith a speed of 48 feet/s, and the other is thrown a second later with a speedof 24 feet/s. Do the balls ever pass each other?” Assume that the acceleration is

a(t) =dv

dt= −32 feet/s2.

Solution: This problem is based on [1, Example 8, pp. 357-358].

Let’s set up a coordinate system for the moving balls. They will move along they-axis, and I will take the origin to be at the bottom of the cliff, with the positivedirection upward, and distance on the axis measured in feet. I will measure timefrom the release of the first ball, and will denote the positions of the first andsecond balls at time t respectively by y1(t) and y2(t).

The motion of the first ball is governed by the equation y′′1(x) = −32. The most

general antiderivative of −32 is −32t + C1 — this is the velocity of the first ball.If we impose the initial velocity condition for that ball, we obtain

48 = y′1(0) = −32(0) + C1,

from which we conclude that C1 = 48, and that the velocity of the first ball is givenby

y′1(t) = −32t + 48 .

From this we conclude that the position of the first ball is given by the equation

y1(t) = −16t2 + 48t + K1 ,


where K is a constant of integration that must be determined. Imposing theinformation we have about the initial position of the ball, we obtain

432 = y1(0) = −16(02) + 48(0) + K1 ,

so K1 = 432, and the position of the first ball is given by

y1(t) = −16t2 + 48t + 432 .

Proceeding to the second ball, its acceleration satisfies the equation y′′2(x) = −32,

so its velocity will bey′

2(t) = −32t + C2 .

When we impose the initial velocity condition for the second ball — but be careful,the release time is t = 1 here — we obtain

24 = y′2(1) = −32(1) + C2 ,

so C2 = 56 andy′

2(t) = −32t + 56

for t ≥ 1; the equation is not valid for t < 1. A second iteration of antidifferentia-tion yields

y2(t) = −16t2 + 56t + K2

and we impose the condition that the ball be at position y2 = 432 when t = 1:

432 = −16(12) + 56(1) + K2

so K2 = 392, and

y2(t) = −16t2 + 56t + 392 (t ≥ 1) .

Finally we are ready to answer the question. The excess of height of the first ballover the second at time t is

y1(t) − y2(t) = −8t + 40 .

Indeed, this difference is 0 once, when t = 5: the balls pass each other 4 secondsafter release of the second ball.

(If this exercise had yielded a value of t between 0 and 1 the answer would have beenNO, since the equation we are using for y2 applies only for t ≥ 1.

Some students observed that the wording is somewhat ambiguous, as it appears to permitone ball to pass the other when the other is stationery, which was probably not the


intention. A more careful solution might proceed as follows: First determine the positionof the first ball:

y1(x) =

432 when t < 0

−16t2 + 48t + 432 when 0 ≤ t ≤ 3+3√

132

0 when t > 3+3√

132

by solving to determine when the ball hits the ground. Then do the same calculationsfor the 2nd ball:

y2(x) =

432 when t < 1−16t2 + 56t + 392 = −8(2t − 13)(t + 3) when 1 ≤ t ≤ 13

20 when t > 13

2

When we work with y1 − y2 we need to restrict t to the interval from

t = 1

tot = the minimum of 3

2(1 +√

13) and 132 ,

i.e. 1 ≤ t ≤ 132 . The value of t = 5 which we found is in that interval, so it is the point

we seek; had it been outside of the interval it would have been irrelevant, and the balls

would not pass in midair.)

Example D.53 A Problem on the First Derivative Test for Global Extrema[1, Review Exercise 50, p. 363]. “Find two positive integers such that the sum of thefirst number and four times the second number is 1000, and the product of the numbersis as large as possible.”Solution: While the problem is phrased in terms of two numbers, we have to express itas a single variable problem, since that is the only calculus machinery we have availablebefore MATH 222. Call the second number u, so the first number will be 1000 − 4u.(Note that this definition is “cleaner” than one that would involve assigning a name tothe first number; but both approaches would lead to the same final answer.)

The function we wish to maximize is (1000 − 4u)u, and I propose to call it P (u).The domain of P for the purpose of the problem is 0 ≤ u ≤ 250, since it would not bemeaningful to permit the function P (u) to be negative. Thus the domain is a closedinterval. We have

P (u) = (1000 − 4u)u = 4(250u− u2) ,

P ′(u) = 4(250 − 2u) , and

P ′′(u) = −8u .


The only critical number is u = 2502

= 125. We apply the Closed Interval Method, andtabulate the function values at the critical point and the end points of the domain:

u 125 0 250P(u) 62,500 0 0

The global maximum occurs at u = 125, so the numbers must be 125 and 1, 000−4(125) =500.



Release Date: Monday, November 27nd, 2006subject to correction

D.24.1 Sketch of Solutions to Problems on One Version of the December,2005 Final Examination






1. BRIEF SOLUTIONS

[2 MARKS EACH] Give the numeric value of each of the following limits if it exists;if the limit is +∞ or −∞, write +∞ or −∞ respectively. In all other cases write“NO FINITE OR INFINITE LIMIT”.

(a) limy→−∞

(

√

y2 + y + y)

=

ANSWER ONLY

Solution:

limy→−∞

(

√

y2 + y + y)

= limy→−∞

(

(

√

y2 + y + y)

·√

y2 + y − y√

y2 + y − y

)


= limy→−∞

(

(y2 + y) − y2

√

y2 + y − y

)

= limy→−∞

(y2 + y) − y2

√

y2(

1 + 1y

)

− y

= limy→−∞

(y2 + y) − y2

√

y2√

1 + 1y− y

= limy→−∞

(y2 + y) − y2

|y|√

1 + 1y− y

(the square root is always non-negative)

= limy→−∞

y

−y√

1 + 1y− y

as y < 0

= limy→−∞

1

−√

1 + 1y− 1

=1

−1 − 1= −1

2.

(b) limx→∞

(

e−x sinh x)

=

ANSWER ONLY

Solution:

limx→∞

(

e−x sinh x)

= limx→∞

e−x · ex − e−x

2

= limx→∞

1 − e−2x

2=

1 − 0

2=

1

2.

(c) limx→8

√x + 8 +

√2x

x2 + 8x=


ANSWER ONLY

Solution:

limx→8

√x + 8 +

√2x

x2 + 8x=

√8 + 8 +

√2 · 8

82 + 8 · 8=

4 + 4

64 + 64=

1

16.

(d) limx→1

(

1

1 − x+

1

ln x

)

=

ANSWER ONLY

Solution: We cannot use the Difference Law because neither summand has alimit as x → ∞. Accordingly we transform the function into one that may bemore amenable; we begin by taking the fractions to a common denominator.

limx→1

(

1

1 − x+

1

ln x

)

= limx→1

ln x − (x − 1)

(1 − x) ln x.

Next we apply l’Hospital’s Rule, since numerator and denominator both ap-proach 0:

limx→1

ln x − (x − 1)

(1 − x) ln x= lim

x→1

1x− 1

− ln x + 1−xx

= limx→1

1 − x

−x ln x + 1 − x

= limx→1

−1

−1 ln x − xx− 1

by a second application of the 0/0 form of L’Hospital’s Rule .

= limx→1

−1

−1 · 0 − 1 − 1=

1

2.

(e) limx→∞

sin ex

cos ex=


ANSWER ONLY

Solution: The quotient is equal to tan ex. As x → ∞, ex → ∞. The tangenthas no limit as its argument becomes large, but takes on all real values as ex

passes through values in an interval of length π.

2. BRIEF SOLUTIONS

[2 MARKS EACH] Evaluate each of the following, and simplify your answers asmuch as possible.

(a)d

dx

(

x2 + 3x

x

)

=

ANSWER ONLY

Solution:

d

dx

(

x2 + 3x

x

)

=d

dx(x + 3) for x 6= 0

= 1 + 0 = 1 .

(The function is not defined for x = 0.)

(b)d

du(uu) =

ANSWER ONLY

Solution: This problem can be solved in different ways. For example,

d

du(uu) =

d

du

((

eln u)u)

since logarithm and exponential are mutual inverses

=d

du

(

eu ln u)



=(

eu ln u)

· d

du(u lnu)

by the Chain Rule

= uu ·(

1 · ln u +u

u

)

= uu · (ln u + 1) .

Equivalently, one could use the method of “logarithmic differentiation”: De-fine y = uu, and take logarithms, to obtain ln y = u lnu; then differentiateboth sides of the equation with respect to u (the left side implicitly), and solvethe resulting equation to obtain d

duy in terms of u alone.

(c) An antiderivative F (x) of f(x) = 5x4 + 2x5 such that F (0) = 3 is

ANSWER ONLY

Solution: All antiderivatives of this polynomial have the form F (x) = x5 +26x6 + C, where C is a constant of integration. Imposing the initial condition

that F (0) = 3 yields 3 = 0 + 0 + C; hence the antiderivative sought is

F (x) = x5 +1

3x6 + 3 .

(d) If f(x) = x3 + 7, its inverse function f−1(x) =

ANSWER ONLY

Solution: Solving the equation y = x3+7 yields a unique solution, x = 3√

y − 7.Thus f−1(y) = 3

√y − 7. But the problem asked that the inverse function be

expressed in terms of an independent variable named x, not y; hence

f−1(x) = 3√

x − 7 .

(e)d

dx

(

|x|4)

=


ANSWER ONLY

We cannot differentiate the absolute value function at x = 0. However, |x|2 =x2, so

d

dx

(

|x|4)

=d

dx

(

x4)

= 4x3.


(a) [6 MARKS] Showing all your work, determine values of the constants a andb that will make the following function continuous everywhere.

f(x) =

(1 + x)1x if x > 0

a + bx if −1 ≤ x ≤ 0sin(x + 1)

x + 1if x < −1

Solution: The three pieces of the function are, indeed, continuous everywhere;but we need to ensure continuity where the pieces are combined, i.e., at thepoints x = −1 and x = 0. We know that lim

x→−1−

sin(x+1)x+1

= 1, and that

limx→−1+

(a + bx) = a − b. The two-sided limit will exist if and only if these two

one-sided limits are equal, i.e., if and only if

a − b = 1 .

As x → 0+, (1 + x)1x → e; as x → 0−, a + bx → a. For the limit as x → 0 to

exist we must thus impose the condition that

a = e .

We can now solve the two equations, to conclude that b = e − 1.

(b) [4 MARKS] Determine whether f is differentiable at x = −1. (For the purposeof this question you may assume that e is approximately 2.72.)

Solution: We must consider the limit as x → −1 of the difference quotient

f(x) − f(−1)

x + 1=

{

(a+bx)−(a−b)x+1

= b = e − 1 if 0 > x > −1sin(x+1)

x+1−(a−b)

x+1=

sin(x+1)x+1

−1

x+1if x < −1


We may apply l’Hospital’s Rule to the lower ratio, obtaining

limx→−1−

sin(x+1)x+1

− 1

x + 1= lim

x→−1−

sin(x + 1) − (x + 1)

(x + 1)2

= limx→−1−

cos(x + 1) − 1

2(x + 1)

= limx→−1−

− sin(x + 1)

2x= 0 6= e − 1 .

Since the limit from the left is different from the limit from the right, the2-sided limit does not exist; the derivative is defined to be that 2-sided limit— consequently the function is not differentiable at x = −1.


[5 MARKS] Let f(x) = x2ex. Prove carefully by mathematical induction that

dnf

dxn(x) =

(

x2 + 2nx + (n − 1)n)

· ex


Solution: Let S(n) denote the nth case of the preceding alleged equality. S(0)states that

f(x) =(

x2 + 0x + 0))

,

which is known to be true, as it is our definition of f . Suppose that it is knownthat

dnf

dxn(x) =

(

x2 + 2nx + (n − 1)n)

· ex .

Then, differentiating with respect to x, we obtain

d

dx

(

dnf

dxn(x)

)

= (2x + 2n) · ex +(

x2 + 2nx + (n − 1)n)

· ex

=(

x2 + 2(n + 1)x + n(n + 1))

· ex

which is precisely S(n + 1). We may conclude, by the Principle of MathematicalInduction, that S(n) is true for all non-negative integers n.

(This topic was not included in the syllabus of MATH 140 2006 09.)


[5 MARKS] Let g(x) = 2x − 3 + cos x. Use Rolle’s Theorem or the Mean ValueTheorem, to prove carefully that there exists exactly one real number x such thatg(x) = 0. (π may be taken to be approximately 3.14.)

Solution:


(a) Since g(0) = 2(0)−3+1 = −2 < 0, and g(π) = 2π−3−1 > 6−3−1 = 2 > 0,and since the function g is continuous (being a sum of functions known to becontinuous), we may conclude by the Intermediate Value Theorem that thereis a point x such that 0 < x < π at which g(x) = 0.

(b) Suppose that there were two such zeros for the function g. Then, by Rolle’sTheorem applied to the interval bounded by those zeros, we could concludethat g′ = 0 at some point. But g′(x) = 2 − sin x, and this function rangesin value between 1 and 3 — it cannot be zero! From this contradiction weinfer that the assumed existence of two zeros for the function is false — thefunction cannot equal zero more than once.

(c) Thus the number of zeros is not less than 1, and not more than 1: we concludethat there is a unique zero.

TO BE COMPLETED AT WEDNESDAY’S LECTURE



Release Date: Wednesday, November 29nd, 2006subject to correction

D.25.1 Sketch of Solutions to Problems on One Version of the December,2005 Final Examination (conclusion)


[10 MARKS] A rectangular poster is to be printed on a rectangular board of min-imum area, leaving margins at the 4 sides. The top and bottom margins are each10 cm, and the side margins are each 4 cm. If the printed area on the poster isfixed at 1,000 cm2, find the best dimensions for the board. Show all of your work,and justify all of your statements. In your solution, you are expected to carefullyapply either the First or the Second Derivative Test, naming the test as you applyit.

Solution: Denote the width and height of the board by w, h, measured in centime-tres; these functions will be assumed to be positive, but we cannot assume anyupper bound for either of them. Then the width and height of the printed areaare, respectively, w − 8 and h − 20, and the constraint on the printed area is that

(w − 8)(h − 20) = 1000 ,

which implies that w = 1000h−20

+8 . The area of the board, which is to be minimized,is

f(h) = wh =

(

1000

h − 20+ 8

)

· h =8(h2 + 105h)

h − 20.

Differentiating f yields, after reduction

f ′(h) =8(h − 70)(h + 30)

(h − 20)2,

hence the only internal extremum is h = 70; (the value h = −30 does not lie in thedomain). For 0 < h < 70 f ′(h) < 0; and for 70 < h f ′(h) > 0. Hence h attains itsonly local minimum at this point, and h = 70 is the global maximum point. Theminimum use of board will occur uniquely when h = 70 and

w =1000

70 − 20+ 8 = 28 .



Let f(x) =√

x2 − 1 .

(a) [1 MARK] State the domain of f .

Solution: The function is defined wherever the argument of the square root isnon-negative; i.e., where |x| ≥ 1.

(b) [1 MARK] State precisely where f is differentiable.

Solution: This function is differentiable everywhere except where x = ±1.While the function is, indeed, defined at those points, it is not differentiablefor two different reasons. First, differentiability requires that the function bedefined on both sides of the point, and f is not defined for x in the interval−1 < x < 1. Secondly, the one-sided limits of the difference quotient on theside where the limit is defined is infinite, as the tangents are approaching thevertical. We do not admit a value of ∞ or −∞ when we speak of a functionas being defined .

(c) [2 MARKS] Define when a line x = a is a vertical asymptote to the graph off .

Solution: A line x = a is a vertical asymptote if the either of the one-sidedlimits of the function is ±∞.

(d) [2 MARKS] Either

i. Find all vertical asymptotes; or

ii. Explain why the graph has no vertical asymptotes.

Solution: This function does not approach infinity, even though the tangentlines to the curve are approaching the vertical.

(e) [2 MARKS] Determine the global maximum value of f , or explain why thereis none.

Solution: As x → ∞, and also as x → −∞, f(x) → ∞. Thus the functionhas no global maximum.

(f) [2 MARKS] Determine the global minimum value of f , or explain why thereis none.

Solution: This function is a square root, and cannot be negative. Since it isequal to 0 at two places — x = ±1 — it attains a global minimum there.


A function f(t) satisfies, for all real numbers t, the equation

t3 + f(t)3 + 6t2 · f(t) = 8 .


(a) [5 MARKS] Find an equation for the tangent to the graph y = f(t) at thepoint (t, y) = (2, 0).

Solution: Differentiating the given equation implicitly with respect to t yields

3t2 + 3f(t)2 · f ′(t) + 12t · f(t) + 6t2 · f ′(t) = 0 ,

which can be solved to yield

f ′ = − t2 + 4tf(t)

(f(t))2 + 2t2.

When t = 2 and f(t) = 0, this yields f ′(2) = −1

2. An equation of the tangent

is y − 0 = −12(t − 2), or t + 2y = 2.

(b) [5 MARKS] Showing all your work, determine the value of f ′′(2).

Solution: Differentiating the ratio obtained for f ′(t) yields

f ′′ = −(2t + 4f(t) + 4tf ′(t)) ((f(t))2 + 2t2) − t(t + 4f(t))(2f(t)f ′(t) + 4t)

((f(t))2 + 2t2)2

which yields

f ′′(2) = −(4 − 4)(

14

+ 8)

− 2(2)(8)

(0 + 8)2=

1

2.

Additional Exercises One student asked about a WeBWorK problem involvinga ladder passing over a fence to touch a wall. The problem was essentially like [1,Exercise 34, p. 338]. I sketched the situation, and stated that there could be severalequivalent approaches. One of them might involve taking the variable to be some anglein the configuration; another — the approach I took — could take the distance from thefoot of the ladder to the fence as the variable. I called that distance x, and obtaineda rather complicated function; when we evaluated the derivative, we obtained a ratiowhose numerator was of the form c(x3 − a3), whose only zero was obvious. Then it wasa matter of applying the First Derivative Test for Global Extrema [1, p. 334]. The useof an angle as the variable would have made the calculations easier: see my solution of[1, Exercise 54, p. 339] beginning on page 2239 of these notes.

Following are some other worked solutions I found in my notes file from last year;they were not discussed at your lecture.

[1, Review Exercise 54, p. 363] “Find the volume of the largest (right) circular conethat can be inscribed in a sphere of radius r.”


Solution: A right circular cone is a cone constructed from a disk by erecting a lineperpendicular to the plane of the disk, at the centre thereof, and joining a point onthat line — the apex of the cone to all points in the boundary of the disk. Supposethat the sphere is formed from the circle x2 + y2 = r2 by revolving the circle aboutthe y-axis. Think of the base of the cone as cutting the xy-plane in a line parallelto the x-axis — say the line y = a, where −r ≤ a ≤ +a; denote the points ofintersection of this line with the circle by A(−

√r2 − a2, a) and B(−

√r2 − a2, a);

the apex of the cone will be the point (0, r) in the xy-plane. The mid-point of ABis the point (0, a), which is the centre of the circular base of the cone. We are nowready to describe the function to be optimized.

The volume of the cone is

1

3× (area of base) × (height) .

The base is a disk of radius√

r2 − a2, so its area is π(r2 − a2); the height of thecone is the distance from the apex C(0, r) to the centre (0, a) of the base, i.e., r−a.Thus we wish to maximize the function

f(a) =1

3· π(r2 − a2)(r − a) =

π

3(r − a)2(r + a)

over the domain − r ≤ a ≤ +r . Differentiation yields

f ′(a) =π

3

(

2(r − a)(−1)(r + a) + (r − a)2)

= −π

3(r − a)(r + 3a) = −π

3

(

r2 + 2ar − 3a2)

f ′′(a) = −2π

3(r + a) .

The only points of the domain that could be critical points of the function are

a = r and a = −r

3, and the value a = r is excluded because it is an end-point of

the domain. By the Second Derivative Test, the fact that

f ′′(

−r

3

)

= −4πr

9< 0

tells us that the point is a local maximum. At the end points of the domain f = 0,

so the global maximum is at the unique critical point, a = −r

3. The problem asks

for the maximum value of the volume; it is

f(

−r

3

)

=32

81πr3 .


[1, Review Exercise 58, p. 363] “A metal storage tank with volume V is to be con-structed in the shape of a right circular cylinder surmounted by a hemisphere.What dimensions will require the least amount of metal.”

Solution: Let r be the radius of the hemisphere and the base of the cylinder, andlet h be the height of the cylinder. Then the volume is

V =1

2· 4π

3r3 + πr2h

and so we may solve to obtain h in terms of r:

h =V

πr2− 2r

3.

The area of sheet metal used (including the base) is

1

2· 4πr2 + 2πr · h + πr2 =

5π

3· r2 +

2V

r

which we shall denote by A(r). The domain of function A is 0 < r < ∞ — not aclosed interval!

The derivatives are

A′(v) =10π

3· r − 2V

r2=

2(5πr3 − 3V )

3r2,

A′′(v) =10π

3+

4V

r3> 0 .

The first derivative vanishes only at

(

3V

5π

)13

; from the second derivative we see

that this is a local minimum point of A. Moreover, if we factorize the formula forA′ as

A′(r) =10π

3r2

(

r −(

3V

5π

) 13

)(

r2 +

(

3V

5π

) 13

r +

(

3V

5π

) 23

)

we see that the last factor is always positive (because it is a quadratic factor withoutreal factors, so its sign is always that of the square term); consequently the deriva-tive changes sign from negative to positive as r passes from left to right through

the value r =(

3V5π

)13 . It follows that the function attains its global minimum at

r =(

3V5π

)13 .


E Assignments from Previous Years

E.1 Fall 1998 Problem Assignments

(For most of these problems the answer was in the back of the (then) text-book [25][27],but the full solution was not in the Student Solution Manual [26][28].)

Assignment Due Exercise NumbersNumber Date

1 25 Sept./98 Chapter 2 Miscellaneous Problems: 2,5, 9, 12, 17, 21, 26, 28, 32, 35, 38, 40,50, 56, 57, 59, 61

2 9 Oct./98 §3.1: 11, 17, 20, 36, 41§3.2: 2, 5, 9, 15, 21, 33, 45, 48, 53§3.3: 14, 15, 17, 44, 50, 53

3 23 Oct./98 §3.4: 45, 50§3.5: 2, 6, 9, 21, 27, 45§3.6: 21, 26, 32§3.7: 62, 68, 74

4 6 Nov./98 §3.8: 17, 21, 26, 29, 35, 47(deferred to §3.9: 3, 8, 11, 23, 27, 50, 51, 59

9 Nov.) Chapter 7 Miscellaneous Problems: 3,11, 17, 21Chapter 8 Miscellaneous Problems: 5, 9

5 23 Nov./98 §4.2: 3, 8, 13, 20, 26, 40, 41, 42§4.3: 15, 23, 29, 33§4.4: 5, 11, 15, 30§4.5: 21, 32§4.6: 31, 49§4.7: 2, 6, 9, 15, 29, 47

Table 13: 1998 Problem Assignments

E.2 Fall 1999 Problem Assignments

E.2.1 First Fall 1999 Problem Assignment, with Solutions

Students were advised as follows in the problem set and its solutions:


• This assignment is intended to help you determine gaps in your background. Itwill not be graded, but the solutions will be made available on the Web. If youhave difficulty with the problems or the solutions which will be posted, you shoulddiscuss these with one of the tutors.

• Do not use a calculator when solving these problems, even for simple arithmetic.(You may, however, wish to use a calculator afterwards to verify whether youranswers are ‘reasonable’.)

• Do not attempt to approximate square roots, or π. Where possible, however,formulæ should always be simplified .

• Unless specifically stated in degrees, all angles should be assumed to be expressedin radians. (Remember that the straight angle, 180◦ is equal to π radians, so anangle of 1◦ is equal to π

180radians, and d◦ is equal to πd

180radians.)

• Caveat lector! As in any duplicated materials, there could be misprints or errors.

1. (a) Give a right-angled triangle in which one angle is α = π3, and use the triangle

to determine the values of sin α, cos α, tan α.

(b) Give a right-angled triangle in which one angle is β = π4, and use the triangle

to determine the values of sin β, cos β, cot β, sec β.

It is not sufficient to state the values: you should explain how you determine themin terms of the lengths of the sides of your triangle.

Solution:

(a) We take the right half of an equilateral triangle, each of whose sides has length2. In ∆DBC ∠C = π

3= α. DB bisects ∠ADC and meets AC in its midpoint,

B. Then

sin α =|BD||CD| =

√3

2

cos α =|BC||CD| =

1

2

tanα =|BD||BC| =

√3

1=

√3

(b) Here we take an isosceles ∆EFG, in which EF = EG = 1, and ∠F = 45◦ = β.

sin β =|EG||FG| =

1√2


��

��

TTTTTTTTTTTTTT

TTT

A B C

D

1 1

2 2√

3

60◦90◦

��E G

F

1

1√

2

45◦90◦

cos β =|EF ||FG| =

1√2

cot β =|EF ||EG| = 1

sec β =|FG||EF | =

√2

1=

√2

(Note: Some students may have been taught never to leave a surd in thedenominator, as in the fraction 1√

2above. This convention derives from the

difficulties, in the days before calculators, of working with numbers like√

2.While it still is useful, for hand calculations, to confine surds to the numerator,you may work with fractions like 1√

2if you are happy with them.)

2. State, without proof,

(a) a formula which expresses sin(x + y) in terms of the sines and cosines of xand y.

(b) a relationship between sin x and sin(−x), and another between cosx andcos(−x).

(c) By applying the results of parts 2b, 2a above, determine a formula that ex-presses sin(x − y) in terms of the sines and cosines of x and y.


The formulæ you state should be valid for all values of x and y.

Solution:

(a)sin(x + y) = sin x cos y + cos x sin y (201)

[25, A-15 (6)]

(b) sin(−x) = − sin x; cos(−x) = − cos x [25, A-15 (4)]

(c) Replacing y by −y throughout (201), one obtains

sin(x − y) = sin x cos(−y) + cos x sin(−y)

= sin x cos y + cos x(− sin y)

= sin x cos y − cos x sin y (202)

3. (a) By substituting x = α = π3

and y = β = π4

in the formula in 2c, determinethe value of sin 15◦.

(b) Using the result of 3a, determine the value of cos π12

.

Solution:

(a)

sinπ

12= sin

(

π

3−

pi4

)

= sinπ

3cos

π

4− cos

π

3sin

π

4

=

(√3

2

)

·(

1√2

)

−(

1

2

)

·(

1√2

)

=

√3 − 1

2√

2=

√6 −

√2

4.

(b) cos2 π12

= 1 − (√

3−1)2

8= 2+

√3

4; hence cos π

12is one of the square roots of 2+

√3

4.

Since the angle is in the first quadrant, the sign is positive. It follows that

cos π12

=√

2+√

34

. This can be seen — you were not expected to see this —

that the square root is√

3+12√

2.

4. State, without proof, formulæ which express cos(x + y) and cos(x− y) in terms ofthe sines and cosines of x and y.


Solution:

cos(x + y) = cos x cos y − sin x sin y (203)

cos(x − y) = cos x cos y + sin x sin y (204)

[25, A-15 (7)]

5. (a) Specialize the formula in Problem 4 (by making a suitable choice for y interms of x) to express cos 2x in terms of sin x and cos x.

(b) Use the identity sin2 x + cos2 x = 1 to express cos 2x in terms of cos x alone;and in terms of sin x alone.

(c) Apply your formulæ in 5b to determine the values of cos π6

and cos π2

from thesines of π

12and π

4.

Solution:

(a) Taking y := x in (203) yields

cos 2x = cos2 x − sin2 x (205)

(b)

cos 2x = cos2 x − (1 − cos2 x) = 2 cos2 x − 1 (206)

= (1 − sin2 x) − sin2 x = 1 − 2 sin2 x (207)

(c)

cosπ

6= cos 2

( π

12

)

= 1 − 2 sin2( π

12

)

= 1 − 2 · (√

3 − 1)2

8= 1 − 2 · 1 + 3 − 2

√3

8=

√3

2

cosπ

2= 1 − 2 sin2 π

4= 1 − 2

(

1√2

)2

= 0

6. Find an equation for each of the following lines in the plane:

(a) The line through the points (−1, 4) and (−2, 1).

(b) The line through the point (4,−1) which is perpendicular to the line

2x + y = −7 . (208)

Solution: (This problem is from the 1998 Final examination in 189-112A.)


(a) The slope of the line will be 1−4(−2)−(−1)

= 3, so an equation is y−4 = 3(x−(−1))or y = 3x + 7.

(b) The slope of the line we seek will be the negative reciprocal of the line y =−2x − 7, which has slope −2; thus the line will have slope 1

2. One equation

will be y − (−1) = 12(x − 4), or x − 2y = 6.

7. (a) It is claimed that the equation (2x + y + 7)2 = 0 represents the same pointsas the equation 2x + y + 7 = 0. Determine whether the claim is correct.

(b) Determine what is represented by the equation (2x + y)2 = (−7)2, i.e. by theequation obtained by squaring both sides of equation (208).

Solution:

(a) The equation of a line is a constraint satisfied by the coordinates of its points,and satisfied by the coordinates of no other points. As (2x + y + 7)2 maybe zero if and only if 2x + y + 7 = 0, this is another equation for the sameline. (Some authors would, however, say that this is “the equation of twocoincident lines”.)

(b) (The symbol ⇔ means that the statements which it connects are logicallyequivalent.)

(2x + y)2 = (−7)2 ⇔ (2x + y)2 − (−7)2 = 0

⇔ (2x + y − 7)(2x − y + 7) = 0

⇔ 2x + y − 7 = 0 or 2x + y + 7 = 0

Thus the given equation is satisfied by the coordinates of the points on eitherof the parallel lines 2x+y−7 = 0 and 2x+y +7 = 0; the equation representsthe union of the sets of points on the two lines.

8. Determine the centre and radius of the circle x2 + y2 + 4x − 6y + 3 = 0.

Solution: (This problem is from the 1998 Final examination in 189-112A.) Wecomplete the squares separately.

x2 + y2 + 4x − 6y + 3 = 0

⇔ (x2 + 4x) + (y2 − 6y) + 3 = 0

⇔(

x2 + 4x +

(

4

2

)2)

−(

4

2

)2

+

(

y2 − 6y +

(−6

2

)2)

−(−6

2

)2

+ 3 = 0

⇔ (x + 2)2 + (y − 3)2 = 4 + 9 − 3 = 10

⇔ (x − (−2))2 + (y − 3)2 = 4 + 9 − 3 = (√

10)2

Hence the centre of the circle is (−2, 3), and its radius is√

10.


9. Simplify completely the following formula, leaving your answer free of negative

exponents or radicals:6√

b ·(

3√

a3b2

√a6b3

)−1

.


6√

b ·(

3√

a3b2

√a6b3

)−1

=6√

b ·√

a6b3

3√

a3b2

=b

16 a3b

32

a1b23

=a3b

16+ 3

2

ab23

=a3b

53

ab23

= a3−1b53− 2

3 = a2b

10. Solve the equation1

x+

1

x − 2=

4

3. (209)

Solution: (This problem is from the 1998 Final examination in 189-112A.) If wemultiply both sides of the equation90 by 3x(x − 2), we obtain the polynomialequation 3(x − 2) + 3x = 4x(x − 2), which reduces to 4x2 − 14x + 6 = 0, then to2x2 − 7x + 3 = 0, which is equivalent to (2x − 1)(x − 3) = 0. The product on theleft can be zero only with at least one of the factors is zero, i.e. when either x = 1

2

or x = 3. These are the only solutions to (209).

11. Determine the “natural” (i.e. largest possible) domain of definition of each of thefollowing functions. Explain your results.

(a) f(x) =

√x

2 − sin x.

(b)1

log3(x + 1).


90We should verify, when we have found the alleged solutions to the equation, that this operation ofmultiplying both sides by the same quantity did not entail the introduction of any values which werenot solutions. For example, if we had multiplied by 3x2(x−2), the subsequent computations would haveproduced a 3rd degree equation whose solutions would be 0, 1

2 and 3 — but x = 0 is evidently not asolution, since the left side is not even defined there. Similarly, if we had multiplied by 3x(x−2)(x+11),we would have obtained x = 11 as one of the solutions of the polynomial equation, but it is certainlynot a solution of the original equation. While the procedure could be made more rigorous, the safestpolicy is to verify that all claimed solutions are indeed solutions of the original equation — particularlywhen, in the course of your solution, you have multiplied both sides of an equation by an expressionthat could be 0.


(a) The denominator is defined for all x. The numerator is defined for all non-negative x. The ratio function is defined only for non-zero denominators,so we must ensure that 2 − sin x is not zero; but this can never happen, as| sin x| ≤ 1. Thus the largest possible domain for this function is the set of allnon-negative real numbers.

(b) Logarithms are defined only for positive numbers, so we must require that x+1 > 0, i.e. that x > −1. But the logarithm here appears in the denominator,which can assume any value except 0. log3(x+1) = 0 precisely when x+1 = 1,i.e. x = 0. Thus the largest possible domain for this function is the union ofthe set −1 < x < 0 with the set x > 0 of positive real numbers.

12. Factorize the following polynomial completely:

(x2 + 4xy + 4y2) − (3x + 6y)

Solution: (This problem is from the 1998 Final examination in 189-112A.) Studentswere expected to observe that the first summand is a perfect square.

(x2 + 4xy + 4y2) − (3x + 6y) = (x + 2y)2 − 3(x + 2y)

= (x + 2y)(x + 2y − 3)

is the desired factorization.

13. Solve the system of equations:

3x − 7y = 1

4x + 3y = 5

Solution: (This problem is from the 1998 Final examination in 189-112A.) Studentswill study the systematic solution of systems of linear equations in courses onlinear algebra and matrices. This problem is simply to detect whether you areable to solve a routine small system, even if your methods are not systematic.Subtracting 3 times the second equation from 4 times the first equation yields−37y = −11, implying that, if there is a solution, y = 11

37; this, substituted into

the first equation, yields x = 3837

. (This solution can be verified by substitutingin the second equation. While that is hardly necessary in the present problem,substitution should be undertaken in large linear systems, unless care has beentaken to avoid the possibility that some constraint has been lost. This is beyondthe present course, but will be studied in courses in linear algebra. What is atissue is the possibility that a system of equations may have no solutions at all, ormay have infinitely many solutions.)


14. Find all solutions to the equation

log2(3x + 2) + log2(x + 1) = 2 (210)


log2(3x + 2) + log2(x + 1) = 2

⇒ log2((3x + 2)(x + 1)) = 2 = log2 22

⇔ 3x2 + 5x + 2 = 4

⇔ 3x2 + 5x − 2 = 0

⇔ (3x − 1)(x + 2) = 0

⇔ x =1

3or x = −2

But the value x = −2 is extraneous, as the logarithm is not defined at −2 + 1 =−1 < 0, or at 3 · (−2) + 2 = −4 < 0. We conclude that the only possiblesolution to (210) is x = 1

3, and verify that it does indeed satisfy the equation, as

log2 3 + log243

= log2 4 = log2 22 = 2.

15. Determine all real numbers x such that

(a) 5 tan2 x − sec2 x = 11

(b) cos x + sec x = 52

Solution:

(a)

5 tan2 x − sec2 x = 11 ⇔ 5 tan2 x − (tan2 x + 1) = 11

⇔ 4 tan2 x = 12

⇔ tan x = ±√

3

The smallest solutions in absolute value are x = ±π3. From the period-

icity of the tangent function we conclude that the set of all solutions is{

(n ± 13)π : n ∈ Z

}

where Z is the set of all integers.

(b)

cos x + sec x =5

2⇔ cos x +

1

cos x=

5

2⇔ 2 cos2 x − 5 cos x + 2 = 0

⇔ (2 cosx − 1)(cos x − 2) = 0


⇔ cos x =1

2or cos x = 2

⇒ cos x =1

2as | cosx| ≤ 1

⇔ x =

(

2n ± 1

3

)

π

16. Show that1

3≤ x2 − x + 1

x2 + x + 1≤ 3 (211)

for any real number x.

Solution: Let us first prove the inequalities

x2 + x + 1

3≤ x2 − x + 1 ≤ 3(x2 + x + 1)

i.e. equivalently, the two inequalities

x2 + x + 1 ≤ 3(x2 − x + 1)

x2 − x + 1 ≤ 3(x2 + x + 1)

The first of these is equivalent to 2(x− 1)2 ≥ 0, which is true because the left sideis a square, hence non-negative; the second is equivalent to 2(x + 1)2 ≥ 0, which isnon-negative for the same reason. In order to pass from these two inequalities to(211) we need to divide by x2 + x + 1. This is a quadratic polynomial, having noreal roots; its sign is that of the leading coefficient x2, i.e. it is positive; this canbe shown by observing that x2 + x + 1 = (x + 1

2)2 + 3

4≥ 0 + 3

4> 0. So dividing

the members of an inequality by this positive quantity preserves the inequalities,thereby yielding the desired pair of inequalities.

A more elegant solution can be found if we define z = x2−x+1x2+x+1

, and transform thisequation to yield a quadratic equation for x in terms of z:

x2 − x + 1 − z(x2 + x + 1) = 0

⇔ (1 − z)x2 − (1 + z)x + (1 − z) = 0

For this equation to admit a solution for every x, the discriminant must be non-negative, i.e.

(1 + z)2 − 4(1 − z)2 ≥ 0

This is equivalent to

[(1 + z) − 2(1 − z)][(1 + z) + 2(1 − z)] ≥ 0


which is equivalent to −(

z − 13

)

(z − 3) ≥ 0. For the product(

z − 13

)

(z − 3) tobe non-positive, z must lie between the roots, i.e. z must be such as to make oneof the factors negative and the other positive. This can be achieved only with13≤ z ≤ 3, as required.

E.2.2 Second Fall 1999 Problem Assignment

1. Determine if limx→1+

f(x), limx→1−

f(x), and/or limx→1

f(x) exist for each of the following

functions, and evaluate those that do exist.

(a) f(x) =√

x2 + x − 2

(b) f(x) =

{

x2 − 3 for x ≤ 12 − x for x > 1

(c) f(x) =

{ √2 − x for x < −3

x3 − x for 2 > x ≥ −3

2. In each of the following cases, evaluate the limit, or show that it does not exist.

(a) limx→3

(

1

x2 − x − 6− 1

x − 3

)

(b) limx→3

(

5

x2 − x − 6− 1

x − 3

)

3. 91In each of the following cases, evaluate the limit, or show that it does not exist.

(a) limx→∞

3x + 2√

x

1 − x

(b) limx→−∞

|2x − 3|x + 2

4. Evaluate limx→0

x +1

x2

2

x2− 3x2

5. 92Find a value for a so that

limx→∞

(

3x + 1 − ax2 + 1

x + 1

)

exists as a finite limit, and evaluate that limit.

91Added 21.09.99: This problem should be omitted. It is based upon [25, §§4.5,4.7], which materialmay not have been discussed in the lectures before the due date of the assignment.

92See footnote 91.


6. Sketch the graph of f(x) =

2 when x ≤ −2x + 4 when −2 < x < 4

20 − x2 when 4 ≤ x.

Find the values of x for which f(x) is not continuous.

7. Find values of a and b which will make the following function continuous:

f(x) =

x2 + 2a when 0 < x < 1bx + a when 1 ≤ x < 22b

x− 3 when 2 ≤ x

.

8. Let f(x) =x2 − 2x − 3

4 − x2. Determine where f(x) is not continuous, and where

f(x) = 0. Then solve the inequalityx2 − 2x − 3

4 − x2≥ 0. [Hint: One method to solve

this problem uses the Intermediate Value Theorem.]

9. Solve the inequality3

x + 1≤ 1 − 4x

x − 1. [Hint: Bring everything to one side of the

equation and simplify first.]

E.2.3 Third Fall 1999 Problem Assignment, with Solutions

Caveat lector! There could be misprints and other errors in these draft solutions.

1. (a) Apply the definition of the derivative as a limit to determine the derivative of

f(x) =

√

7x − 1

3for any x in the domain.

Solution:

f ′(x) = limh→0

√

7(x+h)−13

−√

7x−13

h

= limh→0

√

7(x+h)−13

−√

7x−13

h·

√

7(x+h)−13

+√

7x−13

√

7(x+h)−13

+√

7x−13

= limh→0

7(x+h)−13

− 7x−13

h· 1√

7(x+h)−13

+√

7x−13

= limh→0

7h3

h· 1√

7(x+h)−13

+√

7x−13


= limh→0

7

3· 1√

7(x+h)−13

+√

7x−13

=7

3limh→0

1√

7(x+h)−13

+√

7x−13

=7

6

1√

7x−13

(b) Determine the value of f ′(7) from the result of 1a.

Solution: 7

6√

49−13

= 724

2. You will find below the definitions for a number of functions. If the domain ofdefinition is not stated, then you are to assume the domain to be as large aspossible.

• Determine the derivative of each function wherever it is defined in the domain.

• List any points in the domain where the function does not possess a derivative,and explain precisely why there fails to be a derivative.

• If possible, determine, for those functions where requested, the equation ofthe tangent or the normal to the graph of the function at the given point.

(a) f1(x) =x

5 − 8x; tangent at (1, f1(1)).

Solution: The domain of f1 is R−{58}, the set of all non-zero real numbers x

for which the denominator is non-zero; the function is continuous and differ-entiable at every point of the domain. f ′

1(x) = 1(5−8x)−x(−8)(5−8x)2

= 5(5−8x)2

, at every

point x in the domain, i.e. at all real numbers except x = 58. As f ′

1(1) = 59,

the equation of the tangent at (1,−13) is y− (−1

3) = 5

9(x−1), i.e. 5x−9y = 8.

(b) f2(x) = (5 + 6x2 + x4) − (1 + 2x + x2)2 + 4x(1 + x); normal at (5, f2(5)).

Solution: As f2 is a polynomial, its domain is the entire real line; it is dif-ferentiable (hence continuous) at all points of R also. f ′

2(x) = (12x + 4x3) −2(1 + 2x + x2)(2 + 2x) + (4 + 8x) = −12x2 + 8x; hence f ′

2(5) = −260, and theslope of the normal at the point (x, y) = (5, f2(5)) is 1

260. The equation of the

normal is y − (−396) = 1260

(x − 5).

(c) f3(x) = (1 − x)2.

Solution: f3 is a polynomial, and is both defined and differentiable at all realnumbers. f ′

3(x) = −2(1 − x).


(d) f4(x) = −√

10x − x2; tangent at (1,−3).

Solution: The domain of f4 consists of all real numbers x such that 10x−x2 ≥0, i.e. where the product x(x−10) is non-positive: that is, all x in the interval0 ≤ x ≤ 10. The derivative will be defined at every point in the domain exceptat the end-points of the interval, at one of which the tangent approaches thevertical; f ′

4(x) = −12(10x − x2)−

12 (10 − 2x) = x−5√

10x−x2 . At the point x = 1,

f ′ = −12· 13·8 = −4

3. The equation of the tangent there is y−(−3) = −4

3(x−1),

i.e. 4x + 3y + 5 = 0.

(e) f5(x) = |3 − 2x|; tangent at x =3

2.

Solution: The function is defined for all real x. As

f5(x) =

{

3 − 2x 3 − 2x ≥ 02x − 3 3 − 2x < 0

,

f5(x) =

{

−2 x < 32

2 x > 32

.

The derivative does not exist at the point x = 32, since the difference quotient

approaches +2 on one side of this point, and −2 on the other, so there is no2-sided limit to the difference quotient.

(f) f6(x) =

x3

|x| when x 6= 0

0 when x = 0

; tangent and normal at (0, 0).

Solution: The function has all real numbers in its domain; the only point thatmight have caused difficulty would have been 0, as the given quotient is notdefined there; however, the value at x = 0 is given separately. Differentiatingf within each of the defining half-lines gives

f ′6(x) =

{

2x when x > 0−2x when x < 0

To determine the value of the derivative at x = 0 we have to appeal to first

principles. f ′6(0) = limh→0

h3

|h|−0

h. Since lim

h→0+

h3

|h|−0

h= lim

h→0+

h2−0h

= limh→0+

h = 0,

and the limit from the left is also equal to 0, we can assert that the 2-sidedlimit exists, so f ′

6(0) = 0. The line through the origin with slope 0 is thex-axis, which is tangent to the curve at that point.

(g) f7(x) =

{

x2 − 6x when x < 1−1 − 4x when x ≥ 1

; tangent at (1,−5).

Solution: f7 is defined for all x. The function was created by piecing togethera branch of a parabola for x < 1, and a ray of a straight line for x ≥ 1.


Strictly within each of the constituent half-lines the function is obviouslydifferentiable: for x < 1, f ′

7(x) = 2x − 6; while, for x > 1, f ′7(x) = −4; note

that we were unable to make any assertion about x = 1. Indeed, at x = 1the limit from the right of the difference quotient is again seen to be −4. Thelimit from the left is the value 2(1) − 6 = −4, which can be seen from firstprinciples. As here again the limits from left and right are equal, the limitexists, and the function is differentiable at x = 1. The line through (1,−5)with slope 4 has equation y + 5 = 4(x − 1), or 4x − y = 9.

(h) f8(x) =

x2 − 6x when x < 10 when x = 1

−1 − 4x when x > 1; normal at (1,−5).

Solution: f8 is similar to f7, except for the definition of the function at x = 1.This time, when we attempt to determine the derivative at x = 1 by firstprinciples, we find that the quotient f(1+h)−f(1)

h= f(1+h)

hhas infinite limits —

i.e. does not have a (real) limit from the two sides. As the limit does notexist, the function is not differentiable at x = 1, and there can be no normal.

3. (The functions referred to below were defined in Problem 2 above.)

(a) Determine the value of the function f7 ◦f7 at any point x in its domain. [Thisproblem is technically difficult. Try to understand what makes the problemdifficult. Your solution to this problem will not be graded.]

Solution:

f7◦f7(x) =

(x2 − 6x)2 − (x2 − 6x) if x2 − 6x < 1 and x < 1(−1 − 4x)2 − (−1 − 4x) if x ≥ 1 and −1 − 4x < 1

−1 − 4(x2 − 6x) if x < 1 and x2 − 6x ≥ 1−1 − 4(−1 − 4x) if x ≥ 1 and −1 − 4x ≥ 1

To study the four combinations of inequalities we observe that x2 − 6x =(x − 3)2 − 9. The inequalities simplify to

f7 ◦ f7(x) =

(x2 − 6x)2 − (x2 − 6x) if 3 −√

10 < x < 1(−1 − 4x)2 − (−1 − 4x) if x ≥ 1

−1 − 4(x2 − 6x) if x ≤ 3 −√

10−1 − 4(−1 − 4x) never.

i.e.

f7 ◦ f7(x) =

(x2 − 6x)(x2 − 6x − 1) if 3 −√

10 < x < 12(8x2 + 6x + 1) if x ≥ 1

−4x2 + 24x − 1 if x ≤ 3 −√

10


(b) Determine the value of the functions f1 ◦ f4 and f4 ◦ f1 at all points in theirdomains. Find the derivatives of these functions in two ways:

i. by differentiation of the formulæ you have determined

ii. by applying the chain rule

and show that the results are the same.

Solution:

(f1 ◦ f4)(x) =−√

10x − x2

5 − 8(−√

10x − x2)

= − 1

8 + 5(10x− x2)−12

(f4 ◦ f1)(x) = −

√

10

(

x

5 − 8x

)

−(

x

5 − 8x

)2

= −√

x(50 − 81x)

(5 − 8x)2

Both methods of differentiation should lead to the same results:

(f1 ◦ f4)′(x) =

5(x − 5)

5 + 8√

10x − x2

(f4 ◦ f1)′(x) =

205x − 125

(5 − 8x)2√

x(50 − 81x)

Don’t panic! These computations are much to difficult for an examination!The purpose was to test your perseverence, under conditions where two meth-ods had to lead to the same answer, so you could verify your work.

(c) Suppose that y =1 + s

1 − s, s = t − 1

t, t =

√x. Determine the value of

dy

dxat

x = 2.

Solution:

dy

dx=

dy

ds· ds

dt· dt

dx

=2

(1 − s)2·(

1 − 1

t2

)

· 1

2√

x

When x = 2, t =√

2, s = 1√2. Hence

dy

dx=

2(

1 − 1√2

)2 ·(

1 − 1

2

)

· 1

2√

2=

3√

2 + 4

2.


4. For each of the following functions, and for the given closed interval,

• find all critical points;

• find all local maxima and local minima;93

• find all global maxima and global minima, or explain why none of either exists.

All claims should be supported by careful reasoning; show all your work.

(a) g1(x) = (x − 3)5; interval [2, 4]

Solution: g′1(x) = 5(x − 3)4 for 2 < x < 4. The derivative is thus defined

at all points in the interior of the interval; we cannot speak of a derivativeat the end points, since the behavior of the function must be known on bothsides of a point where a derivative is to be determined; (we could speak of1-sided derivatives, but have not done so in this course). The critical pointswill be those points where the derivative is zero, i.e. only the point x = 3. Todetermine local and global extrema we must consider the value of the functionat the critical point and also at the end-points of the interval. Since g1(3) = 0,g1(2) = −1, and g1(4) = 1, we see that the global maximum, of value 1, occursat x = 4, and that the global minimum, of value −1, occurs at x = 2. Thepoint x = 3 is neither a local minimum nor a local maximum: for x < 3 thefunction value is less than g1(3), while, for x > 3, the function value is greaterthan g1(3).

(b) g2(x) = 6 − 36x + 15x2 − 2x3; interval

[

5

2, 5

]

Solution: The function g2, being a polynomial, is defined at all points. As thedefinition of g2 has been specified as

[

52, 5]

, the function will have a derivativeat all points in the interval

(

52, 5)

; we have no information about g2 for x < 52,

nor for x > 5, so we do not have a derivative at either of those end-points.Where it is defined, g′

2(x) = −36 + 30x − 6x2 = −6(x − 2)(x − 3), whichvanishes at x = 2 and x = 3. But x = 2 is not in the domain of definitionof the function, so this point is irrelevant. We thus have just one criticalpoint, x = 3; there g2(3) = 6 − 108 + 135 − 54 = −21. At the end-points,g2(

52) = −33

2, g2(5) = −49. As g2(

52) > g2(3) > g2(5), the global maximum is

at 52, and the global minimum at 5.

(c) Solution: g′3(x) =

− 1

x2when −1 ≤ x < 0

− 1

x2when 0 < x ≤ 1

. For h 6= 0, the value of

93THIS PROBLEM WAS NOT TO BE GRADED: It was announced in the lectures that this topicwould not be discussed until [25, Chapter 4]; accordingly this part of the question was to be omitted.


the difference quotient g3(h)−0h

is 1h2 , which has no finite limit as h → 0; hence

the function is not differentiable at x = 0. The derivative is not zero anywherein the domain; thus the only critical point is x = 0. Evaluating the functionat this critical point, and at the two end-points, we obtain g3(−1) = −1,g3(0) = 0, g(1) = 1. We may not apply [27, Theorem 3, p. 144] here, becausethis function is not continuous throughout the domain of definition. In fact,it is discontinuous at x = 0. As x → 0−, g3(x) → −∞; and, as x → 0+,g3(x) → ∞; these one-sided limits are not equal real numbers, so no limitexists; for continuity they must be equal, and must equal the function value,here 0. Because the function becomes positively and negatively infinite insidethe domain of definition, it can have neither a global maximum nor a globalminimum. There is no need to evaluate it at the critical point, nor at theend-points of the interval of definition.

(d) Solution: This problem differs from the preceding problem in that the limitas x → 0, while still not existing as a real number, is ∞. In this case therecan be no global maximum. The cited theorem still does not apply, as there isstill a discontinuity at x = 0. The function values at x = ±1 are both 1, and,from examination of the behavior of the function near these points, might besuspected of being global minima. However, the value at x = 0 is lower. Theglobal minimum is thus 0, assumed only at the point x = 0; everywhere elsethe function value is strictly larger.

(e) Solution: The domain of definition of g5 is the whole real line. The derivativeof

√x for x > 0 is 1

2√

x; similarly, the derivative of

√−x for x < 0 is − 1

2√−x

.

Investigation of the difference quotient g5(h)−0h

as h → 0, shows that the limitdoes not exist as a finite number; in fact is it ∞, and the function may besaid to have a vertical tangent. Thus x = 0 is a critical point, because thefunction lacks a derivative there. The function is, however, continuous, so wemay determine the global extrema by studying its behavior at the end-pointsand the critical point: g5(−4) = 2, g5(0) = 0, g5(9) = 3. The global maximumis thus at x = 9, and the global minimum is at x = 0.

5. [29, Examples XLVI.17] (Cambridge Math. Tripos 1930) The graph of the function

h(x) =ax + b

(x − 1)(x − 4)has (2,−1) as a critical point. Determine a and b, and show

that the critical point is a local maximum. (Note: It is intended that this problembe solved without using concepts from Chapter 4 of your textbook (involving higherderivatives.))

Solution: For x different from 1 and 4, h′(x) = a(x2−5x+4)−(ax+b)(2x−5)(x−1)2(x−4)2

. As thederivative exists at x = 2, it can be a critical point only because the derivative is


zero; this implies that a(22−5·2+4)−(a·2+b)(2·2−5) = 0, i.e. b = 0. We are alsotold that h(2) = −1, hence 2a+0

(2−1)(2−4)= −1, so a = 1. Thus h(x) = x

(x−1)(x−4). As

h(2)−h(x) = − (x−2)2

(x−1)(x−4), this difference is non-negative in an interval surrounding

x = 2 (since the numerator is non-negative, and the denominator is negative). Thush(x) ≤ h(2), so 2 is a local maximum.

6. Determine values of the constants a, b, c which will cause the curve y = ax3 +bx2 +cx + d to pass through the points (2, 6) and (−1, 6) and to be tangent to the liney = 3x + 1 at the point (1, 4).

Solution: We are given four kinds of information: three points through which thecurve passes, and the equation of the tangent at one of these points. Imposing thecondition that the curve pass through the points (2, 6), (−1, 6), and (1, 4) yieldsthree linear equations:

8a + 4b + 2c + d = 6

−a + b − c + d = 6

a + b + c + d = 4

Finally, as the tangent at x = 1 has slope 3, we know that 3ax2 + 2bx + c|x=1 = 3,i.e. 3a+2b+c = 3. Solving the 4 linear equations simultaneously yields (a, b, c, d) =(−1, 3, 0, 2).

7. Find all lines with slope −3 which are normal to the curve 64y = x3.

Solution: At the point (x, 164

x3) on the curve the slope of the tangent is 364

x2, sothe slope of the normal is − 64

3x2 . Imposing the condition that this equal −3, weobtain that x = ±8

3. Through the point

(

±83,± 8

27

)

the equation of the line withslope 3 is y ∓ 8

27= −3

(

x ∓ 83

)

, i.e. 3x + y = ±8·2827

.

8. Find the volume of the uncovered box of greatest volume that can be made bycutting equal squares out of the corners of a piece of sheet metal which is 21 cm.× 5 cm., and turning up the sides.

Solution: If the side of the square cut from each corner is of lengh x, where 0 ≤ x ≤52, then the volume obtained after the sides are folded up is x · (21− 2x)(5− 2x) =

4x3−52x2 +105x. Setting the derivative, 12x2−104x+105 (i.e. (2x−15)(6x−7))equal to zero, we find that x = 15

2or x = 7

6. The second of these is the only critical

point; the first is not in the interval of definition of the function. To determine themaximum we compare the value of the volume at x = 7

6with the volume at the

end-points, both of which give volume 0. At x = 76

the volume is 76· 56

3· 8

3= 1568

27

cm.3, which exceeds the value of 0 at the end-points. This is the largest volume.


9. Show that, among all right-angled triangles whose hypotenuse is 10 units long, thetriangle whose area is maximum is isosceles.

Solution: Denote the lengths of the non-hypotenuse sides by a and b. As these areconstrained by the equation x2 + b2 = 102 = 100, we know that b =

√100 − a2.

The area is therefore 12a√

100 − a2. We may take the domain to be 0 ≤ a ≤ 10.

The derivative of the area function is, after reduction, 50−a2√100−a2 , which is zero when

a = ±5√

2. Of these two values, only the positive one is in the domain of definitionof the function. At the end-points of the domain the function is zero; while, atthe critical point we have found, the function value is positive; this, then, is themaximum point. For this value of a, b = a, which was to be proved.

E.2.4 Fourth Fall 1999 Problem Assignment

1. Finddy

dxif

(a) y = ln(

1 + x2)

(b) y = e−x2

(c) y = x√

x

2. Find an equation for the tangent line at x = 0 to the graph of y = ex−e−x .

3. Find the greatest value of f(x) =1

x2ln x .

4. If f(x) = e−x sin x , find the values of x where

(a) f ′(x) = 0 ;

(b) f ′′(x) = 0 .

5. (a) Finddy

dxif y =

sin−1 2x

sin−1 x.

(b) Show that if z = tan−1

(

x + 1

x − 1

)

+ tan−1 x , thendz

dx= 0 .

6. A picture a metres high is placed on a wall with its base b metres abovean observer’s eye. If the observer stands x metres away from the wall, find

(a) the angle α subtended by the picture at the observer’s eye; and

(b) the distance x which will give the maximum value for α .


7. Use the tangent line approximation to find an approximate value for

(a) ln(0.94) ;

(b) cos−1(0.47) .

8. Find an equation for the line tangent to the curve x4 + x2y2 + y4 = 21 at thepoint (1, 2) .

9. If x3 + y3 + 6xy = −5 , findd2y

dx2at the point (−1, 2) .

10. Evaluate limx→1

ln x

x − 1. [Hint: Write down the limit definition of f ′(1) , where

f(x) = lnx .]

E.2.5 Fifth Fall 1999 Problem Assignment

1. For each of the functions,

f(x) =(x2 − 1)2

4x2g(x) =

1 − sin x

cos x

h(x) = xe−x2k(x) =

√ln x

determine all of the information requested below, and sketch a graph of the func-tion. Show all your work.

(a) For any discontinuities, determine whether they are removable.

(b) Where is the function increasing? Where is it decreasing?

(c) Where is the graph concave upwards? Where is it concave downwards?

(d) Where, if any, are the local extrema? In each case you should indicate whetherthe extremum is a local maximum or a local minimum, and provide justifica-tion for your choice.

(e) Where, if any, are the global extrema? Again, justification is expected.

(f) Where, if any, are the intercepts (with the coordinate axes)?

(g) Where, if any, are the inflection points?

(h) Is the graph symmetric about any vertical line x = a? (The graph of F issymmetric about x = a if the equation remains unchanged under the trans-formation x − a → a − x; i.e. if F (x) = F (2a − x). In particular, the graphis symmetric about the y-axis if F (x) = F (−x). Such a function is said to beeven.)


(i) Is the graph symmetric under rotation about the origin? (The graph of F issymmetric under rotation about the origin if the equation remains unchangedunder the transformation (x, y) → (−x,−y); i.e. if F (−x) = −F (x). Such afunction is said to be odd .)

(j) Determine all horizontal and all vertical asymptotes to the graph.

2. (a) Show that at least one of the hypotheses of the Mean Value Theorem fails tohold for the following functions; show also that the conclusion of the theoremfails to hold for each function, where the interval is [a, b] = [−1, 1].

i. f1(x) =

{

1x

when x 6= 00 when x = 0

.

ii. f2(x) = x23 .

(b) [25, Problem 4.3.48] [27, Problem 4.3.48] Show that the function f2(x) de-fined in Problem 2(a)ii above does satisfy the conclusion of the Mean ValueTheorem on the interval [−1, 27].

(c) Using the Mean Value Theorem, show carefully that the equation x7 + x− 12has exactly one real solution. Then show — without using a calculator —that this solution lies between x = 1 and x = 2.

3. Consider the function f3(x) = x cos x.

(a) Determine the intercepts of the graph of f3 on the coordinate axes.

(b) Show that the critical points of f3 occur at points of intersection of y = tan x

with the curve y =1

x.

(c) Show that, wherever it is defined, the function tan x − 1

xis increasing. Use

this fact to show that there is exactly one critical point of f3 in each interval(

(2n − 1)π

2,(2n + 1)π

2

)

, where n is any integer. [Hint: Use a Corollary to

the Mean Value Theorem.]

(d) Sketch the graph of f3.

4. Use the Second Derivative Test in finding all points on the curve y =1

2x2which

are closest to the origin.

5. 94In each of the following cases, evaluate the limit, or show that it does not exist.

94This problem and the next were originally included in Assignment 2, but were subsequently omitted,because they are based upon [25, §§4.5,4.7], which material had not been discussed in the lectures beforethe due date of Assignment 2.


(a) limx→∞

3x + 2√

x

1 − x

(b) limx→−∞

|2x − 3|x + 2

6. Find a value for a so that

limx→∞

(

3x + 1 − ax2 + 1

x + 1

)

exists as a finite limit, and evaluate that limit.

E.3 2000/2001 Problem Assignments, with Solutions

E.3.1 First 2000/2001 Problem Assignment, with Solutions

Instructions

• The basic principle you should follow in any mathematics course is that every statementshould be justified. While it may be that in some cases you are not able to provide aflawless logical argument, that should always be your goal. Getting the right answer isalways desirable, but will usually not be enough.

• Notwithstanding the preceding comment, you should not spend disproportionately longamounts of time on any one question. Solutions will be posted on the Web.

• While the textbook in [11, 1.1] describes four different ways in which to represent a func-tion, we usually regard the representation by an explicit formula as the most desirable.In particular, graphical representation should be used only to assist you in visualizing asolution, not as the final solution; it will normally not be acceptable to explain a step ina proof by reference to a graph — but it is a good policy to make a quick sketch of thegraph of any function you have to work with.

• You should always attempt, in an algebraic representation, to make your functions as“simple” as possible; there may be different “simple” ways of representing the function.Use your good judgment: the simplification is intended to help you!

• Do not approximate numbers unless you are asked to do so. Thus, π should not bereplaced by 3.1415926...; of course, you should know the values of the standard trigono-metric functions at familiar multiples and submultiples of π, and use this information to

simplify your answers, where applicable. For example, you should replace sin π3 by

√3

2 ,but not by 0.8660....

It is usually preferable to leave fractions in the form mn , rather than rewriting as decimal

fractions. This preference applies even when the decimal expansion is finite, as in, for


example, 34 = 0.75. One reason for this preference is that decimal fractions are often

interpreted as approximations, rather than as the exact value; so by writing 0.75 insteadof 3

4 you could be obscuring the fact that your datum is exact.

• You may assume that, over an interval [c, d], the maximum and minimum values of alinear function — that is a function of the form f(x) = ax+b, where a and b are constants— are attained at the end points of the interval.

1. Let a function f be defined by f(x) = |2x| + |2x + 3|.

(a) Find algebraic formulæ which give the value of f(x) without using the absolutevalue function. For this purpose it will be necessary to break the domain upinto several parts, as the formula will be different in different subintervals.

(b) Showing all your work, determine the domain and the range95 of f .

(c) Determine whether or not f has an inverse function. If it does, determine aformula for f−1(x).

Solution:

(a)

Since |2x| =

{

−2x when x < 02x when x ≥ 0

and |2x + 3| =

{

−2x − 3 when 2x + 3 < 02x + 3 when 2x + 3 ≥ 0

or, equivalently |2x + 3| =

{

−2x − 3 when x < −32

2x + 3 when x ≥ −32

,

we may add the inequalities to obtain

|2x| + |2x + 3| =

−2x − 2x − 3 when x < −32

−2x + 2x + 3 when −32≤ x < 0

2x + 2x + 3 when x ≥ 0

=

−4x − 3 when x < −32

3 when −32≤ x < 0

4x + 3 when x ≥ 0

95Many mathematicians prefer to use the term image instead of range.


(b) While the description of the function changes as one moves from one intervalto the next, the function is defined for all x — i.e. the domain is the wholereal line.For the interval −∞ < x ≤ −3

2, the function value ranges between +∞ and

3; it remains at 3 through the middle interval; and, for −32

< x < ∞, againtakes on all values which are ≥ 3. Thus the range is [3,∞).

[How could one prove, rigorously, that the stated intervals do indeed constitute the

range of the function? For example, when x < −32 , we reverse the inequality when

we multiply by −4, to obtain −4x > 6; adding −3 to both sides of the inequality

yields −4x − 3 > 3, showing that the portion of the range of the function for this

part of the domain is contained in the interval (3,∞). And, if y is any real number

such that 3 < y, then 6 < y + 3; so, multiplying both sides of the inequality by −14 ,

and thereby reversing the inequality, we obtain −y+34 < −3

2 . But this tells us that

f(

−y+34

)

= −4(

y+34

)

− 3 = y. We have thus shown that every number y in the

interval (3,∞) is in the range: thus the range of the function corresponding to this

portion of the domain is precisely the interval (3,∞).]

(c) This function is not invertible, since there are values which are attained atmore than one point in the domain. For example, the function takes on thesame value for all x such that −3

2≤ x ≤ 0.

2. Let a function g be defined by

g(x) =√

2x − 5 . (212)

Showing your work, answer each of the following questions for g.

(a) Determine the domain.

(b) Determine the range.

(c) Determine whether the function has an inverse function. If it does, find aformula for g−1(x), and determine the domain and range.

Solution:

(a) g is the composition of two functions. In the first phase x is mapped onto 2x − 5. The second function applied is the square root. The mappingx 7→ 2x − 5 is defined for all x; that is, its domain is the whole real line.However, the mapping to the square root is not defined for negative numbers.Thus we cannot proceed unless 2x − 5 ≥ 0, i.e. unless

x ≥ 5

2. (213)

Inequality (213) gives the domain of g.


(b) Any non-negative real number r is attained as a value of the function g;

specifically, g

(

r2 + 5

2

)

= r. Thus the range of g is the interval [0,∞).

(c) We may square both sides96 of (212) to obtain (g(x))2 = 2x − 5, so x =(g(x))2 + 5

2, showing that x is determined as soon as g(x) is known. Thus g

is invertible, and its inverse is given by x 7→ x2 + 5

2.97

3. A curve is symmetric about the y-axis if the presence of any point (x0, y0) on thecurve implies the presence of (−x0, y0); a curve is symmetric about the x-axis ifthe presence of any point (x0, y0) on the curve implies the presence of (x0,−y0);a curve is symmetric about the origin if the presence of any point (x0, y0) on thecurve implies the presence of (−x0,−y0). A function F is odd if its graph issymmetric about the origin, i.e. for any x in the domain, F (−x) = −F (x) (whichis equivalent to saying that if (x, F (x)) is on the graph, then so is (−x,−F (x))).F is even if its graph is symmetric about the y-axis; i.e., for any x in the domain,F (−x) = F (x) (which is equivalent to saying that if (x, F (x)) is on the graph, thenso is (−x, F (x))).

For each of the following functions F , determine whether

• F is odd.

• F is even.

• the graph of F is symmetric about the x-axis.

• the graph of F is symmetric about the y-axis.

(a) F (x) = | sin x|(b) F (x) = cos x

(c) F (x) = cos(

x − π

2

)

(d) F (x) = 0

(e) F (x) = x3 +1

x

96More precisely, we can multiply both sides of the equation g(x) −√

2x − 5 = 0 by g(x) +√

2x − 5.97Note that, while this formula is meaningful even when x < 0, we are specifically restricting the

domain of this inverse function to be the range of the function g. With that proviso we find that thepair of functions g and g−1 have the property that the domain of each is the range of the other, andvice-versa. Thus the domain of g−1 is [0,∞), and the range is

(

52 ,∞

)


(f) F (x) = x3 − 1

x

(g) F (x) = G(x2), where G is any function.

Solution:

Before beginning our solution, we make the following observations:

• The graph of F is symmetric about the y-axis ⇔ F is even.

• If the graph of F is symmetric about the x-axis, then, for any x0 in thedomain, both points (x0, F (x0)) and (x0,−F (x0)) will lie on the graph. Butthe graph of a function can never meet two distinct points on the same verticalline x = x0. It follows that these points cannot be distinct, i.e. that F (x0) =−F (x0), i.e. that F (x0) = 0 for all points x0 in the domain. Thus only the0 function, or restrictions of this function to subsets of the real line can havethis symmetry property.

• The only function which is both even and odd is the 0 function (or restrictionsof this function to subsets of the real line). For both conditions F (−x) = F (x)and F (−x) = −F (x) must hold for all x, implying that F (x) = 0 for all x.

(a) | sin(−x)| = |−sin x| = | sin x|, so this function is even; (hence, it not being the0 function, it is not odd). By the preceding comments, its graph is symmetricabout the y-axis, but not about the x-axis.

(b) cos(−x) = cos x for all x, so this function is even. It is not odd; for example,cos(

−π4

)

= 1√26= − 1√

2= − cos π

4.

(c) By a well-known property of the cosine function, cos(

x − π2

)

= cos x cos π2

+sin x sin π

2= (cos x) · 0 + (sin x) · 1 = sin x for all x. Hence cos

(

−x − π2

)

=sin(−x) = − sin x = − cos

(

x − π2

)

, so this function is odd. As the functionis not identically 0, it is not even; its graph is not symmetric about eithercoordinate axis.

(d) The 0 function is both even and odd, and its graph is symmetric about bothcoordinate axes.

(e) (−x)3 +1

−x= −

(

x3 +1

x

)

, so this function is odd. It is not even, and its

graph is symmetric about neither coordinate axis.

(f) The function x3 − 1

xhas the same symmetry properties as the preceding

function.


(g) G((−x)2) = G(x2), so this function is even, and its graph is symmetric aboutthe y-axis. The function is not odd, and the graph is not symmetric aboutthe x-axis.

4. Let

f(x) =x2 − 4

x − 1

g(x) =1

f(x − 1)

h(x) = sin(

x − π

4

)

k(x) =1

h(

x − π

4

)

(a) Determine an explicit formula for (f ◦ g)(x).

(b) Describe the domains of the functions f , g, f ◦ g, h and k.

(c) [BONUS QUESTION] Describe the domain of g ◦ f .

Wherever possible in this problem, algebraic expressions should be simplified.

Solution:

(b) f . Since f is defined as a ratio of polynomials, its domain is the set if all realnumbers where the denominator is not zero; as the denominator is x− 1,the domain of f is all real numbers except 1.

g.

g(x) =1

f(x − 1)=

1

(x − 1)2 − 4

(x − 1) − 1

=1

x2 − 2x − 3

x − 2

=x − 2

x2 − 2x − 3

The preceding statement is valid wherever g is defined. That will be theintersection of the set of x such that f(x − 1) is defined, and the setwhere the fraction 1f(x − 1) is defined; i.e., it will be the set of x wheref(x − 1) is defined, from which we must remove all points x such thatf(x − 1) = 0. Since f(x) is defined for x 6= 1, f(x − 1) is defined for allx such that x − 1 6= 1, i.e. such that x 6= 2. And f(x − 1) = 0 preciselywhen (x − 1)2 − 4 = 0, equivalently when x − 1 = ±2, i.e. when x = −1or x = 3. Therefore the domain of g is the set R − {−1, 2, 3}.

f ◦ g. The first function to be applied is g. Its domain consists of R −{−1, 2, 3}. But the application of f requires, further, that we exclude


points x where g(x) = 1, since 1 is not in the domain of f . Thus we

exclude points x such that x2 − 3x− 1 = 0, i.e.3 ±

√13

2. The domain is,

therefore,

R −{

−1, 2, 3,3 +

√13

2,3 −

√13

2

}

(We could also have determined this by excluding from R the roots of thedenominator of the ratio of polynomials that we have determined abovefor f ◦ g.)

h. h(x) is defined for all real numbers x.

g. k(x) is not defined at points where sin(

x − π2

)

is zero, i.e. where x − π2

isan integer multiple of π. Thus the points that have to be excluded from

R are all numbers of the form2n + 1

2π, where n is any integer.

(a)

(f ◦ g)(x) =

(

x − 2

x2 − 2x − 3

)2

− 4

x − 2

x2 − 2x − 3− 1

=(x − 2)2 − 4(x2 − 2x − 3)

(x − 2) − (x2 − 2x − 3)2· 1

x2 − 2x − 3

=4x4 − 16x3 − 9x2 + 52x + 32

(x2 − 3x − 1)(x2 − 2x − 3)

(c) We can also express (g ◦ f)(x) as a ratio of polynomials in x:

(g ◦ f)(x) =f(x) − 2

f(x)2 − 2f(x) − 3

=

x2 − 4

x − 1− 2

(

x2 − 4

x − 1

)2

− 2x2 − 4

x − 1− 3

=((x2 − 4) − 2(x − 1)) (x − 1)

(x2 − 4)2 − 2(x2 − 4)(x − 1) − 3(x − 1)2

=(x2 − 2x − 2)(x − 1)

x4 − 2x3 − 9x2 + 14x + 5


This ratio is not defined at points where the denominator is zero. However,that denominator is a quartic function of x, and its roots are not obvious.98

Instead, let us attack this problem the same way we dealt with f ◦ g above.First we have to exclude the points where f is not defined, i.e., the point x = 1.Next we have to exclude points x such that f(x) is not in the domain of g:

these are the solutions to each of the equationsx2 − 4

x − 1= 3,

x2 − 4

x − 1= 2, and

x2 − 4

x − 1= −1, i.e., the roots of the three quadratic polynomials, x2 − 3x − 1,

x2 − 2x − 2, and x2 + x − 5. It follows that the domain of g ◦ f is

R −{

1,3 +

√13

2,3 −

√13

2,−1 +

√21

2,−1 −

√21

2, 1 +

√3, 1 −

√3

}

5. (a) Showing all your work, and without using tables, computers, slide rules, or acalculator , determine the value of log16 2.

(b) Showing all your work, determine all values of x for which

2x =

(

1

2

)x

. (214)

(c) Showing all your work, determine all values of x for which log2(ln x) = 1.

(d) Showing all your work, simplify cos(x + y) cos(x− y)− cos2 x− cos2 y, wherex and y are any real numbers.

Solution:

(a) By [11, p. 70]

log16 2 =ln 2

ln 16=

1

ln 16

ln 2

=1

log2 16=

1

log2 24=

1

4

(b) By the Laws of Exponents [11, p. 58]

1 = 1x =

(

2 · 1

2

)x

= 2x ·(

1

2

)x

,

hence(

1

2

)x

=1

2x=

20

2x= 2−x .

98Although, in fact, this quartic polynomial does factorize into the product of two quadratic polyno-mials, x2 − 3x − 1 and x2 + x − 5, whose roots can be found in the usual way.


Thus (214)⇔ 2x = 2−x. Applying the inverse function log2 to both sidesof this equation yields the equivalent statement x = −x, which is, in turn,equivalent to x = 0.

(c) Applying the exponential function (to base 2) to both sides of the hypothesislog2(ln x) = 1 yields ln x = 21 = 2. Now apply the exponential function, thistime to base e, to obtain the equivalent statement x = e2.

6. (a) [11, Exercise 14, p. A34] A circle has radius 10 cm. Showing all your work,determine the length of the arc subtended by a central angle of 72◦.

(b) [11, Exercise 34, p. A35] Showing all your work, find the remaining 5 trigono-

metric ratios for the angle θ, if it is known that csc θ = −4

3, and

3π

2< θ < 2π.

(c) (Adapted from [11, Exercise 70, p. A35]) Find all values of x that satisfy theequation cosx + sin 2x = 0.

(d) Determine the value of cos(x+y) cos(x−y)−cos2 x−cos2 y, as x and y rangeover the real numbers.

Solution:

(a) The entire circle has circumference 2π · 10 cm. The angle of 72◦ is 72360

= 15

of the central angle subtending the entire circle. Hence the arc has length15× 20π = 4π cm.

(b) First observe that sin θ =1

csc θ= −3

4.

Now, since θ is in the 4th quadrant, the cosine is positive. This resolves thesign choice when we solve for cos θ in the equation sin2 θ + cos2 θ = 1, so we

have cos θ = +

√

1 − 9

16=

√7

4. The remaining ratios can now be computed

without ambiguity:

tan θ =sin θ

cos θ=

−34√7

4

= − 3√7

cot θ = cos θ · csc θ =

√7

4· −4

3= −

√7

3

sec θ =1

cos θ=

4√7

(c) (There are many ways of approaching problems of this type, so the followingis only one possibility. Of course, all methods lead to the same answers.)

cos x + sin 2x = 0 ⇔ cos x + 2 sinx cos x = 0 (double-angle formula)


⇔ cos x(1 + 2 sin x) = 0

⇔ cos x = 0 or sin x = −1

2

The solution set will, therefore, be the union of the solution sets for the twoequations shown.

cos x = 0. The solutions to this equation are all x of the form x =π

2+ nπ,

where n is any integer.

sin x = −12. One solution is x = −π

6; another solution is x = −5π

6; remember

that the graph of the sine function is symmetric about the line x = 3π2

(as it is about all odd integer multiples of π2): any value which is realized

between 3π2

and 2π will also be attained at a mirror-image point betweenπ and 3π

2. These two solutions each give rise to an infinite set of solutions,

2π units apart. The full set of solutions is all real numbers of either of theforms

(

−16

+ 2m)

π,(

−56

+ 2m)

π or, equivalently of either of the forms(

116

+ 2m)

π,(

76

+ 2m)

π where m is any integer.

The set of solutions to the original equation is, therefore, the union of the twolists above:

{(

1

2+ m

)

π,

(

7

6+ 2m

)

π,

(

11

6+ 2m

)

π

∣

∣

∣

∣

m any integer

}

(d) (There are several possible approaches to this problem. The following maynot be the simplest.)

cos(x + y) cos(x − y) − cos2 x − cos2 y

= (cos x cos y − sin x sin y)(cosx cos y + sin x sin y) − cos2 x − cos2 y

= cos2 x cos2 y − sin2 x sin2 y − cos2 x − cos2 y

= cos2 x cos2 y − (1 − cos2 x)(1 − cos2 y) − cos2 x − cos2 y

= cos2 x cos2 y − (1 − cos2 x − cos2 y + cos2 x cos2 y) − cos2 x − cos2 y = −1

for all x and for all y.

7. (a) [11, Exercise 8, p. A-23] Show that the following equation represents a circle,and determine its centre and radius:

16x2 + 16y2 + 8x + 32y + 1 = 0

(b) [11, Exercise 36, p. A-15] Find an equation of the line through the point(

12,−2

3

)

which is perpendicular to the line 4x − 8y = 1.

Solution:


(a) We first group the terms in x and y separately, as quadratic functions. Thenwe scale to arrange for the coefficients of x2 and of y2 to be 1, and completethe square in each of these:

16x2 + 16y2 + 8x + 32y + 1 = 0

⇔ (16x2 + 8x) + (16y2 + 32y) = −1

⇔(

x2 +1

2x

)

+ (y2 + 2y) = − 1

16

⇔(

x2 +1

2x +

(

1

2· 1

2

)2)

+

(

y2 + 2y +

(

1

2· 2)2)

= − 1

16+

(

1

2· 1

2

)2

+

(

1

2· 2)2

⇔(

x +1

4

)2

+ (y + 1)2 = − 1

16+

1

16+ 1 = 1

which is the equation of the circle with centre(

−14,−1

)

and radius 1.

(b) The slope of the line 4x−8y = 1 is 12; so the line we seek, being perpendicular

to that line, must have slope − 112

= −2. The line through the given point,

with slope −2, has equation

y −(

−2

3

)

= −2

(

x − 1

2

)

which simplifies to 2x + y = 13.

8. (a) [11, Exercise 84(a), p. A36] Showing all your work, determine the exact valueof arctan(−1).

(b) [11, Exercise 86(b), p. A36] Showing all your work, determine the exact valueof arcsin 1.

(c) [11, Exercise 92, p. A36] Showing all your work, determine the exact value ofsin(

arcsin 13

+ sin−1 23

)

.

(d) (Adapted from [11, Exercise 96, p. A36]) Simplify sin (−2 cos−1 x).

Solution:

(a) The inverse tangent function inverts the restriction of the tangent function tothe interval

(

−π2, +π

2

)

. In that interval the only x such that tan x = −1 is−π

4. Hence arctan−1 = −π

4.


(b) The inverse sine function inverts the restriction of the sine function to theinterval

[

−π2, +π

2

]

. In that interval the only x such that sin x = 1 is π2. Hence

arcsin 1 = π2.

(c) Let’s apply the formula for the sine of a sum:

sin

(

arcsin1

3+ sin−1 2

3

)

= sin

(

arcsin1

3

)

· cos

(

sin−1 2

3

)

+ cos

(

arcsin1

3

)

· sin(

sin−1 2

3

)

=1

3· cos

(

sin−1 2

3

)

+ cos

(

arcsin1

3

)

· 2

3

Now sin−1 23

is in the first quadrant, so its cosine is positive, and equal to

+√

1 − 49

=√

53

; arcsin 13

is also in the first quadrant, so its cosine is also

positive, and is equal to 2√

23

. Substituting these values gives

1

3· cos

(

sin−1 2

3

)

+ cos

(

arcsin1

3

)

· 2

3=

√5 + 4

√2

9.

(d)

sin(

−2 cos−1 x)

= − sin(

2 cos−1 x)

= −2 sin(

cos−1 x)

· cos(

cos−1 x)

We know that, for all x, cos cos−1 x = x. Also, since the inverse cosine functiontakes its values in the interval

[

0, π2

]

, and the sine function is non-negative

throughout that interval, we also know that sin cos−1 x = +√

1 − x2. It followsthat

sin(

−2 cos−1 x)

= −2x√

1 − x2 .

E.3.2 Second 2000/2001 Problem Assignment, with Solutions

In all of these problems you were expected to justify every statement you made, and toshow all your work.

1. For the curve y = 2x3, find the slope MPQ of the secant line through the pointsP = (1, 2) and Q = (2, 16), i.e. the points with x = 1 and x = 2.

Solution: The slope of the line joining P and Q is 2·23−2·13

2−1= 14.


2. Given

f(x) =

x3 + 2 if x ≤ −1x2 + x + 1 if −1 < x < 1

x4 + 2 if x ≥ 1

find the following limits, if they exist; or explain why the limit does not exist.Justify your answers.

(a) limx→−1+

f(x)

(b) limx→−1−

f(x)

(c) limx→−1

f(x)

(d) limx→1+

f(x)

(e) limx→1−

f(x)

(f) limx→1

f(x)

Solution:

(a) For x to the right of −1 (but to the left of +1) f(x) = x2 + x + 1; hencelim

x→−1+f(x) = (−1)2 + (−1) + 1 = 1.

(b) For x to the left of −1, f(x) = x3 + 2; hence limx→−1−

f(x) = −13 + 2 = 1

(c) Since the one-sided limits from both left and right exist at −1 and have thesame value, lim

x→−1f(x) exists, and its value is the common value of the one-

sided limits, i.e. 1.

(d) For x to the right of 1, f(x) = x4 + 2; hence limx→1+

f(x) = 14 + 2 = 3.

(e) For x to the left of 1 (but to the right of −1) f(x) = x2 + x + 1; hencelim

x→1−f(x) = 12 + 1 + 1 = 3.

(f) Since the one-sided limits from both left and right exist at 1 and have thesame value, lim

x→1f(x) exists, and its value is the common value of the one-

sided limits, i.e. 3.

3. Given that limx→3

f(x) = 5, limx→3

g(x) = 0, and limx→3

h(x) = −8, find the following

limits, if they exist. If a limit does not exist, explain why.

(a) limx→3

(f(x) + h(x))


(b) limx→3

(x2f(x))

(c) limx→3

(f(x))2

(d) limx→3

f(x)2h(x)

(e) limx→3

g(x)f(x)

(f) limx→3

f(x)g(x)

(g) limx→3

2h(x)f(x)−h(x)

(h) limx→3

3√

h(x)

Solution:

(a) limx→3

(f(x) + h(x)) = limx→3

f(x) + limx→3

h(x) = 5 + (−8) = −3. (We have used the

Sum Law.)

(b) Since we know that the limit of x as x → 3 is 3, we can use the Product Law:

limx→3

(x2f(x)) =(

limx→3

x)2

· limx→3

f(x) = 32 · 5 = 45.

(c) Again by the Product Law, the limit of a square is the square of the limit:

limx→3

(f(x))2 =(

limx→3

f(x))2

= 52 = 25.

(d) First we use the Constant Multiple Law to determine the limit of the de-nominator. Then we use the Quotient Law, since the limit of the denom-inator exists, and is not zero, and the limit of the numerator also exists:

limx→3

f(x)

2h(x)=

5

2(−8)= − 5

16.

(e) Again by the Quotient Law, limx→3

g(x)f(x)

= 05

= 0.

(f) But, in this case, the limit of the denominator is 0. Since the limit of thenumerator exists and is non-zero, the limit of the quotient does not exist.

(g) The Difference Law is used for the denominator; then the Constant Multiple

Law and the Quotient Law: limx→3

2h(x)f(x)−h(x)

= 2(−8)5−(−8)

= −1613

.

(h) By the Root Law, limx→3

3√

h(x) = 3√−8 = −2.


4. Use the Intermediate Value Theorem to show that there is a solution to the equationx3 + 2x2 − 42 = 0 in the interval (0, 3).

Solution: Let f(x) = x3 + 2x2 − 42. This function is continuous everywhere onR. Since f(0) = −42, and f(3) = 33 + 2(9) − 42 = 3, and since −42 < 0 < +3,the function must assume the intermediate value 0 at a point in the open interval0 < x < 3.

5. Given

f(x) =

2x3 + 16 if x ≤ −2x2 + bx + c if −2 < x < 23x4 − 48 if x ≥ 2

determine values for b and c so that f is continuous everywhere. Justify youranswer.

Solution: In each of the intervals (−∞,−2], (−2, 2), [2,∞) the function is a poly-nomial, and is therefore continuous at all points. The only possibly problematicpoints are x = −2 and x = +2. We will determine limits from the left and right ateach of these two points; then we equate the two one-sided values, and obtain con-ditions on b and c which we attempt to satisfy. lim

x→−2−2x3 + 16 = 2(−2)3 + 16 = 0,

limx→−2+

x2 +bx+c = (−2)2 +b(−2)+c = 4−2b+c; limx→2−

x2 +bx+c = 22 +b(2)+c =

4 + 2b + c, limx→2+

3x4 − 48 = 3(2)4 − 48 = 0. We solve the equations 0 = 4 − 2b + c

and 4 + 2b + c = 0, to obtain b = 0, c = −4. With this pair of values — and onlythese values — f is continuous everywhere.

6. The displacement in meters of a particle moving in a straight line is given bys = t3 +sin tπ, where t is measured in seconds. Find the average velocity in metersper second over the time period [1, 5].

Solution: Average velocity =53 + 0 − 13 − 0

5 − 1= 31 meters per second.

7. Find the value of limx→5+

x − 5

|x − 5| .

Solution: For all x > 5, x−5|x−5| = 1. Hence the limit, as x approaches 5 from the

right is 1.

8. Find the value of x at which the curve y = 3xx+7

has a vertical asymptote.

Solution: The domain of the function is R − {−7}: at all points except for −7the function is a ratio of polynomials, each having a limit, where the limit of thedenominator is non-zero. Hence, by the Quotient Law, the limit of the function


exists (as a real number). However, as x → −7−, the denominator approaches 0from the left, while the numerator approaches −21; so the ratio becomes infinitelylarge positively ; this is what we mean when we write lim

x→−7−

3xx+7

= +∞. Similarly,

as x → −7+, the denominator approaches 0 from the right, while the numeratorapproaches −21; so the ratio becomes infinitely large negatively ; this is what wemean when we write lim

x→−7+

3xx+7

= −∞. It is in cases like these, where the one-sided

limits are infinite positively or negatively, that we say that the curve has a verticalasymptote.

9. Find the value of the limit limx→2

x3+8x+2

.

Solution: limx→2

x3 + 8

x + 2=

limx→2

(x3 + 8)

limx→2

(x + 2)=

23 + 8

2 + 2= 16. (Problem: Find the limit

of the given quotient as x → −2. Solution: limx→−2

x3+8x+2

= limx→−2

x2 − 2x + 4 =

22 + 2 · 2 + 4 = 12.)

10. Find the value of limx→−2

(

1x+2

+ 4x2−4

)

.

Solution: limx→−2

(

1x+2

+ 4x2−4

)

= limx→−2

x+2(x+2)(x−2)

= limx→−2

1x−2

= 1−2−2

= −14.

11. Find the value of limx→4

x−4√x−2

.

Solution: For positive x, different from 4, x−4√x−2

= (√

x−2)(√

x+2)√x−2

=√

x + 2. Hence

the limit, as x → 4 is√

4 + 2 = 4.

12. How close to 4 do we have to take x so that 4x + 10 is within1

100of 26?

Solution:

|4x + 10 − 26| < 0.01 ⇔ |4(x − 4)| <1

100

⇔ |x − 4| <1

400

13. At what value of x does the function (x+1)2

x2−1have a removable discontinuity?

Solution: (x+1)2

x2−1= x+1

x−1for all x except x = −1, where the ratio is not defined

because both numerator and denominator of the fraction become 0 there. Every-where except at x = 1 this is a ratio of continuous functions, and the denominator


is non-zero. Only at x = 1 or at x = −1 can this function have any type ofdiscontinuity.

As x → 1−, the function approaches −∞; and, as x → 1+, the function approaches+∞. Thus x = 1 is an infinite discontinuity: it cannot be “removed” by the deviceof defining the function at the point, since no matter what value we would chooseto give the function there, it could still not be continuous.

But a different situation holds at x = −1. Since the function is equal to x+1x−1

near (but not at) x = −1, its one-sided limit from either side of −1 is 0, so thelimit exists at the point. However, the function is not defined at the point, sinceour definition was expressed as a ratio that becomes meaningless there. We can“remove” the discontinuity by extending the function to the domain R − {1} —we simply define the value of the newly extended function to be 0 there. Now thenew function is continuous at every point except −1.

14. At what value of x does the function (x+1)2

x2−1have an infinite discontinuity?

Solution: See the solution to Problem 13.

15. At what value of x does the function |x−2|x−2

have a jump discontinuity?

Solution: This function is equal to +1 when x > 2, and to −1 when x < 2; it isnot defined at x = 2. The limits from the left and right are different at x = 2,so the function has no limit there, and is said to be discontinuous. But, as bothone-sided limits exist, the discontinuity is a jump discontinuity.

16. Find the distance between the two values of x at which the function 1x2−3x+2

isdiscontinuous.

Solution: 1x2−3x+2

= 1(x−1)(x−2)

. Expressed as a ratio of two polynomials — whichare continuous everywhere — this function is also continuous everywhere, exceptpossibly at points where the denominator is zero. That happens precisely at x = 1and x = 2. At either of these points the function is not even defined — hence it isdiscontinuous at both of these points. The distance between the two discontinuitiesis, therefore, |2 − 1| = 1.

17. Find the values of x where the function f(x) = (x2−1)(x2+3x+2)(x2−1)2(x+2)(x−3)

has a removablediscontinuity.

Solution: This function is a ratio of polynomials, and polynomials are continuouseverywhere. This function will also be continuous everywhere, except where it is notdefined. That can happen only when the denominator is zero. So the discontinuitiesof the function are at x = −1, 1,−2, 3. We arrived at this information without evenconsidering the factorization of the numerator.


Let us now factorize both numerator and denominator:

(x2 − 1)(x2 + 3x + 2)

(x2 − 1)2(x + 2)(x − 3)=

(x − 1)(x + 1)2

(x − 1)2(x + 1)2(x − 3).

We can cancel equal non-zero factors in numerator and denominator. Thus

f(x) =1

(x − 1)(x + 3)

provided x is distinct from 1, −1, and −2. As x → 1, and also as x → −3, thisratio becomes infinite; so the limits as we approach these two points are infinite;in fact the function approaches +∞ from one side of each of these points, and −∞from the other side. So 1 and −3 are infinite discontinuities . But, as x → −2, theratio does not become infinite: the limit is 1

3from either side. Nor, as x → −1,

does the ratio become infinite: the limit is 14

from either side; but the function isnot defined at either x = 2 or x = −1. We can “remove” these discontinuities bydefining a new function which takes on the values 1

3at x = −2, 1

4at x = −1, and,

elsewhere, behaves like f(x).

18. Find the real number(s) c for which the function f(x) =

{

−3x if x ≤ 1(x − c)(x + c) if x > 1

is continuous on (−∞, +∞).

Solution: For x < 1 the function is defined by the upper line of the array; asx → 1−, f(x) approaches the value (−3) · 1 = −3. For x > 1 it is the secondline of the array which defines the function. Here, as x → 1+, (x − c)(x + c) →(1 − c)(1 + c). To make the function continuous at x = 1 it is necessary andsufficient that −3 = (1 − c)(1 + c), i.e. that c = 2 or c = −2.

19. Evaluate limx→∞

√

8x + 3x2

13x2 − 9.

Solution:

limx→∞

√

8x + 3x2

13x2 − 9= lim

x→∞

√

1x2

1x2

· 8x + 3x2

13x2 − 9

= limx→∞

√

1x2 (8x + 3x2)1x2 (13x2 − 9)

= limx→∞

√

8x

+ 3

13 − 9x2


= limx→∞

√

8x

+ 3√

13 − 9x2

=lim

x→∞

√

8x

+ 3

limx→∞

√

13 − 9x2

(Quotient Law)

=

√

limx→∞

(

8x

+ 3)

√

limx→∞

(

13 − 9x2

)

(Root Law)

=

√0 + 3√13 − 0

(Sum Law)

=

√3√13

=

√

3

13

20. Determine whether there exists a real number x which is exactly 10 more than its5th power.

Solution: (cf. [13, Solution to Exercise 2.6.59]). The problem may be paraphrasedas asking whether there exists a solution x to the equation x5 + 10 = x; or, equiv-alently, whether the function f(x) = x5 − x + 10 is ever zero. Now f is continuouseverywhere; f(0) = 10 > 0, and f(−2) = −32 + 2 + 10 < 0. By the IntermediateValue Theorem there will exist a point in the interval (−2, 0) at which f(x) = 0.

E.3.3 Third 2000/2001 Problem Assignment, with Solutions

1. Let f and g be functions whose domain is R, and which possess derivatives at everypoint in R. Suppose also that the following data are given about f , g and theirderivatives.

x f(x) g(x) f ′(x) g′(x)

1 1 −1 4 −22 −2 3 0 23 5 0 5 6

Showing all your work, determine each of the following, or explain why either

• it does not exist; or

• you do not have enough information to find the value.

(a)d

dx

(

f(x)

g(x)

)∣

∣

∣

∣

x=1


(b)d

dt(f(t) + g(t))

∣

∣

∣

∣

t=2

(c)d

dx

(

1

g(x)

)∣

∣

∣

∣

x=3

(d)

(

d

dx(f(x) + g′(x))

)

(2)

(e)d

dx

(

ex

f(x)

)∣

∣

∣

∣

x=1

(Do not attempt to approximate e.)

(f)d

dx

(

f(x)

f(x)

)∣

∣

∣

∣

x=2

Solution:

(a) By the Quotient Rule,

d

dx

(

f(x)

g(x)

)∣

∣

∣

∣

x=1

=f ′(1) · g(1) − f(1) · g′(1)

g(1)2=

4 · (−1) − 1 · (−2)

(−1)2= −2 .

(b) By the Sum Rule,

d

dx(f(x) + g(x)) (2) = f ′(2) + g′(2) = 0 + 2 = 2 .

(c) Since g(3) = 0, the function1

g(x)is not defined at x = 3. The derivative of

g at x = 3 is defined in terms of g(3); so, if g is not defined there, it cannothave a derivative there either.

(d) We do not know whether the function g′ has a derivative at x = 2 — equiv-alently, whether the function g has a second derivative at x = 2. Withoutthis information, and the actual value of that derivative if it exists, we cannotevaluate the derivative of the sum.

(e) By the Quotient Rule,

d

dx

(

ex

f(x)

)∣

∣

∣

∣

x=1

=ddx

ex∣

∣

x=1· f(1) − e1 · f ′(1)

f(1)2

=e1 (f(1) − f ′(1))

f(1)2= e(1 − 4) = −3e


(f) This case can also be attacked using the Quotient Rule, and we find that

d

dx

(

f(x)

f(x)

)

(2) =f ′(2) · f(2) − f(2) · f ′(2)

f(2)2=

4 · 1 − 1 · 412

= 0 .

Alternatively, one can observe that the functionf(x)

f(x)is defined wherever

f(x) 6= 0; in particular, it is defined at x = 2; wherever it is defined, its valueis 1. Thus this is a constant function! The derivative of any constant functionis zero.

2. Suppose that a particle moves in a straight line with its position at time t givenby the formula f(t) = et(sin t + cos t).

(a) Find the velocity, the speed, and the acceleration of the particle at time t.

(b) Determine the average velocity during the time interval from t = 0 to t = 2π.

(c) Determine the smallest positive value of t — if any — when the particlereturns to the origin.

(d) Determine the smallest positive value of t — if any — when the particle isstationary — i.e. the velocity is 0.

(e) Determine whether there is a maximum distance that the particle attainsaway from the origin. If there is a maximum distance, determine what it is.

[The velocity is given by the derivative of f(t) (which is called the displacement ;the speed is the magnitude of the velocity. The acceleration is the derivative of thevelocity with respect to time.]

Hints: You may wish to use the identities:

sin(

x +π

4

)

=1√2

(sin x + cos x)

cos(

x +π

4

)

=1√2

(− sin x + cos x)

which are consequences of the addition formulæ and the known values for the sineand cosine of π

4. Do not attempt to simplify complicated formulæ involving π.

Solution:

(a) The velocity at time t is f ′(t) = et (cos t − sin t) + et(sin t + cos t) = 2et cos t.The speed is the magnitude of the velocity, i.e., 2et| cos t|. (We may removethe exponential factor outside the absolute signs since exponentials are always


positive.) The acceleration is the derivative of the velocity, i.e.d

dtf ′(t) =

2(

et(− sin t) + et cos t)

= 2et(− sin t + cos t).

(b) Using one of the given identities, we see that f(t) =√

2 · et · sin(

x + π4

)

. Theaverage velocity from t = 0 to t = 2π is

f(2π) − f(0)

2π − 0

=1

2π

(√2e2π · sin

(

2π +π

4

)

−√

2 · e0 · sin(π

4

))

=1

2π

√2(

e2π − 1)

sinπ

4=

e2π − 1

2π

(c) We seek the smallest positive value t when f(t) = 0, i.e. when et√

2 sin(

t + π4

)

=0. Since et is never zero, this equation is equivalent to sin

(

t + π4

)

= 0, whosesolutions are t + π

4= nπ, where n is any integer. The smallest positive value

will then be 34π.

(d) 2et cos t will be zero precisely when the cosine is zero, i.e. when t is an oddinteger multiple of π

2.

(e) We have seen above that f(t) is a product of et (which is never 0), the con-stant

√2, and the factor sin

(

t + π4

)

, which oscillates in value between +1 and−1. Because of the exponential factor et the amplitude of the oscillations isincreasing as t becomes large. The limit lim

t→∞f(t) does not exist. We can see

this by exhibiting values of t where the function is arbitrarily large positively,and others where it is arbitrarily large negatively. For example, if we take

t = 2nπ +π

4, we find the values oscillating between et and −et, and we know

that the exponential function approaches infinity.

3. Determine the derivative of each of the following functions. You may not use theChain Rule, but you may use any of the Rules and Theorems in [11, §§3.1–3.4],including the General Power Rule.

(a) 12

(

ex +1

ex

)

(b) 12(ex − e−x)

(c)x4 + 2x2 − x − 5√

x

(d) xe − ex


Solution:

(a)

d

dx

(

1

2

(

ex +1

ex

))

=1

2

d

dx

((

ex +1

ex

))

Constant Multiple Rule

=1

2

(

dex

dx+

0 · ex − 1 · ex

e2x

)

Quotient Rule

=ex − e−x

2

(b) Analogously to the preceding,ex + e−x

2.

(c)

x4 + 2x2 − x − 5√x

= x72 + 2x

32 − x

12 − 5x− 1

2

⇒ d

dx

(

x4 + 2x2 − x − 5√x

)

=d

dx

(

x72 + 2x

32 − x

12 − 5x− 1

2

)

=7

2x

52 +

3

2· 2x 1

2 − 1

2x− 1

2 −(

−1

2

)

5x− 32

=7

2x

52 + 3x

12 − 1

2x− 1

2 +5

2x− 3

2

(d)d

dx(xe − ex) = e · xe−1 − ex, by the Generalized Power Rule and properties of

the exponential.

4. Showing all your work, determine the derivative of each of the following functionsfrom first principles: that is, you are to evaluate a limit in each case, and are notto use any of the “Rules” for evaluating derivatives.

(a) f(x) = 4x − 6

(b) g(x) =√

2x + 3

(c) h(x) =1

x3. [Hint: Remember the factorization a3− b3 = (a− b)(a2 +ab+ b2).]

Solution:

(a) For all x,

f ′(x) = limh→0

(4(x + h) − 6) − (4x − 6)

h= lim

h→0

4h

h= 4


(b) For all x,

g′(x) = limh→0

√

2(x + h) + 3 −√

2x + 3

h

= limh→0

√

2(x + h) + 3 −√

2x + 3

h·√

2(x + h) + 3 +√

2x + 3√

2(x + h) + 3 −√

2x + 3

= limh→0

(2(x + h) + 3) − (2x + 3)

h(√

2(x + h) + 3 +√

2x + 3)

= limh→0

2√

2(x + h) + 3 +√

2x + 3

=1√

2x + 3

(c) For all x,

h′(x) = limh→0

1

(x + h)3− 1

x3

h

= limh→0

x3 − (x + h)3

hx3(x + h)3= lim

h→0

−3x2h − 3xh2 − h3

hx3(x + h)3

= limh→0

−3x2 − 3xh − h2

x3(x + h)3

=limh→0

(−3x2 − 3xh − h2)

limh→0

(x3(x + h)3)=

−3x2

x6= −3x−4

5. Determine precisely where the function f , defined below, is differentiable.

f(x) =

−x − 2 if x ≤ −1−x2 if −1 < x < 01 if x = 0

−x2 if 0 < x ≤ +24 − 4x if x > +2

Solution: The function is a polynomial in each of the intervals (−∞,−1], (−1, 0),(0, 2), and (2,∞). As polynomials are differentiable [11, §3.1], the only placeswhere differentiability can fail is at the points x = −1, 0, 2. We check each of themseparately.


x = −1: We have to determine whether the limit limx→−1

f(x) − f(−1)

x − (−1)exists. By

[11, Theorem 2.3.1, p. 107], the two one-sided limits must exist and be equal.But

limx→−1−

f(x) − f(−1)

x − (−1)= lim

x→−1−

(−x − 2) − (1 − 2)

x − (−1)

= limx→−1−

−x − 1

x + 1= −1 while

limx→−1+

f(x) − f(−1)

x − (−1)= lim

x→−1+

−x2 − (1 − 2)

x − (−1)

= limx→−1+

−x2 + 1

x + 1= lim

x→−1+(−x + 1) = 2 6= −1

As the one-sided limits are not equal, the function is not differentiable at thispoint.

x = 0: Here we have to consider the existence of limx→0

f(x)−f(0)x

, that is limx→0

−x2−1x

.

But this limit cannot exist, since the ratio becomes infinitely large close to0, as the numerator approaches −1, but is divided by an arbitrarily smalldenominator. Thus the limit does not exist, and the function fails to bedifferentiable at x = 0.

We could have argued this alternatively by appealing to [11, Theorem 2.9.4,p. 169], since the function is discontinuous at x = 0, so it cannot possibly bedifferentiable there.

x = 2: This time we have to consider the existence of limx→2

f(x)−f(2)x−2

. We observe

from the definition that f(2) is defined to be −22 = −4. Then

limx→2−

f(x) − f(2)

x − 2= lim

x→2−

−x2 − (−4)

x − 2= lim

x→2−(−x − 2) = −4 ; and

limx→2+

f(x) − f(2)

x − 2= lim

x→2+

(4 − 4x) − (−4)

x − 2= lim

x→2+−4 = −4

As the limits from the left and right are equal, the function is, indeed, differ-entiable at the point x = 2.

6. The curve y = 3x5 − 20x3 − 675x + 12 has some points with horizontal tangents.Showing all your work, find all such points.


Solution: Settingd

dx

(

3x5 − 20x3 − 675x + 12)

equal to zero, we find that 15x4 −60x2 −675 = 0, or, equivalently, (x2)2 −4x2 −45 = 0. This is a quadratic equation(in x2), and its solutions are x2 = −5 and x2 = 9. There are no real solutions tox2 = −5, so we may confine ourselves to the second equation, whose solutions arex = ±3. Thus there are exactly 2 points with horizontal tangents: (3,−1824) and(−3, 1848).

7. Determine the value of each of the following limits, or explain why they do notexist. Do not use l’Hopital’s Rule.

(a) limx→0

tan3 πx

x3

(b) limθ→0

(cot 5θ)2 · (sin 3θ) · tan(−2θ)

Solution:

(a)

limx→0

tan3 πx

x3= lim

x→0

(

(

sin πx

πx

)3

·( π

cos πx

)3)

=

(

limx→0

sin πx

πx

)3(

limx→0

π

cos πx

)3

(Product Rule)

=

(

limx→0

sin πx

πx

)3(limx→0 π

limx→0 cos πx

)3

(Quotient Rule)

=

(

limy→0

sin y

y

)3(limx→0 π

limx→0 cos πx

)3

(where y = πx)

= 13 ·(π

1

)3

= π3

(b)

limθ→0

(

(cot 5θ)2 · (sin 3θ) · tan(−2θ))

= limθ→0

(

cos2 5θ · 1

25θ2·(

5θ

sin 5θ

)2

·(

sin 3θ

3θ

)

· 3θ ·(

tan(−2θ)

−2θ

)

· 2θ)

= − 6

25limθ→0

cos2 5θ · limθ→0

(

(

5θ

sin 5θ

)2

·(

sin 3θ

3θ

)

·(

sin(−2θ)

−2θ

)

)

· limθ→0

1

cos(−2θ)

= − 6

25· 1 · 12 · 1 · 1 · 1 = − 6

25


8. Determine all points (x, cos2 x) on the graph of the function cos2 x (i.e. cos x ·cos x)at which

(a) the tangent to the curve is parallel to the line y = x + 1.

(b) the tangent to the curve is horizontal

(c) the tangent to the curve is vertical

(d) the tangent to the curve is parallel to the line y = 2x − 3

(e) the normal to the curve passes through the origin, and |x| > 1.

[Hints: The first and last parts of the problem are more difficult than the others.Remember the identities involving sin 2x and cos 2x. The normal to the curve ata point is the line through the point which is perpendicular to the tangent.]

Solution:

(a) The derivative of (cos x)2 is cos x·(− sin x)+(− sin x)·cos x = −2 sin x·cos x.99

We impose the condition that this be equal to the slope of the line y = x + 1,i.e., that

−2 sin x · cos x = 1 (215)

The points we seek will be the solutions to equation (215). A simple attackis to observe that this equation is equivalent to

sin 2x = −1 . (216)

The general solution to the last equation is 2x =(

32

+ 2n)

π, where n is anyinteger; hence x =

(

34

+ n)

π, where n is any integer. Thus there are infinitelymany points where the tangent has slope 1: the x-coordinates are as indicated;the y-coordinates are cos2

(

34

+ n)

π = 12. But note that the line y = 1

2cuts

the curve also in infinitely many points where the tangent has slope −1.

(b) We have to solve the equation −2 sin x ·cosx = 0, or, equivalently, −2 sin 2x =0. The solutions to this equation are 2x = nπ, i.e. x = nπ

2, where n is any

integer.

(c) The function is differentiable everywhere; that means that the derivative hasa (finite) value at every point, and so the tangent cannot be vertical. Thereare no such points on the curve.

(d) The value of the derivative is −2 sin x · cos x, or − sin 2x. As a sine cannothave magnitude exceeding 1, the slopes of tangents to this curve can neverequal 2. There are no points of this type!

99We are avoiding using the Chain Rule, as the assignment was due before the Rule had been fullydiscussed in the lectures.


(e) The slope of the normal at the point (t, cos2 t) is − 1

− sin 2t. The equation of

the normal is

y − cos2 t =1

sin 2t(x − t)

The normal will pass through the origin if the coordinates of the origin satisfythe equation, that is if t = cos2 t sin 2t. But this implies that t is the productof two factors, neither of which can exceed 1 in magnitude, so there are nopoints with this property outside the interval −1 ≤ t ≤ 1.

E.3.4 Fourth 2000/2001 Problem Assignment, with Solutions

Solutions to most of the problems can be found in the Student Solutions Manual [13].

1. (a) [11, Exercise 3.5.3, p. 221] Write the composite function cos(tanx) in the formf(g(x)). [Identify the “inner” function u = g(x), and the “outer” function

y = f(u).] Then find the derivativedy

dx.

(b) Find an equation for the tangent to the curve y = cos(tanx) at the point

x =π

4.

(c) Find an equation for the normal to the curve y = cos(tanx) at the point

x =5π

4.

Solution:

(a) In [13, Exercise 3.5.3, p. 82] the derivative is shown to have value − sin tanx ·sec2 x.

(b) The slope of the tangent at(π

4, cos 1

)

is − sin 1 ·√

22. Hence an equation of

the tangent is

y − cos 1 = −2 sin 1 ·(

x − π

4

)

.

(c) The slope of the tangent at

(

5π

4, cos(1)

)

is − sin(1) · (−√

2)2 = −2 sin 1;

hence the slope of the normal is1

2 sin 1. An equation of the tangent is

y − cos 1 =1

2 sin 1·(

x − 5π

4

)

.

or x − 2 sin 1 · y = − sin 2 +5π

4.


2. Find the derivative of each of the following functions:

(a) [11, Exercise 3.5.19, p. 222] (2x − 5)4(8x2 − 5)−3

(b) [11, Exercise 3.5.21, p. 222] xe−x2

(c) [11, Exercise 3.5.29, p. 222] 5−1x

Solution:

(a) [13, Exercise 3.5.19, p. 82]

(b) [13, Exercise 3.5.21, p. 82] Note that the original function is odd ; and thederivative is even. This is an instance of [11, Exercise 3.5.73, p. 224].

(c) [13, Exercise 3.5.29, p. 82]. (This problem could be solved from first princi-ples. Simply observe that 5 = eln 5, apply the usual rules for exponents, anddifferentiate an exponential [now to base e] using the Chain Rule.)

3. [11, Exercise 3.5.56, p. 222] A table of values for f , g, f ′, g′ is given:

x f(x) g(x) f ′(x) g′(x)

1 3 2 4 62 1 8 5 73 7 2 7 9

(a) If F (x) = (f ◦ f)(x), determine F ′(2).

(b) If G(x) = (g ◦ g)(x), determine G′(3).

Solution:

(a) F ′(2) = f ′(f(2)) · f ′(2) = f ′(1) · f ′(2) = 4 · 5 = 20.

(b) G′(3) = g′(g(3)) · g′(3) = g′(2) · g′(3) = 7 · 9 = 63.

4. In each of the following cases, finddy

dxby implicit differentiation.

(a) [11, Exercise 3.6.11, p. 230]y

x − y= x2 + 1

(b) [11, Exercise 3.6.13, p. 230]√

xy = 1 + x2y

(c) [11, Exercise 3.6.17, p. 230] cosx − y = xex

Solution:

(a) Two different approaches are given in [13, Exercise 3.6.11, p. 86].


(b) [13, Exercise 3.6.13, p. 86]

(c) [13, Exercise 3.6.18, p. 86]

5. (a) [11, Exercise 3.6.23, p. 230] Regarding y as the “independent” variable, and xas the “dependent” variable (i.e. regarding x as a function of y) apply implicitdifferentiation to the equation

y4 + x2y2 + yx4 = y + 1

to determinedx

dy.

(b) Use the same equation to determinedy

dx, and verify that

dy

dx· dx

dy= 1 (a fact

which follows from the Chain Rule by interpreting y as a function of x which,in turn, is a function of y, i.e. taking the point of view y = y(x(y))).

Solution:

(a) [13, Exercise 3.6.23, p. 86]

(b) Differentiating both sides of the given equation with respect to x yields

4y3 · y′ +(

2xy2 + x2 · 2yy′)+(

y′x4 + y · 4x3)

= y′ + 0

⇒(

4y3 + 2x2y + x4 − 1)

y′ = −2xy2 − 4yx3

⇒ y′ =2xy2 + 4yx3

1 − 4y3 − 2x2y − x4

6. (a) Determine the derivative of cot−1(

x −√

1 + x2)

(b) [11, Exercise 3.6.41, p. 231] Determine the derivative of arcsin(x2).

Solution:

(a)

d

dx

(

cot−1(

x −√

1 + x2))

= − 1

1 +(

x −√

1 + x2)2 · d

dx

(

x −√

1 + x2)

= − 1

1(

2 + 2x2 − 2x√

1 + x2) ·(

1 − x√1 + x2

)

= − 1

2√

1 + x2(√

1 + x2 − x) ·

√1 + x2 − x√

1 + x2= − 1

2(1 + x2)


(b) [13, Exercise 3.6.41, p. 88]

7. (a) [11, Exercise 3.6.35] Find all points on the curve 2(x2 + y2)2 = 25(x2 − y2)where the tangent is horizontal. (You may assume that the origin is not oneof these points.)

(b) For the same curve, determine all points wheredx

dy= 0, i.e. where the tangent

is vertical. (You may assume that the origin is not one of these points.)

Solution:

(a) [13, Exercise 3.6.35, p. 88] (It is shown that the four points with horizontaltangents lie on a certain circle centred at the origin. To determine the pointsone must solve the equation of this circle with the equation of the given curve.)

(b) One finds by implicit differentiation that

dx

dy=

25y + 4y(x2 + y2)

25x − 4x(x2 + y2)

This derivative vanishes when y = 0 or 4(x2 +y2) = −25. The latter situationcannot occur, as a sum of squares cannot be negative; hence y = 0. Thatis, the points with vertical tangents are at points where the curve meets the

x-axis: these are the points where x = 0 or x = ± 5√2. But being on the x-axis

is a necessary condition for having a vertical tangent, and is not sufficient. Atthe origin there are two tangents, and neither of them is vertical. The only

points with vertical tangents are(

± 5√2, 0)

.

8. (cf. [11, Exercise 3.7.53, p. 239]) Find constants ci (i = 0, 1, 2) such that thefunction f (t) = c2t

2 + c1t1 + c0 has the properties that f(−1) = 0, f ′(−1) = 7,

f ′′(−1) = 10.

Solution: Imposing the given conditions yields the three equations

c2 − c1 + c0 = 0

−2c2 + c1 = 7

2c2 = 10

from which we conclude that c0 = 22, c1 = 27, c2 = 5.

9. [11, Exercise 3.8.27, p. 246] Find the domain of the function f , and determine itsderivative, where f(x) = x2 ln(1 − x2).

Solution: [13, Exercise 3.8.27, p. 94]


10. Determine the first and second derivatives of the function f(x) = | ln(sec x+tanx)|.Solution:

d

dxln(sec x + tanx) =

1

sec x + tanx· d

dx(sec x + tanx)

=1

sec x + tanx· (sec x · tanx + sec2 x) = sec x

d2

dx2ln(sec x + tanx) =

d

dxsec x = sec x · tanx

The problem referred to the derivatives of f(x) = | ln(sec x+tan x)|; the logarithmwill be negative precisely when sec x+tan x < 1; this last inequality may be shown,using trigonometric identities, to be equivalent to

cos(x

2+

π

4

)

· sin(x

2

)

< 0 .

For the range 0 < x < 4π, we find, examining the signs of the two factors as0 < x

2< π

4, ... 7π

4< x

2< 2π, that the product is negative except when 0 < x

2< π

4

or π < x2

< 5π4

. Thus, if we define s(x) to be +1 when 2nπ < x <(

2n + 12

)

π,where n is any integer, and −1 everywhere else, f ′(x) = s(x) · sec x, and f ′′(x) =s(x) · sec x tan x; neither function is defined at odd integer multiples of π

2.

11. Using “logarithmic differentiation”, or otherwise, determine the derivative of

(a) [11, Exercise 3.8.37, p. 246] y =sin2 x · tan4 x

(x2 + 1)2.

(b) y = xln x

Solution:

(a) [13, Exercise 3.8.37, p. 95]

(b) Since ln y = (ln x)2,1

y· y′ = 2 lnx · 1

x. Hence y′ = 2x(ln x)−1 ln x.

12. (a) [11, Exercise 3.7.19, p. 238] Find the first, second, and third derivatives ofg(t) = t3e5t.

(b) [11, Exercise 3.7.13, p. 238] Find the second, third, and fourth derivatives ofx

1 − x.

(c) [11, Exercise 3.7.29, p. 238] Find y′′ where y is defined implicitly by theequation x3 + y3 = 1. Simplify your answer as much as possible.

Solution:


(a) [13, Exercise 3.7.19, p. 90]

g′′′(t) =d

dt

(

(25t3 + 30t2 + 6t)e5t)

=(

75t2 + 60t1 + 6t0)

e5t + 5(

25t3 + 30t2 + 6t)

e5t

=(

125t3 + 225t2 + 90t1 + 6t0)

e5t

(b) [13, Exercise 3.7.13, p. 90] This problem can be attacked naıvely, with re-peated applications of the Quotient Rule. Another approach is to observe

thatx

1 − x= −1 +

1

1 − x. Hence

d

dx

x

1 − x= 0 +

d

dx

(

(1 − x)−1)

= (−1) · (1 − x)−2 · d

dx(1 − x) = (1 − x)−2

d2

dx2

x

1 − x=

d

dx

(

(1 − x)−2)

= (−2) · (1 − x)−3 · (−1) = 2 · (1 − x)−3

d3

dx3

x

1 − x= 2

d

dx

(

(1 − x)−3)

= 2(−3) · (1 − x)−4 · (−1) = 3 · 2 · (1 − x)−4

d4

dx4

x

1 − x= 3 · 2 d

dx

(

(1 − x)−4)

= 3 · 2 · (−4) · (1 − x)−5 · (−1)

= 4 · 3 · 2 · (1 − x)−5

(c) [13, Exercise 3.7.29, p. 91]

13. Evaluate the following limits, if possible; you may assume that limn→∞

(

1 +x

n

)n

= ex.

In each case it is suggested that you make a substitution of the form m = kn, wherek is a constant that you choose, and then replace lim

n→∞by lim

m→∞when k > 0, or by

limm→−∞

when k < 0.

(a) limn→∞

(

1 +2

n

)n

.

(b) limn→∞

(

1 − x

3n

)n

.

Solution:

(a) Define m =n

2, so n = 2m. Then

limn→∞

(

1 +2

n

)n

= limm→∞

(

1 +1

m

)2m


= limm→∞

(

((

1 +1

m

)m)2)

=

(

limm→∞

(

1 +1

m

)m)2

Product Law

= e2

(b) Define m = −3n, so n = −m

3. Then

limn→∞

(

1 − 1

3n

)n

= limm→−∞

3

√

(

1 − x

m

)m

= 3

√

limm→−∞

(

1 − x

m

)m

Root Law

= 3

√

limt→∞

(

1 +x

t

)−t

(t = −m)

= 3

√

limt→∞

1(

1 + xt

)t

=1

3

√

limt→∞

(

1 +x

t

)t=

13√

ex= e−

x3

E.3.5 Fifth 2000/2001 Problem Assignment, with Solutions

1. [11, Exercise 3.9.12, p. 251] Prove the following identity about hyperbolic functions,using only the definitions of these functions in terms of exponentials:

cosh(x + y) = cosh x · cosh y + sinh x · sinh y

Solution:

cosh x · cosh y + sinh x · sinh y

=ex + e−x

2· ey + e−y

2+

ex − e−x

2· ey − e−y

2

=(ex+y + ex−y + e−x+y + e−x−y) + (ex+y − ex−y − e−x+y + e−x−y)

4

=2ex+y + 2e−x−y

4= cosh(x + y)


2. [11, Exercise 3.9.21, p. 251] If tanhx =4

5, find the values of the other 5 hyperbolic

functions at x.

Solution: [13, Exercise 3.9.21, p. 96]. (The problem could also be solved by usingthe given information to show that e2x = 9, hence ex = +3, and thence determiningthe values of the other functions.)

3. Use the definitions of the hyperbolic functions to find the following limits:

(a) [11, Exercise 3.9.23(b), p. 252] limx→−∞

tanh x

(b) [11, Exercise 3.9.23(c), p. 252] limx→∞

sinh x

(c) [11, Exercise 3.9.23(e), p. 252] limx→∞

sechx

(d) [11, Exercise 3.9.52, p. 252] limx→∞

sinh x

ex

Solution:

(a) [13, Exercise 3.9.23(b), p. 96]

(b) [13, Exercise 3.9.23(c), p. 96]

(c) [13, Exercise 3.9.23(e), p. 96]

(d) Sincesinh x

ex=

1

2

(

1 − e−2x)

, and limx→∞

e−x = 0, the limit as x → ∞ is1

2.

4. Determine the derivative of each of the following functions:

(a) [11, Exercise 3.9.32, p. 252] g(x) = sinh2 x

(b) [11, Exercise 3.9.38, p. 252] f(t) = ln(sinh t)

(c) [11, Exercise 3.9.45, p. 252] x sinh−1 x

3−

√9 + x2

Solution:

(a) g′(x) = 2 sinh x·cosh x, which could be expressed more compactly as sinh(2x).

(b)d

dtf(t) =

1

sinh t· cosh t = coth t

(c) [13, Exercise 3.9.45, p. 98]

5. (a) [11, Exercise 3.10.7, p. 257] A street light is mounted at the top of a 15-foot-tall pole. A man 6 feet tall walks away from the pole with a speed of 5feet/second along a straight path. How fast is the tip of his shadow movingwhen he is 40 feet from the base of the pole?


(b) The man extends his arms horizontally so that the distance between his fin-gertips is 6 feet. What is the rate of increase of the shadow of his extendedarms on the ground when his feet are 40 feet from the base of the pole? Youmay assume that his shoulders are 5 feet above the ground.

Solution:

(a) [13, Exercise 3.10.7, p. 99]

(b) We can consider two sets of similar triangles similar to the triangles consideredin part (a). Let A denote the top of the lamppost, and B its foot; let C bea point on the man, at shoulder-height, and let D be a point on the man’sfeet; let G be the point where the line through A and C meets the ground.

Then triangles ABG and CDG are similar, so15

5=

|AB||CD| =

|AG||CG| =

|BG||DG| ,

so |BG| = 3|DG|, |AG| = 3|CG|, and |BD| = 2|DG|.Now let E be the end of the man’s horizontally-extended left arm, and let Fbe the end of the shadow cast by that arm on the ground. Here triangles AGF

and ACE are similar, so|AG||AC| =

|GF ||CE| , so, since |AG| = 3|CG|, |GF |

|CE| =3

2;

since |CE| = 3, |GF | =9

2. The shadow of the extended arms is 2 × 9

2= 9

feet long. As this shadow is of constant length, its rate of change is 0.

6. [11, Exercise 3.10.25, p. 258] Boyle’s Law states that, when a sample of gas iscompressed at a constant temperature, the pressure P , and volume V satisfy theequation PV = C, where C is a constant depending on the sample. Suppose that,at a certain instant, the volume is 600 cm3, the pressure is 150kPa, and the pressureis increasing at a rate of 20 kPa/minute. At what rate is the volume decreasing atthis instant?

Solution: [13, Exercise 3.10.25, p. 100]

7. Use differentials (or, equivalently, a linear approximation) to estimate the followingnumbers:

(a) [11, Exercise 3.11.34, p. 265]

(b) [11, Exercise 3.11.35, p. 265]

(c) [11, Exericse 3.11.36, p. 265]

Solution:

(a) Let y = f(x) = x6, so dy = 6x5 dx. (1.97)6 = (2 − 0.03)6 = f(2 − 0.03) ≈f(2) + f ′(2) · (−0.03) = 26 + 6 · 25 · (−0.03) = 58.24.


(b) [13, Exercise 3.11.35, p. 104]

(c) Let y = g(x) = ln x. Then dy = g′(x) dx =dx

x. Then ln(1.07) = g(1+0.07) ≈

g(1) + g′(1) · 0.07 = ln 1 +0.07

1= 0.07.

8. [11, Exercise 4.2.32, p. 294] Prove the identity

2 arcsin x = arccos(1 − 2x2) (x ≥ 0)

by using the method of [11, Example 6, p. 2902].

Solution: Define f(x) = 2 arcsinx − arccos(1 − 2x2). Then f ′(x) =1√

1 − x2−

−1√

1 − (1 − 2x2)2·(0−4x) = 2

1√1 − x2

− 4x√

22|x|2(1 − x2)= 2

1√1 − x2

− 4x

2x√

1 − x2=

0; note that we have used the hypothesis that x > 0. Hence, by [11, Theorem 4.2.5,p. 291], f(x) =constant. We can determine the value of the constant by select-

ing a “convenient” value; for example, take x = 1: then, since arcsin 1 =π

2, and

arccos(−1) = π, we find that the constant is equal to 0.

Note that this argument does not hold for x = 0. There, however, we can simplyobserve that f(0) = 2 arcsin 0 − arccos 1 = 2 · 0 − 0 = 0.

9. [11, Exercise 4.2.6] Let f(x) = (x − 1)−2. Show that f(0) = f(2), but there is nonumber c such that 0 < c < 2 and f ′(c) = 0. Why does this not contradict Rolle’sTheorem?

Solution: f(0) = (−1)−2 = 1; f(2) = 1−2 = 1. Since f ′(x) = −2(x − 1)−3,wherever it is defined, the value of the derivative ranges, as 0 ≤ x < 1, from 2 to∞; similarly, as 1 < x ≤ 2, the derivative ranges between −∞ and −2. Thus thederivative never takes values between −2 and 2.

This is not a counterexample to Rolle’s Theorem, since that theorem requiresthat the function be differentiable throughout the open interval, and the presentfunction fails to be differentiable at one point in the interval — namely the pointx = 1. That failure to be differentiable, even though it is only at a single point,renders Rolle’s Theorem inapplicable.

10. [11, Exercise 4.2.17, p. 293] Show that the polynomial x5 +10x+3 has exactly onereal root.

Solution: f(0) = 3 > 0, but f(−1) = −12 < 0; by the Intermediate Value Theoremthis function — which, being a polynomial, is continuous everywhere, in particular


in the interval −1 < x < 0 — must take on the value 0 at least once between −1and 0. However, the Intermediate Value Theorem does not exclude the possibilitythat there is more than one place where f = 0. If, however, f were to vanish atdistinct points x1 and x2, then, by Rolle’s theorem, f ′ would be zero somewherebetween x1 and x2. But f ′(x) = 5x4 +10; being the sum of a square and a positivenumber, this cannot be zero.

11. For each of the following functions

• find the “critical numbers”;

• find the vertical and horizontal asymptotes;

• find the intervals of increase or decrease;

• find the local maximum and minimum values;

• find the intervals of concavity and the inflection points;

• use the information you have determined above to sketh the graph.

(A “critical number” of a function is a point in the domain of the function whereeither the derivative is not defined, or the derivative is defined and equal to zero.This term is sometimes used (by other authors) with a slightly different definition.)

(a) (cf. [11, Exercise 4.3.39, p. 304]) f1(x) =√

x2 + 1 − x

(b) [11, Exercise 4.3.43, p. 304] f2(x) =1 + x2

1 − x2

(c) [11, Exercise 4.3.45, p. 304] f3(x) =√

x2 + 1 − x

(d) [11, Exercise 4.3.47, p. 304] f4(x) = ln(1 − lnx)

(e) [11, Exercise 4.3.49, p. 304] f5(x) = e−1

x+1

Solution:

(a) (cf. [13, Exercise 4.3.39, p. 132]) Since f1(x) = x

(

1 +3

x

)

, and since

(

1 +3

x

)

→1 as either x → ∞ or x → −∞, lim

x→±∞f1(x) = ±∞, so there are no horizontal

asymptotes. All limits of f1(x) as x approaches any (finite) point are finite,so there are no vertical asymptotes either. (However, the graph of f1 doeshave a vertical tangent at x = −3.)

(b) [13, Exercise 4.3.43, p. 133]. The derivative is defined everywhere in thedomain of the function, and vanishes when x = 0, so x = 0 is the unique“critical number”.


(c) [13, Exercise 4.3.45, p. 134]. The derivative is defined everywhere, and nevervanishes, so there are no “critical numbers”.

(d) [13, Exercise 4.3.47, p. 134]. The derivative is defined everywhere in thedomain of the function, and never vanishes, so there are no “critical numbers”.

(e) [13, Exercise 4.3.49, p. 134]. The derivative is defined everywhere in thedomain of the function, and never vanishes, so there are no “critical numbers”.

E.3.6 Sixth 2000/2001 Problem Assignment, with Solutions

1. In each of the following cases you are to find the limit, if it exists. Since the subjectof [11, §4.4] is l’Hopital’s Rule, you should try to use that rule, if it is possible.However, if l’Hopital’s Rule cannot be applied, explain why. Mathematicians usu-ally regard the avoidance of l’Hopital’s Rule as a challenge; try to find another wayto evaluate those limits where you use the Rule.

(a) [11, Exercise 4.4.9, p. 312] limx→0

ex − 1

sin x

(b) [11, Exercise 4.4.15, p. 312] limx→∞

ln x

x

(c) [11, Exercise 4.4.17, p. 312] limx→0+

ln x

x

(d) [11, Exercise 4.4.27, p. 312] limx→0

1 − cos x

x2

(e) [11, Exercise 4.4.37, p. 312] limx→0

2x − arcsin x

2x + arccos x

(f) [11, Exercise 4.4.38, p. 312] limx→0

2x − arcsin x

2x + arctan x

(g) [11, Exercise 4.4.51, p. 312] limx→∞

(

x −√

x2 − 1)

(h) [11, Exercise 4.4.57, p. 312] limx→0

(

(1 − 2x)1x

)

Solution: We shall follow the notation of the Student Solution Manual [13], and

writeH= when we are applying l’Hopital’s Rule.

(a) [13, Exercise 4.4.9, p. 138]

limx→0

ex − 1

sin x

H= lim

x→0

ex

cos x


=limx→0

ex

limx→0

cos x=

e0

cos 0= 1

limx→0

ex − 1

sin x= lim

x→0

ex − 1

x· x

sin x

= limx→0

ex − e0

x· lim

x→0

x

sin x(Product Law)

=d

dxex

∣

∣

∣

∣

x=0

· 1

= e0 = 1

(b) [13, Exercise 4.4.15, p. 138].

(c) L’Hopital’s rule cannot be used, since the limit of the numerator is (negatively)infinite, while the limit of the denominator is 0. As x → 0+, ln x → −∞,

while1

x→ +∞, so the product ln x · 1

xapproaches −∞.

(d) [13, Exercise 4.4.27, p. 138]. Here is one way to avoid using l’Hopital’s Rule:

limx→0

1 − cos x

x2= lim

x→0

(

1 − cos x

x2· 1 + cos x

1 + cos x

)

= limx→0

(

1 − cos2 x

x2· 1

1 + cos x

)

= limx→0

(

(

sin x

x

)2

· 1

1 + cos x

)

=

(

limx→0

(

sin x

x

))2

· 1

limx→0

(1 + cosx)

= 12 · 1

1 + 1=

1

2

(e) [13, Exercise 4.4.37, p. 138].

(f)

limx→0

2x − arcsin x

2x + arctan xH= lim

x→0

2 − 1√1 − x2

2 +1

1 + x2

=2 − lim

x→0 1

√1 − x2

2 + limx→0

1

1 + x2


=2 − 1

2 + 1=

1

3

It is possible to avoid the Rule, but the result looks very much like what wehave proved above:

limx→0

2x − arcsin x

2x + arctan x= lim

x→0

2 − arcsin x − arcsin 0

x − 0

2 +arctan x − arctan 0

x − 0

=2 − lim

x→0

arcsin x − arcsin 0

x − 0

2 + limx→0

arctan x − arctan 0

x − 0

=

2 − d

dxarcsin x

∣

∣

∣

∣

x=0

2 +d

dxarctan

∣

∣

∣

∣

x=0

=

2 − 1√1 − 02

2 +1

1 + 12

=1

3

(g) [13, Exercise 4.4.51, p. 139].

(h) [13, Exercise 4.4.57, p. 139].

2. Discuss each of these functions under the following headings, using the guidelinesof the same names in [11, pp. 315–316]

A. Domain

B. Intercepts

C. Symmetry

D. Horizontal or Vertical Asymptotes. (Do not attempt to investigate slantasymptotes.)

E. Intervals of Increase or Decrease

F. Local Maximum and Minimum Values

G. Concavity and Points of Inflection


Then sketch the graph of the function.

(a) [11, Exercise 4.5.5, p. 321] f1(x) = x4 + 4x3

(b) [11, Exercise 4.5.17, p. 321] f2(x) =1

x3 − x

(c) [11, Exercise 4.5.29, p. 321] f3(x) = x +√

|x|(d) [11, Exercise 4.5.47, p. 322] f4(x) = ln(x2 − x)

(e) [11, Exercise 4.5.39, p. 322] f5(x) = sin(2x) − 2 sin x

Solution:

(a) [13, Exercise 4.5.5, p. 142]

(b) [13, Exercise 4.5.17, p. 145]

(c) [13, Exercise 4.5.29, p. 147]

(d) [13, Exercise 4.5.47, p. 150]

(e) [13, Exercise 4.5.39, p. 148]

3. Before attempting these problems, try [11, Exercise 4.7.7, p. 335]; then compareyour solution with that in [13, Exercise 4.7.7, p. 162].

(a) [11, Exercise 4.7.11, p. 335] If 1200 cm2 of material is available to make a boxwith a square base and an open top, find the largest possible volume of thebox.

(b) [11, Exercise 4.7.15, p. 335] Showing all your work, use the calculus to findthe point on the line y = 4x+7 that is closest to the origin. (Use the calculus,even though you may know methods for solving this problem that require nocalculus.)

(c) [13, Exercises 4.7.25 and 4.7.27, p. 162] A right circular cylinder is inscribed ina sphere of radius r. Showing all your work, find the largest possible volumeand the largest possible surface area of such a cylinder. (You may assumethat the cylinder has been inscribed so that its axis passes through the centreof the sphere.)

Solution:

(a) [13, Exercise 4.7.11, p. 162]

(b) [13, Exercise 4.7.15, p. 163]


(c) The solutions in [13, Exercises 4.7.25 and 4.7.27, p. 164] use, as the variable,either the half-height or the radius of the base of the inscribed cylinder. Youmight wish to try to solve the problem another way, using, as your variable,the angle subtended at the centre of the sphere by the radius of the base.

4. Showing all your work, find the most general antiderivative of the following func-tions. Check your answer by differentiation.

(a) [11, Exercise 4.10.3, p. 356] g1(x) = 1 − x3 + 5x5 − 3x7

(b) [11, Exercise 4.10.7, p. 356] g2(x) =√

x + 3√

x

(c) [11, Exercise 4.10.15, p. 356] g3(x) = 2x + 5 (1 − x2)− 1

2

Solution:

(a) [13, Exercise 4.10.3, p. 175]

(b) [13, Exercise 4.10.7, p. 175]

(c) [13, Exercise 4.10.15, p. 175]

5. Showing all your work, find the functions that have the listed properties. Checkyour answers by differentiation and substitution in the differential equation.

(a) [11, Exercise 4.10.19, p. 356] h′′1(x) = 6x + 12x2

(b) [11, Exercise 4.10.27, p. 356] h′2(x) = 3

√x − 1√

x, h2(1) = 2.

(c) [11, Exercise 4.10.39, p. 357] h′′3(x) =

1

x3, x > 0, h3(1) = 0 = h3(2).

Solution:

(a) [13, Exercise 4.10.19, p. 175]

(b) [13, Exercise 4.10.27, p. 176]

(c) [13, Exercise 4.10.39, p. 176]

F Some Tests and Quizzes from Previous Years

F.1 Fall 1998 Class Quiz, with Solutions

(This quiz was given to both sections of 189-140A in the middle of October, 1998. Nogrades were recorded. The solutions were posted on the Web.)


1. Determine the global maxima and global minima of the functionf(x) = 5x2/3 − x5/3 on the interval −2 ≤ x ≤ 4. (You may take 1.6 as an ap-

proximation to 413 . You may find it helpful to make a rough sketch, but your

solutions must not depend on that sketch.)

Solution:

f ′(x) = 5 · 2

3· x− 1

3 − 5

3· x 2

3

=5(2 − x)

3x13

,

provided x 6= 0. When x = 0, the function lacks a derivative. The derivativevanishes (i.e. is equal to zero) if and only if x = 2. Thus the list of critical points100

for this function, whose domain is the closed interval [−2, 4], is

Points where there is no derivative:

(0, f(0)) = (0, 0)

Points where the derivative vanishes:

(2, f(2)) = (2, 3 · 2 23 )

End-points of the domain of definition:

(−2, f(−2)) = (−2, 7 · 4 13 )

(4, f(4)) = (4, 1 · 4 23 )

Both the global maximum and the global minimum must be attained at pointsamong these 4. Comparing the function values — if necessary students could haveused the approximation of 1.6 for 4

13 — shows that the maximum occurs at the

point (−2, 7 · 4 13 ), and the minimum at the point (0, 0).

(A graph of this function may be found in [25, Example 6 of §3.5].)

2. A rectangle, with its base on the x-axis, is constructed so that its upper two verticesare on the semi-circle x2 + y2 = 4, y ≥ 0. Determine the maximum possible areafor such a rectangle, and determine where the maximum value is attained.

Solution: There are several ways of approaching this problem; we present two.

100The definition [25, p. 144] of critical point in the text-book is unclear as to whether or not theend-points of the interval of definition are to be considered critical . If you choose to consider end-pointsas not being included in this definition, then you must include them with the (other) critical points inyour list of candidates for extrema.


(a) A rectangle inscribed in the semicircle must have its upper side parallel to thex-axis, so the coordinates of the upper vertices may be taken to be (±x, y);as x2 + y2 = 4, and y is non-negative, y =

√4 − x2. Thus the vertices of the

rectangle will be (±x,√

4 − x2) and (±x, 0). The area, which we may denoteby A(x) is then given by

A(x) = |(2x) ·√

4 − x2| .

(The absolute signs are needed since the intention of the problem is thatthe area should be non-negative.) We can now approach the problem in twoequivalent ways.

i. We may take the domain of A(x) to be −2 ≤ x ≤ 2, since the circle hasradius

√4 = 2. Differentiating yields

A′(x) = 2√

4 − x2 + 2x · 1

2

1√4 − x2

· (−2x)

=4 − 2x2

√4 − x2

However, this is valid only for x > 0, because of the absolute signs; forx < 0

A′(x) =2x2 − 4√

4 − x2.

Note that there is no derivative at x = 0.Thus the critical points are

(0, A(0)) = (0, 0)

(−√

2, A(−√

2)) = (−√

2, 4)

(√

2, A(√

2)) = (√

2, 4)

and the end points,

(−2, f(−2)) = (−2, 0)

(2, f(2)) = (2, 0)

Comparing the values, we find the maximum of A = 2 to be attained atboth of the points x = ±

√2. The minimum value of 0 is attained at 3

points: x = ±2 and x = 0.

ii. The preceding approach can be modified, in that we can restrict thedomain to be 0 ≤ x ≤ 2, by the symmetry of the function. In this


approach the only critical points are

(√

2, A(√

2)) = (√

2, 4)

and the end points,

(0, A(0)) = (0, 0)

(2, A(2)) = (2, 0)

and we obtain the same extreme values.

(b) A more elegant approach would observe that all points on the semi-circle havecoordinates of the form (2 cos t, 2 sin t), where 0 ≤ t ≤ π. Here again we couldtreat two cases, according as we permit the domain to be the full intervalstated, or only 0 ≤ t ≤ π

2. In either case the area is α(t) = |4 cos t · 2 sin t| =

|4 sin 2t|. For convenience we will take the domain to be 0 ≤ t ≤ π2. Then,

observing that α(t) = 4 sin 2t, we have α′(t) = 4(cos 2t) · 2, which vanishesonly where 2t = π

2, so t = π

4. Other than the end-points of the interval, this

is the only critical point, and the area there is α(π4) = 4 sin π

2= 4. At the

end-points we have α(0) = sin 0 = 0, and α(π2) = 0. Thus the maximum value

of 4 occurs only at t = π4, and the minimum value of 0 occurs at t = 0 and

t = π2.

F.2 Last Three Tutorial Quizzes in 2000/2001 (many versions)

F.2.1 Fourth 2000/2001 Tutorial Quizzes

T01

Distribution Date: Monday, October 23rd, 2000 — 13:30 to 14:30 h.

1. Evaluate the following limit, but do not use L’Hopital’s rule:

limx→1

sin(x2 − 1)

(x − 1).

2. Finddy

dxif y = 3x2

.

3. 101Find an equation for the tangent line to the graph of exy = x− 1

y+ e at the

point(

e, 1e

)

.

101corrected, 7 November 2000


T02, T03, T05


1. Evaluate the following limit, but do not use L’Hopital’s rule:

limx→0

x + tan x

sin x.

2. Finddy

dxif y = tan(3x).

3. Find an equation for the tangent line to the graph of x2(x2 + y2) = y2 at the

point(

1√2, 1√

2

)

.

T04, T06


1. Use the limit definition of the derivative to find f ′(0) for

f(x) =

{

x2 sin(

1x

)

for x 6= 00 for x = 0

.

2. Finddy

dxif y = sec(5x2

) .

3. If the line y = 3x − 1 is tangent to the graph of y = f(x) at the pointwith x = 1 , find the equation of the tangent line to the graph of y = [f(x)]2

at the point with x = 1 .

T07

Distribution Date: Wednesday, October 25th, 2000 — 13:30 to 14:30 h.


f(x) =

{

1 − cos x

xwhen x 6= 0

0 when x = 0.

2. Finddy

dxif y = 2

12x .

3. If g(x) = 2x · f(x2 − 2x + 2) and f(1) = 3 , find g′(1) .


T08, T09, T11


1. Use the limit definition of the derivative to find f ′(8) for f(x) = x1/3 .

2. Finddy

dxif y = x tan

(

1

3x

)

.

3. If f(x) = ex·g(x) and g(0) = 5 , find f ′(0) .

T10, T12, T15


1. Use the limit definition of the derivative to find f ′(1) for f(x) = x−1/3 .

2. Finddy

dxif y = x sec−1

(

12x

)

.

3. Find the equation(s) of the line(s) through the origin, tangent to the graph ofy = ex.

T16


1. Use the limit definition of the derivative to find f ′(0) for f(x) = sec x .

2. Finddy

dxif y = tan−1

(

2x

1 − x2

)

.

3. Find equations(s) of the line(s) through the origin, tangent to the graph of

y = x2 + 9 .

T13

Distribution Date: Thursday, October 26th, 2000 — 16:00 to 17:00 h.

1. Use the limit definition of the derivative to find f ′(0) for f(x) = ex sin x .

2. Finddy

dxif y = tan−1

(

ex − e−x

2

)

.x

3. Find the equations(s) of the line(s) through the origin, tangent to the graph of

y = ex2

.


T14

Distribution Date: Thursday, October 26th, 2000 — 17:00 to 18:00 h.


f(x) = e3x(1 − cos x) .

2. Finddy

dxif y = sin−1

(

x√1 + x2

)

.

3. Find equations(s) of the line(s) through the origin, tangent to the graph of

y = e2x .

F.2.2 Fifth 2000/2001 Tutorial Quizzes

T01

Distribution Date: Monday, November 13th, 2000 — 13:30 to 14:30 h.

1. Find dydx if y = x

xx−1 .

2. Water is pouring into a leaky tank at the rate of 10m3/h. The tankis a cone with the vertex down, 9m deep and 6m in diameter at thetop. The surface of the water is rising at a rate of 0.2m/h when thedepth is 6m. How fast is the water leaking out at that time?

T02, T03, T05



√x

x−1 .

2. How fast must you let out the line if the kite you are flying is 30mhigh, 40m horizontally away from you, and moving horizontally awayfrom you at 10m/min?


T04, T06


1. Find dydx if y = (

√x + 1)x.

2. A ferris wheel you are riding has diameter 20m and is rotating at 1revolution per minute. How fast are you rising or falling when youare 6m horizontally away from the vertical line through the centre ofthe wheel?

T07

Distribution Date: Wednesday, November 13th, 2000 — 13:30 to 14:30 h.

1. Find dydx if y = (x2 + 1)1/x.

2. A policeman stationed at a fixed distance from a highway aims aradar gun at a car. When the gun is pointing at an angle of 45◦ tothe highway, the radar gun records the rate at which the distance ofthe car from the gun is increasing at 100km/h. How fast is the cartravelling?

T08, T09, T11


1. Find dydx if y = (xx + 1)1/2.

2. [CORRECTED] A balloon is released and rises vertically from pointA and is tracked from point B, 100m horizontally away. If the balloonis rising at the rate of k m/sec., how fast is the angle of elevation ofthe balloon, as observed at point B, increasing when it is 200m aboveA?

T10, T12, T15


1. Find dydx if y = xlnx.

2. A man 6ft tall walks towards a building at the rate of 5ft/sec. Ifthere is a light on the ground 50ft away from the building, how fastis the man’s shadow shrinking when he is 30ft from the building?


T16



1

2x .

2. A spherical balloon is being inflated so that its volume is increasing

at the rate of 5m3/min. At what rate is the surface area increasingwhen the radius is 6m?

(Volume = 43πr3, Area = 4πr2, where r is the radius.)

T13

Distribution Date: Thursday, November 14th, 2000 — 16:00 to 17:00 h.

1. Find dydx if y = (x2x + 1)1/3.

2. A 10m long ladder has one end on the ground, and is supported partway along its length by a fence 3m high, so that part of the ladderprojects past the fence. If the end on the ground is 4m from the baseof the fence and is being dragged away at 0.2m/sec, how fast is thevertical height of the other end of the ladder changing?

T14


1. Find dydx if y =

(

12x

)x.

2. A light shines from a pole 50ft high. A ball is dropped from the sameheight from a point 30ft away from the light. In t secs the ball falls16t2 ft. How fast is the shadow of the ball moving along the groundafter 0.5 secs?


F.2.3 Sixth 2000/2001 Tutorial Quizzes

T01


Let f(x) = (2x2 + x + 1)ex .

1. [5 MARKS] Determine the domain of f .

2. [5 MARKS] Determine all horizontal asymptotes and all vertical asymptotes of thegraph of f .

3. [5 MARKS] Determine the intervals where f increases, and the intervals where itdecreases.

4. [5 MARKS] Determine all local maxima and all local minima of f .

5. [5 MARKS] Determine the intervals where the graph is concave upwards, and theintervals where it is concave downwards.

6. [5 MARKS] Determine all inflection points of the graph.

7. [5 MARKS] Sketch the graph.

(You may use, without proof, the fact that limx→∞

xa

ex= 0 for any real number a.)

T02, T03, T05


Let f(x) = (2x2 − 3x + 2)ex .










xa


T04, T06


Let f(x) = (6x2 + x + 5)ex .









xa


T07


Let f(x) = (2x2 − 5x + 4)ex .










xa


T08, T09, T11


Let f(x) =1

x − 1− 64

x + 1.








T10, T12, T15


Let f(x) =64

x − 1− 1

x + 1.









T16


Let f(x) =1

x − 1− 64

x + 8.








T13


Let f(x) =1

4x + 1− 16

x + 2.









T14


Let f(x) = ln |x + 2| − 4 ln |x − 1| .









F.3 First 2005/2006 Written Assignment W1, with Sketch ofSolutions

F.3.1 The assignment question





MONDAY, 26 Sept., 2005 1 − x2 1x

WEDNESDAY, 28 Sept., 2005√

2x − 1 x2 + 1THURSDAY, 29 Sept., 2005 1

x−1sin2 x

FRIDAY, 30 Sept., 2005 cos x 1 + 1x

F.3.2 Solutions

Marking Scheme: 1. 2 MARKS for each of the 4 formulæ for the compositions

2. 2 MARKS for each of the 6 domains. But, where the domain is not all of R,the student must provide an acceptable explanation in order to receive fullmarks.

3. THE CERTIFICATE OF ORIGINALITY must be signed. If the student hasnot signed it, insist on seeing it signed before returning the graded assignment.

TOTAL = 20 MARKS

MONDAY: 1. Since f is a polynomial, there are no real numbers where it is unde-fined. Domain(f) = R

2. Since g(x) = 1x, it is defined only where the fraction is defined. In the devel-

opment of the real number system we do not assign a meaning to 10; we can’t!

No matter what real number r we would want to choose as the value of 10, we

find that one of our other basic properties of the real number system wouldbe violated if we used r as the value of this quotient. We resolve this issue byrestricting fractions a

bto cases where b 6= 0. Thus Domain(g) = R − {0}.

3. f(f(x)) = 1− (1−x2)2 = 2x2 −x4. Since f is defined for all x, we may applythe function again to f(x). It follows that Domain(f ◦ f) = R.


4. f(g(x)) = 1 −(

1x

)2= 1 − 1

x2 . Since g(x) is not defined for x = 0, 0 willbe excluded from the domain of the composition. As f is defined for all x,the composition is also defined for all x except this one excluded value. ThusDomain(f ◦ g) = R − {0}.

5. g(f(x)) = 11−x2 . There is no restriction when the first function, here f , is

applied. But we must exclude the possibility that its value be 0, since g willnot be defined there. Thus we exclude all x such that 1−x2 = 0, equivalently,that x = ±1. It follows that Domain(g ◦ f) = R − {−1, 1}.

6. g(g(x)) = 11x

= x. One might be tempted to say that the domain of the

composition is R. But we would be forgetting that the first application of gwas restricted to x 6= 0. So the composition g ◦ g differs from the function xonly in the fact that its domain excludes the point 0: Domain(g◦g) = R−{0}.

WEDNESDAY: 1. The square root√

2x − 1 is defined only where the argument isnot negative, i.e., where 2x− 1 ≥ 0, equivalently, where x ≥ 1

2. Domain(f) =

[

12, ∞).

2. Since g is a polynomial, it is defined for all real numbers x. Thus Domain(g) =R.

3. f(f(x)) =√

2√

2x − 1 − 1. The first application of f requires that x ≥ 12.

The result of this application is a non-negative number, which, when doubledand added to −1, must be non-negative also. We thus require 2

√2x − 1 ≥ 1,

which implies that x ≥ 58. Domain(f ◦ f) =

[

58, ∞).

4. f(g(x)) =√

2(x2 + 1) − 1 =√

2x2 + 1. The first function yields a numberno less than 1, which, when doubled and the result decreased by 1, yieldsa positive number, which certainly has a non-negative square root. HenceDomain(f ◦ g) = R.

5. g(f(x)) =(√

2x − 1)2

+ 1 = |2x − 1| + 1. The application of f requires thatx ≥ 1

2. Thus 2x − 1 ≥ 0, and we may remove the absolute sign and further

simplify to g(f(x)) = 2x. But, unlike the polynomial 2x, this function hasa more restricted domain, because of the restriction on f : Domain(g ◦ f) =[

12, ∞).

6. g(g(x)) = (x2 + 1)2. This is no restriction on the applicability of g, and so

the 2nd application is also unrestricted. Domain(g ◦ g) = R.

THURSDAY: 1. The reciprocal 1x−1

is defined only where x − 1 6= 0, i.e., wherex 6= 1. Domain(f) = R − {1}. (See solution MONDAY.2.)

2. Since g is a power of the sine function, which is defined for all x, Domain(g) =R.


3. f(f(x)) = 11

x−1−1

= x−12−x

. This rational expression conceals one restriction in

the domain, so let’s examine the steps in computing this composition. Inthe first application of f we must avoid the value x = 1. No matter whathappens in the second application, the value x = 1 has now been irrevocablyremoved from the domain of the composition. In the second application wehave to ensure that 1

x−1− 1 6= 0. But this is equivalent to x 6= 1, 2. Thus

Domain(f ◦ f) = R − {1, 2}.4. f(g(x)) = 1

sin2 x−1= − sec2 x. When we first apply g there is no restriction, as

the sine function and its square are applicable everywhere. However, we haveto ensure that the result of the application of g is different from 1, i.e., thatsin x 6= ±1, i.e., that x 6= n

(

π2

)

, where n is any odd integer. Thus

Domain(f ◦ g) = R −{

all odd integer multiples ofπ

2

}

.

5. g(f(x)) = sin2(

1x−1

)

. The first step in calculating f is to take the reciprocalof a non-zero real number; we need to restrict x to be different from 1, andno other restriction is needed. The subsequent application of g imposes nofurther restriction, as the sine function is defined for all real numbers. HenceDomain(g ◦ f) = R − {1}.

6. g(g(x)) = sin2(

sin2 x)

. There are no restrictions on x: Domain(g ◦ g) = R.

FRIDAY: 1. The cosine function is defined for all x. Domain(f) = R.

2. The only restriction in evaluating g is that 1x

must be well defined, so x 6=0. This restriction is not affected by the subsequent addition of 1 to thereciprocal. Domain(g) = R − {0}.

3. f(f(x)) = cos(cosx). The cosine function is defined everywhere, so one ap-plication can follow another without restriction. Domain(f ◦ f) = R.

4. f(g(x)) = cos(

1 + 1x

)

. In applying g first we need to ensure that x 6= 0. Thesubsequent application of f is unrestricted, as its domain is all of R. HenceDomain(f ◦ g) = R − {0} .

5. g(f(x)) = 1 + 1cos x

= 1 + sec x. The first step in calculating f is to evaluate acosine. That is possible for all x. However, we cannot proceed to the secondstep of the calculation if that cosine is equal to 0, and that is the only obstacleto completing the calculation. It follows that

Domain(g ◦ f) = R −{

all odd integer multiples ofπ

2

}

.


6. g(g(x)) = 1+ 11+ 1

x

= 2x+1x+1

. In the first application of g we must avoid the value

x = 0 in order to take the reciprocal of x. But, for a second application to bepossible, we need to ensure that g(x) 6= −1, i.e., that 1

x6= −1, equivalently

that x 6= −1. Domain(g ◦ g) = R − {−1, 0}.

F.4 Second 2006/2006 Written Assignment W2, with Sketch of

Solutions

Your completed solution to this assignment should be submitted, with a copy of thisquestion sheet at your Tutorial, during the week October 17-21, 2005. Please slip theassignment into your folded quiz answer paper. All materials must bear your nameand/or student number. No other method of submission is acceptable.

F.4.1 The assignment questions with solutions

Prove the following statement by using the Principle of Mathematical Induction.

Students whose tutorial is on Monday:

12 + 22 + 32 + . . . + n2 =n(n + 1)(2n + 1)

6


Solution: Define S(n) to be the preceding equation.

“Base” or “Anchor” Case: Since

1(1 + 1)(2 + 1)

6=

6

6= 1 = 12 ,

S(1) is true.

Induction Step: Suppose that it is known that S(k) is true, where k is anypositive integer. Then

12 + 22 + 32 + . . . + k2 + (k + 1)2

=(

12 + 22 + 32 + . . . + k2)

+ (k + 1)2

=k(k + 1)(2k + 1)

6+ (k + 1)2


=k + 1

6· (k(2k + 1) + 6(k + 1)) =

k + 1

6· (2k2 + 7k + 6)


=k + 1

6· (2k + 3)(k + 2)

=(k + 1)(k + 2)(2k + 3)

6,

=(k + 1)((k + 1) + 1)(2(k + 1) + 1)

6

which is S(k + 1).

Conclusion: We may conclude that S(n) is true for all positive integers n.

Students whose tutorial is on Wednesday:

If n is a positive integer, then 5n + 3 is divisible by 4.

Solution: Let S(n) denote the preceding statement.

“Base” or “Anchor” Case: When n = 1, 51 + 3 = 8, which is equal to 2 × 4.Thus 51 + 3 is divisible by 4, and S(1) is true.

Induction Step: Suppose it is known that S(k) is true, i.e., that 5k + 3 is amultiple of 4, i.e., that there exists an integer b such that 5k + 3 = b · 4. Then

5k+1 + 3 = 5 · 5k + 3


= 4(

5k)

+(

5k + 3)

since 5 = 4 + 1

= 4(

5k)

+ b · 4 by the Induction Hypothesis

=(

5k + b)

· 4 ,

which is a multiple of 4. We have thus proved that S(k + 1) is true.


Students whose tutorial is on Thursday: A sequence of real numbers a1, a2, . . . , an,. . . is defined102 as follows:

a1 =1

2,

For n = 1, 2, 3, . . . an+1 = an +1

(n + 1)(n + 2).

Prove that1

22+

1

32+

1

42+ . . . +

1

(n + 1)2< an for all n ≥ 1.

Solution: Let the preceding inequality be designated as S(n).

102We call a definition of this type a recursive definition.


“Base” or “Anchor” Case: The case n = 1 reads1

22< a1. Since

a1 =1

2>

1

4=

1

22,

statement S(1) is true.

Induction Step: Suppose it is known that S(k) is true, i.e., that

1

22+

1

32+

1

42+ . . . +

1

(k + 1)2< ak .

Then

1

22+

1

32+

1

42+ . . . +

1

((k + 1) + 1)2

=

(

1

22+

1

32+

1

42+ . . . +

1

(k + 1)2

)

+1

(k + 2)2

< ak +1

(k + 2)2


< ak +1

(k + 1)(k + 2)

since the denominator is reduced, so the fraction is increased

= ak+1 by the definition of ak+1

proving Sk+1.


Students whose tutorial is on Friday: A sequence of functions f0, f1, . . . , fn, . . . isdefined by

f0(x) = 2x + 1 ,

For n = 0, 1, 2, . . . fn+1 = f0 ◦ fn .

Prove thatfn(x) = 2n+1x + 2n+1 − 1 .

Solution: Let S(n) denote the preceding sentence.

“Base” or “Anchor” Step: When n = 0,

2n+1x + 2n+1 − 1 = 2x + 2 − 1 = 2x + 1 = f0(x)

so S(0) is true.


Induction Step: Suppose it is known that S(k) is true. Then

fk+1(x) = (f0 ◦ fk) (x) by definition of fk+1

= f0 (fk(x)) by definition of ◦= f0

(

2k+1x + 2k+1 − 1)


= 2(

2k+1x + 2k+1 − 1)

+ 1 by definition of f0

= 2(k+1)+1x + 2(k+1)+1 − 1

proving Sk+1.

Conclusion: Hence, by the Principle of Induction, S(n) is true for all n ≥ 0.

F.4.2 Solution to the “Fun Problem” (Student were not asked to submit asolution.):

What is wrong with the following “proof” by induction?

Claim: All birds are of the same colour.

Fallacious Proof: Let Sn be the statement

“any n birds are of the same colour”.

The statement is clearly true for n = 1, because all members of a set of one bird havethe same colour. Assume that it is always true that any k birds are of the same colour.(We call this the Induction Hypothesis.) Now take k + 1 birds. Put one aside. Thereare k birds left; by the Induction Hypothesis, the k birds are of the same colour. Forsimplicity, say they are all black. Replace the bird previously taken aside into the group,and take out a different bird. Again, there are k birds remaining, which by assumptionmust have the same colour. In particular, the bird that was taken out in the first placemust now have the same colour as all the other birds, namely black. The bird taken outwas also black. Therefore, the k + 1 birds must all be black. But that means that wehave proved the statement Sn for all natural numbers by induction. Yet, the statementis obviously false. Where is the fallacy in this proof?The Fallacy: We have given what appears to be a general proof that Sk implies Sk+1.That general proof is almost correct. The problem is that it is not valid in the caseS1 ⇒ S2. In that one induction step, the assumption that the kth case implies the nextis defective. Consider a set containing exactly 2 birds. If you remove 1 bird, then theresulting set of 1 bird does, indeed, consist of birds of the same colour. But the removalof another element of the set of 2 — i.e., of the only possible other element — producesa set of 1 bird which is totally disjoint from the one we discussed earlier. These 2 sets


of size 2 − 1 do not overlap; so, while they each consist of elements of the same colour,there is no justification in concluding that those colours are the same for both sets ofsize 2 − 1.

F.5 Third 2005/2006 Written Assignment W3

Your completed solution to this assignment should be submitted, with a copy of thisquestion sheet at your Tutorial, during the week October 31 – November 4, 2005. Pleaseslip the assignment into your folded quiz answer paper. All materials must bear yourname and/or student number. No other method of submission is acceptable.

F.5.1 Certificate




and assert that my work submitted for W3, R7, and R8 does not violateMcGill’s regulations concerning plagiarism.


F.5.2 Instructions

• This assignment consists of problems taken from YOUR VERSIONS of WeBWorKassignments R6 and R7.

• There is no place in this assignment where the use of a calculator is requiredor permitted.

• The tutor may grade only part of your solutions, but may check whether you havesubmitted solutions to all of the problems.

• You are expected to write FULL SOLUTIONS to the following problems; it is not suf-ficient to get the correct answer — you should show a full explanation, comparable tosolutions in the Student Solutions Manual to the textbook.


• You are expected either to attach a printout of the questions from yourassignments, or to copy the questions verbatim on to your answer sheet beforebeginning the solution.

F.5.3 The assignment questions

1. Problem 1 on R6: DO NOT USE A CALCULATOR; YOU MAY EXPRESSPORTIONS OF YOUR SOLUTION USING INVERSE TRIGONOMETRIC FUNC-TIONS.

2. Problem 4 on R6

3. Problem 14 on R6: Do not use l’Hospital’s Rule anywhere in your solutions.

4. Problem 16 on R6

5. Problem 9 on R7

F.6 Fourth 2005/2006 Written Assignment W4

Your completed solution to this assignment should be submitted, with a copy of thisquestion sheet at your Tutorial, during the week November 14 – November 18, 2005.Please slip the assignment into your folded quiz answer paper. All materials must bearyour name and/or student number. No other method of submission is acceptable.

F.6.1 Certificate




and assert that my work submitted for W4, R9, and R10 does not violateMcGill’s regulations concerning plagiarism.



F.6.2 Instructions

• This assignment consists of 2 problems taken from YOUR version of WeBWorK as-signment R9. One of the problems is extended beyond the version on your WeBWorKassignment.


• The tutor may grade only part of your solutions, but may check whether you havesubmitted solutions to all of the problems.

• You are expected to write FULL SOLUTIONS to the following problems; it is not suf-ficient to get the correct answer — you should show a full explanation, comparable tosolutions in the Student Solutions Manual to the textbook.

• You are expected either to attach a printout of the questions from yourassignments, or to copy the questions verbatim on to your answer sheet beforebeginning the solution.

F.6.3 The assignment questions

1. Problem 6 on R9:

(a) Solve the problem as stated.

(b) Find the normal line to the curve at the point of contact of the tangent.

2. Problem 19 on R9. (This problem resembles Exercise 23, p. 261 in your text-book.)

F.7 Fifth 2005/2006 Written Assignment W5

Your completed solution to this assignment should be submitted, with a copy of thisquestion sheet at your Tutorial, during the week November 28 – December 2, 2005.Please slip the assignment into your folded quiz answer paper. All materials must bearyour name and/or student number. No other method of submission is acceptable.

F.7.1 Certificate





and assert that my work submitted for W5, R11, and R12 does notviolate McGill’s regulations concerning plagiarism.


F.7.2 Instructions

• This assignment consists of an extension of one problem taken from YOUR version ofWeBWorK assignment R11. Note that the problem given below asks more than theWeBWorK problem.

Solution of the problem may require the use of L’Hospital’s Rule, which you may nothave seen yet in your lectures. Part of your assignment is to read ahead in your textbook,and learn enough about that rule that you may use it if you need to. Example 2 on page309 of the textbook, may prove helpful.


• The tutor may grade only some parts of your solutions, but may check whether you havesubmitted solutions to all of the parts.

• You are expected to write FULL SOLUTIONS; it is not sufficient to report the correctanswer — you should show a full explanation, comparable to solutions in the StudentSolutions Manual to the textbook.

F.7.3 The assignment question

Consider the function f(x) = x2eax with the value of a in Problem 13 of your WeB-WorK assignment R11. Show all your work! It is not sufficient to make unsubstan-tiated statements.

1. Find an equation for each horizontal asymptote to the graph of f .

2. Find an equation for every vertical asymptote, or show that there is none.

3. Determine all critical numbers.

4. Determine the global maximum of the function, if there is one, or show that thereis none.


5. Determine the global minimum of the function, if there is one, or show that thereis none.

6. Determine the locations of all inflection points of this graph. You are expectedto refer to the textbook definition of inflection point and to demonstrate that thepoints you have given are, indeed, inflection points.

7. Give the intervals where the graph is concave downward, and the intervals wherethe graph is concave upward.

8. Give a rough sketch of your graph, showing the approximate locations of the in-flection points and the intervals of upward and downward concavity. (The sketchmay be freehand, and need not be to scale.)

G Examinations from Previous Years

G.1

1. These examinations have been taken from the original versions in computer files.Occasionally an examination has been changed or corrected in the examinationroom: such changes may not have been incorporated into these versions.

2. Listed below are only the texts of the problems on most of the examinations from1996-May 2005; the actual formats of most of the examinations can be seen at thefollowing URL:


3. It is possible that the form and format of the examination in MATH 140 willundergo further changes in December, 2006. Further information may be given atthe lectures.

4. The syllabus on which the following examinations was based has undergone onlyminor changes over the years. Some changes derive from the use of different text-books. The examinations can serve as a circumstantial evidence of what a studentin Math 140 was expected to know by examination time. Notwithstanding theforegoing, you are advised not to base your preparation for your examination on astudy of these examinations. The best way to prepare for your examination is towork through your textbook, using the Student Solutions Manual and consultationswith the tutors to help you evaluate the quality of your solutions to problems.

5. Solutions to these examinations are not available: in some cases you may be ableto find similar, solved problems in the Student Solutions Manual.


G.2 December 1996 Final Examination in 189-122A

1. [4 MARKS] Evaluate limx→2

1x− 1

2

x − 2.


x − 9√x − 3

.


x cot 3x .

4. [5 MARKS] Find an equation for the straight line which is normal to the graph off(x) = x2 at x = −3 .

5. [8 MARKS] The surface area of a sphere is increasing at a rate of 10 square metresper hour. At what rate is the volume increasing, when the radius is 2 metres?(Note: The volume of a sphere of radius r is 4

3πr3 ; the surface area is 4πr2 .)

6. [8 MARKS] Show that the equation 6x4 − 7x + 1 = 0 does not have more thantwo distinct real roots.

7. [10 MARKS] A right circular cylinder is inscribed in a right circular cone of heightH and radius R. Determine the dimensions of the cylinder with the largest possiblevolume. What is that largest volume?

8. [8 MARKS] Show that the function f(x) = 2 + (1 − x3)15 has an inverse. Deter-

mine f−1(x) .

9. The function f(x) = x − cos x (−π ≤ x ≤ π) is differentiable.

(a) [2 MARKS] Show that this function has an inverse.

(b) [4 MARKS] Calculate the derivative of the inverse function.

(c) [4 MARKS] Evaluate the derivative of f−1(x) at all points x such that f(x) =−1 .

10. [8 MARKS] Use the mean value theorem to show that, whenx > 1 ,

x − 1

x< ln x < x − 1 .

11. Let f(x) = ln

(

x4

x − 1

)

.

(a) [3 MARKS] Specify the domain of f .


(b) [3 MARKS] Determine the interval(s) where f increases, and the interval(s)where f decreases.

(c) [3 MARKS] Determine the concavity of the graph of f , and find the pointsof inflection.

(d) [3 MARKS] Sketch the graph, using the information determined above.

12. Let f(x) = xe−x .

(a) [3 MARKS] Specify the domain of f .

(b) [3 MARKS] Determine the interval(s) where f increases, and the interval(s)where it decreases.

(c) [3 MARKS] Determine the concavity of the graph of f , and find the pointsof inflection.

(d) [3 MARKS] Sketch the graph, using the information determined above.

13. [7 MARKS] Given the curve x3 + y3 = 1 + 3xy2 , verify that the point (x, y)

= (0, 1) is on the curve. Finddy

dxand

d2y

dx2at (x, y) = (0, 1) .


1. [8 MARKS] For the function f(x) =x2 − 4

|x − 2|

(a) Find the left-hand and right-hand limits at x = 2 .

(b) Determine whether the two-sided limit exists at x = 2 .

(c) Sketch the graph of y = f(x) .

2. [2 MARKS] Multiple Choice: Circle the correct answer (A, B, C, D, or E.)

limx→0

√9 + x −

√9 − 3x

x

A B C D E

= 0 =2

3= 1 does not exist (or =

∞, or = −∞).is a real number r dif-

ferent from 0,2

3, 1.


3. [8 MARKS] Apply the intermediate value property of continuous functions to provethat the equation x3 − 4x2 + 1 = 0 has at least three different solutions. (Hint:Denoting f(x) = x3 − 4x2 + 1 , we have f(+1) = −2.)

4. [8 MARKS] At time t, the radius of a leaking spherical balloon is r =60 − t

12centimetres. Determine the rate (in cm.3/second) at which the volume is decreasingwhen t = 30 . (Hint: You may assume that the volume of a sphere of radius r is43πr3.)

5. [8 MARKS] Determine an equation for the straight line that passes through thepoint (1, 5) and is tangent to the curve y = x3 .


limx→0

x − 4 tanx

sin x

A B C D E

= 0 =2

3= 1 does not exist (or =

∞, or = −∞).is a real number r dif-

ferent from 0,2

3, 1.

7. [8 MARKS] Showing all your work, determine the maximum area of a rectanglewith a base that lies on the x-axis, and with two upper vertices that lie on thegraph of the equation y = 4 − x2.

8. [8 MARKS] A tank is in the shape of an inverted right circular cone of height 800cm., whose top is a disk of radius 160 cm. Water is running out of a small holeat the vertex (apex) of the cone, which is at the bottom. Showing all your work,determine the rate of change of volume V with respect to height h, at a time whenthe height is 600 cm. (Hint: You may assume that the volume of a right circularcone of height h, whose base has radius r, is 1

3πr2h.)


If y = sin 2x cos 3x , the value ofdy

dxwhen x =

π

2is

A B C D E

02

31 −1 different from 0,

2

3, 1, −1.


10. [8 MARKS] A covered rectangular box is to be constructed with volume 576 cubiccentimetres, with its bottom twice as long as it is wide. Determine the dimensionsof the box that will minimize its total surface area (including the cover).

11. [5 MARKS] Showing all your work, determine the maximum and minimum valuesof f(x) = 3 − |x − 2| on the interval [1, 4] .


The function f(x) =(

2x2 − 3x)

e−x has a global maximum on the half-line x ≥ 0 .The maximum value is

A B C D E

−√

e − 1√e

9

e2

9

8e−

34 none of the preceding

4 values.

13. [8 MARKS] Showing all your work, sketch the graph of the function

f(x) =e2x

e2x + 3,

identifying asymptotes, critical points, and inflection points. Show clearly wherethe graph is concave upward and where it is concave downward.

14. [8 MARKS]

(a) State the Mean Value Theorem.

(b) Given that f(x) = |x| for −1 ≤ x ≤ 1 , does it follow from the Mean ValueTheorem that there exists a point c such that −1 ≤ c ≤ 1 and f ′(c) = 0 ?Explain your answer.


If x5 − y5 = 2x2y2 , the value ofdy

dxwhen (x, y) = (1,−1) is

A B C D E

−1 1 −5

3

5

3none of −1, 1, −5

3,

5

3.


16. [10 MARKS] Showing all your work, sketch the graph of the function

f(x) =x2 − x − 2

(1 − x)2− 1 ,

identifying asymptotes, critical points, and inflection points. Show clearly wherethe graph is concave upward and where it is concave downward.


1. [4 MARKS] Showing your work, find limx→3

x3 − 27

x − 3.

2. [5 MARKS] Given that a function f has the property that

|f(x) − 2| ≤ (x − 1)2 ,

determine limx→1

f(x).

3. (a) [4 MARKS] Define v(x) = x3 − 4x2 + x + 3 . Use the Intermediate ValueTheorem to show that the equation v(x) = 0 has a solution between x = 1and x = 2.

(b) [4 MARKS] By examining the behavior of v(x) as x → ∞ and as x → −∞,or otherwise, discuss the existence of other solutions to v(x) = 0.

4. [8 MARKS] Find equations for all straight lines which are both normal to the curvey =

√x − 3 and parallel to the straight line y = −2x + 11.

5. For the function g(x) = |1 − x2|,

(a) [3 MARKS] Sketch the graph of g.

(b) [3 MARKS] Show that g is continuous at x = 1.

(c) [2 MARKS] Determine whether g is differentiable at x = 1.

6. (a) [3 MARKS] Determine h′(x) , if h(x) =sin x

1 − 2 cos x.

(b) [3 MARKS] If u(x) = ln ((2x − 1)3), determine u′(x).

7. [12 MARKS] Determine the slope of the curve

(x + y)2 − (x − y)2 = x4 + y4

at all point(s), other than the origin, where the curve meets the line x = y.


8. [6 MARKS] A child is building a snowman by rolling a snowball on the ground; itsvolume is increasing at the rate of 8 cubic centimetres per minute. Find the rateat which the radius is increasing when the snowball is 75 centimetres in diameter.(The volume of a sphere of radius r is 4

3πr3.)

9. [8 MARKS] The domain of the function F is to be taken to be the interval[

−2, 12

]

.On that interval,

F (x) = x3 + x2 − x + 1 .

Determine the global maximum ( = absolute maximum), global minimum ( = abso-lute minimum), local maxima ( = relative maxima), and local minima ( = relativeminima).

10. [10 MARKS] A closed box with a square base is to have a volume of 2, 000 cubiccentimetres. The material for the top and bottom of the box costs $3 per squarecentimeter, while the material for the sides costs $1.50 per square centimetre.Determine the dimensions of the least expensive box.

11. For the function f(x) = x2

x2 − 4,

(a) [2 MARKS] Determine the (largest possible) domain.

(b) [4 MARKS] Determine all local extrema, and all points of inflection.

(c) [2 MARKS] Determine precisely where the function is increasing, decreasing,concave upward, concave downward.

(d) [2 MARKS] Determine all horizontal and all vertical asymptotes, if any.

(e) [1 MARK] Sketch the graph.

12. It is given that the function f , defined by f(x) = x + x3 has an inverse, denotedby f−1.

(a) [3 MARKS] Determine the value of f−1(2).

(b) [3 MARKS] Determine the value of(

ddx

(f−1))

(2).

13. [4 MARKS] Determine the derivative of the function

m(x) = xx .

14. [4 MARKS] Determine the value of the following limit, if it exists:

limx→0

ex3 − 1

x − sin x


G.5 May 1999 Supplemental Examination in 189-140A

1. (a) Evaluate limx→2

x2 − 4

x − 2.

(b) Let f be the function with domain ∞ < x < ∞ given by

f(x) =

{

x2 − 2x if x ≤ 1x − 2 if x > 1

i. Is f continuous at x = 1? Explain (justify) your answer.

ii. Is f differentiable at x = 1? Explain (justify) your answer.

2. Find the derivative of each of the following functions. (You need not simplify youranswers.)

(a) F (x) =sin(2x2 − 1)

(x2 + 1)3

(b) G(x) = tan−1

(

1√x3 + 1

)

(c) H(x) = x2e−2x ln x

[a, b], and theorem. g′(x) > 0

3. It is given that the function f , defined by f(x) = x + x3 has an inverse, denotedby f−1.

(a) Determine the value of f−1(2).

(b) Determine the value of(

ddx

(f−1))

(2).

4. (a) Find an equation for the line tangent at the point (1, 1) to the curve y2 =x3(2 − x).

(b) A woman who is 1.75 metres tall walks at a rate of 1.5 metres per secondaway from a lamp that is at the top of a 4-metre high lamp post. At whatrate is her shadow lengthening when she is 30 metres from the lamp post.

5. Find the area of the largest rectangle which can be inscribed in a semicircle ofradius 1, where one side of the rectangle lies on the diameter of the semicircle, andthe other two vertices lie on the semicircle.

6. Consider the function h(x) = 4x3 − 15x2 + 12x + 7, with domain −∞ < x < ∞.


(a) Find all points at which h has a local maximum, a local minimum, or a pointof inflection. Justify all of your answers.

(b) Find the global (absolute) maximum and the global (absolute) minimum of hon the interval [0, 3].

7. Sketch the curve y =2x2

x2 − 1, indicating any horizontal or vertical asymptotes.

8. Let the function u be defined by u(x) = ln(2x + 1).

(a) What is the (maximum possible) domain of u. (Explain.)

(b) Sketch the graph of u(x) = ln(2x + 1).


1. Let f(x) =

x + sin 3x

tan 4xx > 0

(x − a)2

4x ≤ 0

, where a is a constant, to be determined.

(a) [6 MARKS] Determine each of limx→0−

f(x), limx→0+

f(x) or explain why either or

both do not exist.

(b) [3 MARKS] Use the information of part (a) to determine all values, if any, forthe constant a, which will make f continuous at x = 0.

2. [8 MARKS] Let m(x) = xx2 − (x2)−3

. Determine the value of m′(1).

3. [8 MARKS] If y is defined implicitly as a function of x by 2x2 − 3xy + 5y2 = 10,

determine the value ofd2y

dx2when (x, y) = (1,−1).

4. For the function h(x) = arctan2 + x

1 − 2x,

(a) [1 MARK] State the (largest possible) domain.

(b) [5 MARKS] Determinedh

dxfor all points x where the derivative exists.

(c) [2 MARKS] Give an example of a function different from h which has exactlythe same domain and exactly the same derivative as h.

5. The function u is defined by u(x) =x

1 + x2, for −∞ < x < ∞.


(a) [3 MARKS] Determine the intervals where the function u is increasing, andthose where it is decreasing.

(b) [3 MARKS] Find all critical points. In each case determine whether the pointis a maximum or minimum point, or neither.

(c) [3 MARKS] Determine the intervals where the graph of u is concave upwards,and those where it is concave downwards.

(d) [3 MARKS] Determine all inflection points of the graph.

(e) [2 MARKS] Determine all horizontal or vertical asymptotes of the graph.

(f) [4 MARKS] Sketch the graph.

You may assume that u′(x) = −(x − 1)(x + 1)

(x2 + 1)2, and that u′′(x) =

2x(x2 − 3)

(x2 + 1)3.

(For each of parts (a) through (e) you are expected to show all your work and yourresults, clearly marked by the question number, e.g., 5(c); it is not sufficient toprovide information only on your graph.)

6. [8 MARKS] Use the Intermediate Value Theorem and/or the Mean Value Theorem

and/or properties of G′(x) to show that the function G(x) = x2 − e1

1+x assumesthe value 0 for exactly one real number x such that 0 < x < 2. Show all your work.[Hint: You may assume that e

13 < 2.]

7. [8 MARKS] Triangle OBC, in the first quadrant, has vertex O at the origin, vertexB on the x-axis, and vertex C on the y-axis. If the vertices are constrained so thatthe line joining B and C passes through the point (2, 3), determine the minimumarea for triangle OBC. Show all your work.

8. Showing all your work, evaluate the following limits, if they exist:

(a) [8 MARKS] limx→∞

(√

x2 + x −√

x2 − x).

(b) [8 MARKS] limx→0

tan x − sin x

x3.

9. [8 MARKS] Showing all your work, determine all lines with slope 3 which arenormal to the curve 64y + x3 = 0, (i.e. which are perpendicular to the tangent ateach point where they meet the curve).

10. [9 MARKS] Showing all your work, determine the (global) maxima and minimaof the function R(x) = 3x4 + 4x3 − 6x2 − 12x on the closed interval −2 ≤ x ≤ 2.[Hint: x3 + x2 − x − 1 = (x2 − 1)(x + 1).]


G.7 December 1999 Special Final Examination in 189-140A

1. [4 MARKS] Evaluate limx→1−

1 − x + ln x

1 + cos πx.

2. [6 MARKS] When f(x) = x−2x2, evaluate f ′(1).


2x

tan 3x.

4. [5 MARKS] Find an equation for the straight line which is normal to the graph off(x) = x3 at x = −2 .

5. [10 MARKS] The volume of a sphere is increasing at a rate of 10 cubic metres perhour. At what rate is the surface area increasing, when the radius is 2 metres?(Note: The volume of a sphere of radius r is 4

3πr3 ; the surface area is 4πr2 .)

6. [10 MARKS] Show that the equation 3x4 − 28x + 8 = 0 does not have more thantwo distinct real roots.

7. [10 MARKS] Determine the value ofdy

dxand

d2y

dx2at the point (x, y) = (1,−1) when

x and y are related by the equation x3 + xy + y4 = 1.

8. [10 MARKS] A box with a square base and an open top is to have volume 62.5cubic centimetres. Neglect the thickness of the material used to make the box, andfind the dimensions that will minimize the amount of material used.

9. Let f(x) = ln

(

x3

x2 − 1

)

.

(a) [4 MARKS] Specify the domain of f , and any points of discontinuity.

(b) [5 MARKS] Determine all horizontal asymptotes and all vertical asymptotes.

(c) [5 MARKS] Determine the interval(s) where f increases, the interval(s) wheref decreases, and all local extrema.

(d) [3 MARKS] Determine the concavity of the graph of f , and find all points ofinflection.

(e) [3 MARKS] Sketch the graph, using the information determined above.

10. [5 MARKS] Showing all your work, determine sin−1

(

sin8π

3

)

.


G.8 December 2000 Final examination in 189-140A

1. (a) [5 MARKS] Find the derivative of ln cosh(x2) .

(b) [5 MARKS] Find the derivative of arctan

(

2t

1 + t2

)

. You are expected to

simplify your answer.

2. [10 MARKS] Determine the equation of the tangent to the curve

x2 + y sin x + tan2 y = 1

at the point (1, 0) .

3. [10 MARKS] Use a linear approximation to estimate the value of

3√

0.97 +5√

0.97 .

4. [10 MARKS] Showing all your work, determine the domain of the following func-tion, and all vertical or horizontal asymptotes to its graph.

f(x) =2x + 5√x2 − 36

.

5. [10 MARKS] Let f(x) = |2x + 5|. Show that there is no value of c such that

f(−4) = f(0) + f ′(c) · (−4 − 0) .

Explain why this does not contradict the Mean Value Theorem.

6. [10 MARKS] A farmer wants to fence an area of 20,000 square metres in a rectan-gular field, and then divide it into three parts by fences parallel to one of the sidesof the rectangle. What shape of field will minimize the cost of the fencing? Showall your work.

7. For the function h(x) = x2e2x, showing all your work

(a) [5 MARKS] Determine the intervals where h is increasing, and where it isdecreasing.

(b) [5 MARKS] Determine all local maxima, and all local minima.

(c) [5 MARKS] Determine intervals of concavity, and all inflection points.

(d) [5 MARKS] Sketch a graph of the function.


8. (a) [5 MARKS] Find the function f that satisfies all of the following conditions:

f ′′(x) = 3ex + 5 sin x , f(0) = 1 , f ′(0) = 2 .

(b) [5 MARKS] Determine all values of the constant c that will make thefollowing function continuous everywhere:

f(x) =

{

cx + 1 if x < 5cx2 − 1 if x ≥ 5

G.9 May 2001 Supplemental/Deferred Examination in 189-140A

1. (a) [5 MARKS] Showing all your work, evaluate limx→∞

(√

x + 2 −√x) .

(b) [5 MARKS] Showing all your work, evaluate limx→−1

sin(x2 − 1)

x + 1.

2. (a) [5 MARKS] Find the derivative ofx2 + 1

x2 − 2.

(b) [5 MARKS] Find the derivative of cos

(

1√x + 1

)

.

(c) [5 MARKS] Find the derivative of x sin−1 x .

3. [10 MARKS] Showing all your work, determine the greatest and least values at-

tained by the function f(x) =sin x

(7 − 2 cos x)2.

4. [10 MARKS] If x3 − xy + y3 = 1 , determinedy

dxand

d2y

dx2when x = 1

and y = 0.

5. [10 MARKS] The illumination I of an object by a light source is given by the

formula I =S

d2units, where S is the strength of the light source, and d is the

distance of the object from the light source. If two light sources, one 8 times asstrong as the other, are placed 3 units apart, where should an object be placed onthe line between them so as to receive the least illumination?

6. [10 MARKS] Use a “linear” or “tangent-line” approximation at x = 4 to compute

an approximate value for f(x) =√

x +1√x

at x = 3.97.

7. For the function f(x) = x5 + x6 , showing all your work


(a) [5 MARKS] Determine the intervals where f is increasing, and those where itis decreasing.

(b) [5 MARKS] Determine all local maxima, and all local minima.

(c) [5 MARKS] Determine intervals of concavity, and all inflection points.

(d) [5 MARKS] Sketch a graph of the function.

8. (a) [5 MARKS] If f ′(x) = x − 1 and f(1) =1

5, find f(x).

(b) [5 MARKS] If f ′(x) =1

1 + x3and f(0) = 0, use the Mean Value Theorem

to show that

f(2) − f(0) =2

1 + c3

for some c such that 0 < c < 2, and deduce that2

9< f(2) < 2. (Use the

Mean Value Theorem — do not determine an exact formula for f(x).)


A TOTAL OF 137 MARKS ARE AVAILABLE ON THIS EXAMINATION.

1. In each of the following cases evaluate the limit, or explain why the limit does notexist. Show all your work. Do not use l’Hopital’s Rule.

(a) [5 MARKS] limx→2−

x − 2

|x − 2|


sin x − sin(3x)

x2 + 6x

(c) [5 MARKS] limx→−4

x + 4

x2 + 5x + 4

2. In each of the following cases evaluate the limit, or explain why the limit does notexist. Show all your work. If you wish, you may use l’Hopital’s Rule.

(a) [10 MARKS] limx→0

ex2 − 1

sec x − 1.

(b) [5 MARKS] limx→∞

sinh(x + 1)

cosh(x). (Hint: Express the functions in terms of

exponentials.)


3. Showing your work, find the derivative of each of the following functions at thepoints indicated, or explain why the function fails to be differentiable there. Inall cases you are expected to simplify your answers as much as possible, but theexaminers are aware that you do not have a calculator.

(a) [5 MARKS] a(x) = ex sin x at x = −π

2.

(b) [5 MARKS] b(x) =

√x + 5√x − 5

at x = 4 .

(c) [5 MARKS] c(x) =eln(2x)

ln(e3x)at x = 4 .

4. A function K(x) is defined as follows, where α and β are constants to be evaluated:

K(x) =

{

α + x − x2 if x < 2x2 − β(x − 2) − 4 if x ≥ 2

(a) [8 MARKS] Showing all your work, determine all values of α and β — if any— that will make K continuous at x = 2.

(b) [7 MARKS] Showing all your work, determine all values of α and β — if any— that will make K differentiable at x = 2.

5. Suppose that y = 1 − xy4 .

(a) [10 MARKS] Showing all your work, determine the values ofdy

dxand

d2y

dx2when x = 0 .

(b) [5 MARKS] Find an equation for the tangent to the curve y = 1 − xy4

at the point where x = 0 .

6. Let f(x) =1

x− 1

x − 1− 1.

(a) [3 MARKS] Determine the domain of f .

(b) [5 MARKS] Determine the intervals where f is increasing, and the intervalswhere it is decreasing.

(c) [5 MARKS] Determine the intervals where the graph of f is concave upwards,and the intervals where it is concave downwards.

(d) [5 MARKS] Determine the local extrema of f , or prove that there are nolocal extrema.


(e) [5 MARKS] Determine the inflection points of the graph of f , or prove thatthere are no inflection points.

(f) [4 MARKS] Determine the horizontal asymptotes (if any) and the verticalasymptotes (if any) to the graph of f .

(g) [3 MARKS] Sketch the graph of f .

7. A line ℓ with positive slope m is drawn through the point (−4, 9) in the plane.

(a) [1 MARK] Find an equation for ℓ.

(b) [2 MARKS] ℓ intersects the x-axis in a point A and the y-axis in a point B.Find the coordinates of A and B.

(c) [12 MARKS] Showing all your work, determine which values of m minimizethe area of the triangle AOB (where O denotes the origin).

8. (a) [6 MARKS] Showing all your work, find a linear approximation to 4√

10020 .

(b) [2 MARKS] Carefully state the Mean Value Theorem.

(c) [7 MARKS] Showing all your work, apply the Mean Value Theorem to showthat the function arctanx − x is equal to zero only at x = 0 .

(d) [2 MARKS] Prove that the curves y = x and y = arctan x intersect onlyat the origin.

G.11 May 2002 Supplemental/Deferred Examination in 189-140A

1. Compute the derivatives of the functions

(a) [4 MARKS] f(x) = 2xx3

(b) [4 MARKS] g(x) = sinh(√

x4 + 1)

(c) [4 MARKS] h(x) =sec x

arctanx

2. Suppose that the function f satisfies

limx→1

f(x) − 3

x2 − 1= 4

(a) [6 MARKS] Find limx→1

f(x).

(b) [6 MARKS] Given that f is continuous at x = 1, use the definition of thederivative to show that f is differentiable at x = 1 and find f ′(1).


3. The equationy5 + xy2 + x3 = 4x + 3

defines y implicitly as a function of x near the point (2, 1).

(a) [6 MARKS] Determine the values of y′ and y′′ at this point.

(b) [4 MARKS] Use the tangent line approximation to estimate y when x = 1.98.

4. Let f(x) =x3

x2 − 4.

(a) [3 MARKS] Compute and simplify f ′(x) and f ′′(x).

(b) [3 MARKS] Determine, if any, all vertical and horizontalasymptotes of f .

(c) [3 MARKS] Determine the intervals where f is increasing, and the intervalswhere f is decreasing.

(d) [3 MARKS] Determine all local maxima and all local minima.

(e) [3 MARKS] Determine all intervals where (the graph of) f is concave upwardand all intervals where (the graph of) f is concave downward.

(f) [3 MARKS] Determine all inflection points.


5. Evaluate the following limits, using l’Hopital’s rule or otherwise:


tan(3x)

sin(5x)


(

xe1/x − x)

6. A rectangle is inscribed with its base on the x-axis, its upper left vertex on they-axis and its upper right vertex on the graph of the function y = e−2x.

(a) [6 MARKS] Find the dimensions of the rectangle that maximize its area. Fullyjustify your answer!

(b) [6 MARKS] Find the dimensions of the rectangle that minimize its circum-ference. Fully justify your answer!

7. [10 MARKS] Use the mean value theorem, or properties of the derivative, to showthat the graphs of y = arcsin x and y =

√2x2 intersect exactly once for x

between√

22

and 1.


8. In each of the following problems, find the function that satisfies all the statedconditions:

(a) [6 MARKS] f ′(x) = e2x − sin x, f(0) = 0.

(b) [6 MARKS] g′′(x) =1

x2, g′(1) = 0, g(1) = 0.

G.12 December 2002 Final Examination in MATH 140 2002 09

A TOTAL OF 140 MARKS ARE AVAILABLE ON THIS EXAMINATION.

1. In each of the following cases evaluate the limit, or explain why the limit does notexist. Show all your work. Do not use l’Hopital’s Rule.

(a) [5 MARKS] limu→−5

5 + u

u2 − 25.


x2

sin2 3x

(c) [10 MARKS] limx→∞

x(cosh ln x − sinh ln x).

2. In each of the following cases evaluate the limit, or explain why the limit does notexist. Show all your work. If you wish, you may use l’Hopital’s Rule.


2 − x2 − 2 cos x

x4 .


x (2 arctanx − π) .

3. In each of the following problems you are expected to show your work, and tosimplify your answer as much as possible. The examiners are aware that you donot have the use of a calculator.

(a) [5 MARKS] Find the value of the derivative of a(x) = tan(1 + x2) at x = 0 .

(b) [8 MARKS] Find the value of the derivative of f(x) = 2 · xln x at x = e3 .

(c) [7 MARKS] Find a function f(x) such that f(−1) = 7 and, for x < 0 ,

f ′(x) =2

x.

4. Let A be the point (6,−3) on the curve C with equation x2 = y2(y + 7) .

(a) [12 MARKS] Showing all your work, determine an equation for the tangentto the curve at A .


(b) [8 MARKS] A point P is moving along the curve C . Let t represent time.

Determine the value ofdy

dtwhen P is at position A on the curve, if it is

known that, at that moment,

dx

dt= 4 m/s.

5. Let f(x) = x(4 − x2)12 .

(a) [3 MARKS] Determine the (largest possible) domain of f .

(b) [5 MARKS] Determine the intervals where f is increasing, and the intervalswhere it is decreasing.

(c) [5 MARKS] Determine the local extrema of f , or prove that there are nolocal extrema.

(d) [5 MARKS] Prove that f ′′(x) = 2x(x2 − 6)(4− x2)−3/2 . Then determine theintervals where the graph of f is concave upwards, and the intervals whereit is concave downwards.

(e) [5 MARKS] Determine the inflection points of the graph of f , or prove thatthere are no inflection points.

(f) [4 MARKS] Determine every horizontal asymptote to the graph of f , or provethat there is none; determine every vertical asymptote to the graph of f , orprove that there is none.


6. (a) [8 MARKS] Determine the derivative of the function

h(x) = arctanx + arctan 1 − arctan1 + x

1 − x.

for x 6= 1 . You are expected to simplify your answer.

(b) [7 MARKS] Use your solution to question 6.(a) to determine the value ofh(−5) . Only a solution using the previous result will be accepted. Reduceyour answer as much as possible; the examiners are aware that you do nothave the use of a calculator.

7. Let f(x) = sin3 x +√

3 · cos3 x .

(a) [10 MARKS] Showing all your work, determine all local maxima (= relativemaxima) and all local minima (= relative minima) x of f such that

−π

2< x <

π

2.


(b) [5 MARKS] Determine the global maximum (= absolute maximum) andglobal minimum (= absolute minimum) value of f on the interval

−π

2≤ x ≤ π

2.

(c) [5 MARKS] Showing all your work, use a linear approximation to estimatethe value of

sin3 62◦ +√

3 · cos3 62◦ .

You may assume that the sine of 30◦ is1

2, and that the sine of 60◦ is

√3

2.

G.13 May 2003 Supplemental/Deferred Examination in MATH140 2002 09

1. In each of the following cases, evaluate the limit or explain why the limit does notexist. Do not use L’Hopital’s Rule!

(a) [3 MARKS] limx→∞

2x3 − 3x2 + 7

−x3 + 5x − 1


√x + 4 − 3

x − 5

(c) [3 MARKS] limx→0

(tan 2x) · (cos 2x)

x

2. Let a be a real number, and f(x) =

{

2x − 4a2 if x ≤ 2a(x2 − 6x + 8) if x > 2

.

(a) [5 MARKS] Determine the value(s) of a — if any — for which f is continuous.

(b) [5 MARKS] Explain why f is not differentiable at x = 2 if a = −2.

3. Compute the derivatives of the following functions. Simplify your answers!

(a) [4 MARKS] f(x) =sin x

ex.

(b) [4 MARKS] g(x) = arctan

(

1

x

)

.

(c) [4 MARKS] h(x) = ln(

x +√

1 + x2)

.

4. (a) [5 MARKS] Determine the function f if it is known that f ′′(x) = 6x − 6,f(0) = −2 and f(1) = 3.


(b) [5 MARKS] Let g(x) =3√

1 − x2. Find an antiderivative G of g such that

G

(√3

2

)

= 0.

5. Consider the curve C with equation 2x3 + 2y3 = 9xy .

(a) [2 MARKS] Verify that the point P = (1, 2) lies on C.

(b) [4 MARKS] Determine the equation of the tangent line to C at P .

(c) [6 MARKS] Determined2y

dx2when x = 1. Is C concave upward or concave

downward at P ?

6. [10 MARKS] Use the Mean Value Theorem to prove that x = 0 is the only solutionof the equation sinh x = x. Carefully explain your reasoning!

7. [10 MARKS] A particle is moving along the parabola 4y = (x + 1)2 in such away that its x-coordinate is increasing at the constant rate of 5 units per second.Determine how fast the distance from the particle to the origin is changing at theinstant the particle is at the point (3, 4).

8. [10 MARKS] A rectangle with sides parallel to the coordinate axes has one vertexat the origin, one on the positive x-axis, one on the positive y-axis, and its fourth

vertex in the first quadrant on the curve y =1

1 + x2. What is the maximum

possible area of such a rectangle? Fully justify your answer!

9. Let f(x) = (x2 + 4x + 4)e−x.

(a) [2 MARKS] Determine all x — if any — for which f(x) = 0 .

(b) [4 MARKS] Determine all horizontal asymptote(s) — if any — to the graphof f .

(c) [4 MARKS] Find all local extrema — if any — of f ; determine which arelocal maxima and which are local minima. Determine the intervals where fis increasing, and the intervals where f is decreasing.

(d) [4 MARKS] Find all inflection points — if any — of f . Determine the intervalswhere the graph of f is concave upward, and the intervals where the graph isconcave downward.

(e) [3 MARKS] Sketch the graph of f .


G.14 December 2003 Final Examination in MATH 140 2003 09






1. BRIEF SOLUTIONS

[2 MARKS EACH] Give the limit in each of the following cases. If the limit doesnot exist, or is +∞ or −∞, write “DOES NOT EXIST”, +∞, or −∞ respectively.

(a) limx→3−

15 − 2x − x2

x − 3=

ANSWER ONLY

(b) limx→−∞

7x5 + 2x3 − x2 + 11

x5 − 3x4 + 2=

ANSWER ONLY

(c) limx→0

tan 5x

sin 4x=

ANSWER ONLY


(d) limx→−5−

30 + 6x

|5 + x| =

ANSWER ONLY

(e) limx→∞

(√

x2 + x −√

x2 − 4x) =

ANSWER ONLY

(f) limx→−∞

x2ex =

ANSWER ONLY

2. BRIEF SOLUTIONS


(a)d

dtesin t2 =

ANSWER ONLY

(b)d

dxln(e4x) =

ANSWER ONLY

(c)d

dx|x4| =

ANSWER ONLY


(d)d

dxarcsin

√x =

ANSWER ONLY

(e) If y(x) satisfies the equation(y(x))5 + x (y(x))2 + 9x4 = 1 , then y′(0) =

ANSWER ONLY

(f) If g(x) =(

x6)

x2 , g′(0) =

ANSWER ONLY

3. BRIEF SOLUTIONS

[3 MARKS EACH]

(a) Give equations for all of the vertical asymptotes of the graph of

f(x) =x

x2 + x − 2.

If there is none, write “NONE”.

ANSWER ONLY

(b) Determine all values of the constant c that will make the function

f(x) =

{

−4x + c when x < 0(x + c)2 when x ≥ 0

continuous from the right at x = 0. If there is none, write “NONE”.


ANSWER ONLY

(c) Determine all horizontal asymptotes to the following curve; if there is none,write “NONE”.

y = arctan(x2)

ANSWER ONLY

(d) If tanh x =3

5, sinh x =

ANSWER ONLY

4. BRIEF SOLUTIONS

[4 MARKS EACH]

(a) Determine all points on the curve

y3 − x2 = 4

where the tangent is horizontal. If there is none, write “NONE”.

ANSWER ONLY

(b) On the interval −2 < x < 0, determine the values of x at which the functionf has a local (=relative) minimum, if it is known that

f ′(x) = (2x + 1)(x + 1)2(x + 3) .

If there is no local minimum, write “NONE”.


ANSWER ONLY

(c) Find f(2) if it is known that

f ′(x) =x2 − x + 1

x, and

f(1) = 5 .

ANSWER ONLY

The examiners are aware that you do not have a calculator.


[3 MARKS EACH]

For each of the following descriptions of a function or functions, either

• give an example of functions with the properties stated, or

• name or state a theorem, law, or rule which can be used to show that no suchfunction or functions exist.

(a) f(3) = 0, f(1) = −4, f ′(x) ≤ 1 for all x.

(b) f is continuous on [−4, 4], f(−3) = −1, f(3) = 2, f(x) 6= 0 for all x.

(c) limx→4

f(x) = 3, f(

limx→4

x)

= 2 .

(d) f ′(1) = 5 and f is not continuous at x = 1 .


[12 MARKS] A balloon leaves the ground 100 metres from an observer, and risesvertically at the rate of 40 metres per minute. Determine the rate at which theangle of inclination of the observer’s line of sight (the angle between the line ofsight and the horizontal) is increasing at the instant when the balloon is exactly100 metres above the ground?



Let f(x) = x√

x + 3 .







Let t represents time measured in seconds, and let C be a constant that is tobe determined. A particle moves so that its position on the x axis at time t isx(t) = t3 − 12t2 + 36t + C.

(a) [3 MARKS] Determine the acceleration of the particle as a function of time.

(b) [6 MARKS] The speed of the particle is the absolute value of its velocity.Determine the time intervals when the speed is increasing.

(c) [3 MARKS] Determine the value of C if it is known that the particle is at theorigin the whenever the acceleration is 0.

G.15 May 2004 Supplemental/Deferred Examination in MATH140 2003 09







1. BRIEF SOLUTIONS

[2 MARKS EACH] Find each of the following. If the item requested does not exist,or is +∞ or −∞, write “DOES NOT EXIST”, +∞, or −∞ respectively.

(a) If g(x) = 2 + x + ex, find g−1(3).

ANSWER ONLY

(b) Find a formula for the inverse of the function

r(x) = ln(x + 2) .

ANSWER ONLY

(c) Find limt→0

√t2 + 4 − 2

t2.

ANSWER ONLY

(d) limx→−∞

(

x +√

x2 + 3x)

=

ANSWER ONLY

(You may do rough work in this space, or on the back of the preceding page.)

2. BRIEF SOLUTIONS


(a)d

dt

(

23t)

=


ANSWER ONLY

(b)d

dx

(

ex

x2

)

=

ANSWER ONLY

(c)d

dx

(

(x2 + 1)50(x4 + 1)3)

=

ANSWER ONLY

(d)d

dxsin(sin 2x) =

ANSWER ONLY


Full explanations are required for all of your statements.

(a) [4 MARKS] Determine the domain of the function

r(x) =√

x −√−x .

(b) [4 MARKS] Determine whether the function

h(x) =

{

x − 2 for x < 35 − x for x ≥ 3

is continuous at x = 3.

(c) [4 MARKS] Determine whether the function f(x) = |x − 4|5 is differentiableat x = 4.



[6 MARKS] A plane flying horizontally at an altitude of 2 km and a speed of 500km/h passes directly over a certain radar station. Determine the rate at which the(shortest) distance from the plane directly to the station is increasing when theplane is 3 km away from the station.

5. [8 MARKS] SHOW ALL YOUR WORK!

Showing all your work, and naming any theorems that you use, prove that theequation x5 − 7x + 5 = 0 has exactly one root in the closed interval [−1, +1].


(a) [6 MARKS] Find an equation for the line tangent to the curve y2 = x3(2−x)at the point (1, 1).

(b) [8 MARKS] A particle is moving on the x axis so that its position at time tis given by

x(t) = 2t3 − 7t2 + 4t + 1 .

Find

i. The times when the velocity is 0.

ii. The acceleration at each time when the velocity is 0.


(a) [2 MARKS] State the Mean Value Theorem.

(b) [8 MARKS] For the function

f(x) =

{

2|x + 1| when x ≤ −1x(1 − |x|)(1 + |x|) when x > −1

decide whether f satisfies the hypotheses of the Mean Value Theorem on theinterval [−2, 3]. Then answer the appropriate one of the following questions:

I. If f satisfies the conditions of the Mean Value Theorem on the interval[−2, 3], explain precisely what the conclusion of the theorem is for thisfunction.

II. If f does not satisfy the conditions of the Mean Value Theorem, explainprecisely where it fails to satisfy those conditions.



Let x and y = f(x) be related by the equation

ey + x · y = e .

Showing all your work,

(a) [2 MARKS] determine f(0);

(b) [3 MARKS] determine f ′(0);

(c) [5 MARKS] determine f ′′(0).


Let f(x) = x√

4 − x2 .






10. [9 MARKS] SHOW ALL YOUR WORK!

A metal box with a square base and an open top must have a volume of 500 m3.Find the dimensions of the box that minimize the total amount of metal used.

G.16 December 2004 Final Examination in MATH 140 2004 09(One of several versions)







1. BRIEF SOLUTIONS

[3 MARKS EACH] Give the numeric value of each of the following limits if it exists;if the limit is +∞ or −∞, write +∞ or −∞ respectively. In all other cases write”NO FINITE OR INFINITE LIMIT”.

(a) limy→−∞

|y|y

=

ANSWER ONLY

(b) limu→−∞

sin u

u=

ANSWER ONLY

(c) limn→∞

(

1 +1

n

)2n

=

ANSWER ONLY

(d) limx→0

(

1

x− 1

sin x

)

=

ANSWER ONLY

(e) limt→∞

(

ln(3t2) − ln(t2 + 7))

=


ANSWER ONLY

(f) limx→8

√x + 8 −

√2x

x2 − 8x=

ANSWER ONLY

2. BRIEF SOLUTIONS

[3 MARKS EACH] Determine each of the following derivatives, and simplify youranswers as much as possible.

(a)d

dx

(

x2 + 3x

x

)

=

ANSWER ONLY

(b)d

dt

(

t−3t)

=

ANSWER ONLY

(c)d

ds

(


=

ANSWER ONLY

(d)d

dy

(

cosh2(3y))

=


ANSWER ONLY

(e)d

dX

(

cos2 X − cos(X2))

=

ANSWER ONLY

3. BRIEF SOLUTIONS

(a) [5 MARKS] Find an equation for a line through the point (−2, 0) which istangent to the curve y = x2 and is not horizontal.

ANSWER ONLY

(b) [5 MARKS] Determine values of the constants a and b that will make thefollowing function continuous at x = −6.

f(x) =

√x + 31 if x < −6a + b if x = −6

a(x + 5) if x > −6

ANSWER ONLY

(c) [5 MARKS] Determine values of the constants k and ℓ that will make thefollowing function differentiable at x = 1.

g(x) =

{

kx2 + ℓ if x ≤ 16x − 4 if x > 1


ANSWER ONLY


(a) [6 MARKS] Coffee is draining from a conical filter of depth 10 cm and diameter10 cm (at the top) into a cylindrical coffee pot of diameter 12 cm, at the rateof 100 cm3/min. Determine how fast, in cm/min, the level of coffee in the potis rising when the coffee in the filter is 3 cm deep?

(b) [6 MARKS] You are given that y = y(t) is a function of t satisfying t3y+ty3 =2. Assuming that y(1) = 1, determine the values of y′(1) and y′′(1).





[10 MARKS] Using the calculus carefully, determine how to express 8 as the sumof 2 nonnegative real numbers such that the sum of the square of the first and thecube of the second is as small as possible.


Let f(x) =

(

x

x + 3

)2

.







[6 MARKS] Consider the function f(x) = 5x + 9 near x = −1. Is it, or is it nottrue that f is continuous at x = −1? If the statement is true, prove it carefully,using the ǫ-δ definition. If it is false, prove that carefully.

G.17 May 2005 Supplemental/Deferred Examination in MATH

140 2004 09






1. BRIEF SOLUTIONS


(a) limx→4

(

2x

x − 4− 8

x − 4

)

=

ANSWER ONLY

(b) limu→∞

1 − cos u

u2=

ANSWER ONLY


(c) limn→∞

(

1 − 1

n

)−n

=

ANSWER ONLY

(d) limx→π

2

(

1

x− 1

sin x

)

=

ANSWER ONLY

(e) limx→−2

x + 2√−x −

√2

=

ANSWER ONLY

2. BRIEF SOLUTIONS

[3 MARKS EACH] Determine each of the following derivatives or antiderivatives,and simplify your answers as much as possible. If the derivative or antiderivativedoes not exist, write “DOES NOT EXIST”.

(a)d

dx

(

x2 + 3x

x

)

=

ANSWER ONLY

(b)d

dt

(

t−3t)

=

ANSWER ONLY


(c)d

ds(sinh(ln s) − ln(sinh s)) =

ANSWER ONLY

(d) At the point y = −1,d

dy

(

2y · | − y| + 3 · |y|4)

=

ANSWER ONLY

(e) A function f such that f ′′(θ) = cos θ − sin θ, f(0) = 4, f ′(π) = −3.

ANSWER ONLY

3. BRIEF SOLUTIONS

(a) [5 MARKS] Find all points — if any — on the graph of the function

f(x) = 2 sin x + sin2 x (−π ≤ x ≤ π)

at which the tangent line is horizontal.

ANSWER ONLY

(b) [5 MARKS] Determine values of the constants a and b that will make thefollowing function continuous at x = 24.

f(x) =

√49 − x if x < 24a − b if x = 24

b(x − 19) if x > 24


ANSWER ONLY

(c) [5 MARKS] Determine the domain of the function

sin(arcsin x) − arcsin(sin x) .

ANSWER ONLY


(a) [6 MARKS] You are given that y = y(x) is a function of x satisfying

2x3 − 3xy4 + 5xy − 10 = 0 .

Express the value of y′(x) in terms of x and y.

(b) [9 MARKS] Suppose the lengths of the sides of a triangle are a, b and c. Defines = 1

2(a + b + c). Heron’s formula for the area T of the triangle is

T =√

s(s − a)(s − b)(s − c) .

If the lengths of the sides are all increasing at the rate of 1 cm/min, determine— using logarithmic differentiation or otherwise — the rate of change of the

area, T , when a=3 cm, b=4 cm, c= 5 cm. (Hint: First findds

dt.)






[10 MARKS] Determine all points on the curve y = 1 − 40x3 + 3x5 at which theslope of the tangent line has a local maximum, and all points where the slope ofthe tangent line has a local minimum. Then discuss whether the slope has globalmaxima and global minima.


Let f(x) = e−1x .






[6 MARKS] Let f(x) = 5x − 8. Given a real number ǫ > 0, find a real number δ— depending on ǫ — such that |f(x) − 7| < ǫ whenever 0 < |x − 3| < δ.

G.18 December, 2005, Final Examination in MATH 140 2005

09 (one version)







1. BRIEF SOLUTIONS


(a) limy→−∞

(

√

y2 + y + y)

=

ANSWER ONLY

(b) limx→∞

(

e−x sinh x)

=

ANSWER ONLY

(c) limx→8

√x + 8 +

√2x

x2 + 8x=

ANSWER ONLY

(d) limx→1

(

1

1 − x+

1

ln x

)

=

ANSWER ONLY

(e) limx→∞

sin ex

cos ex=

ANSWER ONLY


2. BRIEF SOLUTIONS


(a)d

dx

(

x2 + 3x

x

)

=

ANSWER ONLY

(b)d

du(uu) =

ANSWER ONLY

(c) An antiderivative F (x) of f(x) = 5x4 + 2x5 such that F (0) = 3 is

ANSWER ONLY

(d) If f(x) = x3 + 7, its inverse function f−1(x) =

ANSWER ONLY

(e)d

dx

(

|x|4)

=

ANSWER ONLY



(a) [6 MARKS] Showing all your work, determine values of the constants a andb that will make the following function continuous everywhere.

f(x) =

(1 + x)1x if x > 0

a + bx if −1 ≤ x ≤ 0sin(x + 1)

x + 1if x < −1

(b) [4 MARKS] Determine whether f is differentiable at x = −1. (For the purposeof this question you may assume that e is approximately 2.72.)


[5 MARKS] Let f(x) = x2ex. Prove carefully by mathematical induction that

dnf

dxn(x) =

(

x2 + 2nx + (n − 1)n)

· ex



[5 MARKS] Let g(x) = 2x − 3 + cos x. Use Rolle’s Theorem or the Mean ValueTheorem, to prove carefully that there exists exactly one real number x such thatg(x) = 0. (π may be taken to be approximately 3.14.)


[10 MARKS] A rectangular poster is to be printed on a rectangular board of min-imum area, leaving margins at the 4 sides. The top and bottom margins are each10 cm, and the side margins are each 4 cm. If the printed area on the poster isfixed at 1,000 cm2, find the best dimensions for the board. Show all of your work,and justify all of your statements. In your solution, you are expected to carefullyapply either the First or the Second Derivative Test, naming the test as you applyit.


Let f(x) =√

x2 − 1 .




(c) [2 MARKS] Define when a line x = a is a vertical asymptote to the graph off .

(d) [2 MARKS] Either

i. Find all vertical asymptotes; or

ii. Explain why the graph has no vertical asymptotes.

(e) [2 MARKS] Determine the global maximum value of f , or explain why thereis none.

(f) [2 MARKS] Determine the global minimum value of f , or explain why thereis none.


A function f(t) satisfies, for all real numbers t, the equation

t3 + f(t)3 + 6t2 · f(t) = 8 .

(a) [5 MARKS] Find an equation for the tangent to the graph y = f(t) at thepoint (t, y) = (2, 0).


G.19 May, 2006, Supplemental/Deferred Examination in MATH

140 2005 09







1. BRIEF SOLUTIONS

[2 MARKS EACH] Give the value of each of the following limits if it exists; if thelimit is +∞ or −∞, write +∞ or −∞ respectively. In all other cases write “NOFINITE OR INFINITE LIMIT”.

(a) limx→∞

sin e−x

cos e−x=

ANSWER ONLY

(b) If a and b are real numbers, limy→∞

(

√

y2 + ay −√

y2 − by)

=

ANSWER ONLY

(c) limx→−∞

(

e−x cosh x)

=

ANSWER ONLY

(d) limx→π

6

sin x + sin π6

x + π6

=

ANSWER ONLY

(e) limx→0+

x3

x − sin x=

ANSWER ONLY


2. BRIEF SOLUTIONS


(a) An antiderivative G(x) of g(x) = ex + 4x − 1 such that G(0) = 5 is

ANSWER ONLY

(b)d

dy

(

y(cos 2y)

cos2 y − sin2 y

)

=

ANSWER ONLY

(c)d

dx

(

x−x)

=

ANSWER ONLY

(d)d

dx(x arctan(4x)) =

ANSWER ONLY

(e) If A(t) = ln (t3 + 7), its inverse function A−1(t) =

ANSWER ONLY



(a) [5 MARKS] Showing all your work, determine all values of the constants aand b that will make the following function continuous everywhere.

f(x) =

{

x(2a − x) if x ≤ 0x(x2 − 6x + 12) + b if x > 0

(b) [5 MARKS] By examining one-sided limits, determine all values of the con-stants a and b that will make f differentiable at x = 0.


[5 MARKS] Two sides of a triangle are 3 m. and 4 m. in length, and the anglebetween them is increasing at the rate of 1 degree/second. Find the rate at whichthe area of the triangle is increasing when the angle between the sides of fixed

length isπ

3. [Remember that the area of the triangle is one-half the product of the

lengths of the base and the altitude (or height).]


[5 MARKS] Use Rolle’s Theorem or the Mean Value Theorem, to prove carefullythat there exists no function f such that f(1) = 5, f(4) = 14, and f ′(x) ≤ 2 for−3 ≤ x ≤ 8.


[10 MARKS] Showing all your work, determine the points on the curve y = −3x5 +40x3 − 1 where the tangent line has the largest positive slope. In your solution,you are expected to carefully apply either the First or the Second Derivative Test,naming the test as you apply it.


Let f(x) = x√

x2 − 1 .



(c) [4 MARKS] Investigate whether the graph of f has horizontal and/or verticalasymptotes, and determine them all, if there are any.

(d) [4 MARKS] Investigate whether the graph of f has global extremum points,and determine them all, if there are any.



A function y = f(x) satisfies, for all real numbers x, the equation

ey + xy = e .

(a) [5 MARKS] Find an equation for the tangent to the graph y = f(x) at thepoint (x, y) = (0, 1).



H WeBWorK

H.1 Frequently Asked Questions (FAQ)

H.1.1 Where is WeBWorK?

WeBWorK is located on Web servers of the Department of Mathematics and Statistics,and is accessible at the following URL’s:

http://msr04.math.mcgill.ca/webwork/m140f06or

http://msr05.math.mcgill.ca/webwork/m140f06

If your student number ends with an odd digit — 1, 3, 5, 7, or 9 — you should accessthe URL

http://msr05.math.mcgill.ca/webwork/m140f06;

if your student number ends with an even digit — 0, 2, 4, 6, or 8 — you should access

http://msr04.math.mcgill.ca/webwork/m140f06.

If you access WeBWorK through WebCT, the link on your page will have been pro-grammed to take you to the correct WeBWorK server automatically.

H.1.2 Do I need a password to use WeBWorK?

You will need a user code and a password.

Your user code. Your user code will be your 9-digit student number.

Your password. The WeBWorK system is administered by the Mathematics andStatistics Department, and is not accessible through the myMcGill Portal; your initialpassword will be different from your MINERVA password, but you could change it tothat if you wish. Your initial password will be your 9-digit student ID number. You willbe able to change this password after you sign on to WeBWorK.103

103If you forget your password you will have to send a message to Professor Brown so that the systemadministrator may be instructed to reset the password at its initial value.



Your e-mail address. The WeBWorK system requires each user to have an e-mailaddress. After signing on to WeBWorK, you should verify that the e-mail addressshown is the one that you prefer. You should endeavour to keep your e-mail address upto date, since the instructors may send messages to the entire class through this route.

We suggest that you use either your UEA104 or your po-box address. You may be ableto forward your mail from these addresses to another convenient address, (cf. §1.8.1.)

H.1.3 Do I have to pay an additional fee to use WeBWorK?

WeBWorK is available to all students registered in the course at no additional charge.

H.1.4 When will assignments be available on WeBWorK?

Each assignment will have a begin date and a due date. The assignment is available toyou after the begin date; solutions will be made available soon after the due date.

H.1.5 Do WeBWorK assignments cover the full range of problems that Ishould be able to solve in this course?

The questions on the WeBWorK assignments (A1 through A7) are a sampling of sometypes of problem you should be able to solve after successfully completing this course.Some types of calculus problems do not lend themselves to this kind of treatment, andmay not appear on the WeBWorK assignments. Use of WeBWorK does not re-place studying the textbook — including the worked examples, attendinglectures and tutorials, and working exercises from the textbook — using theStudent Solutions Manual [3] to check your work. Students are cautioned notto draw conclusions from the presence, absence, or relative frequencies of problems ofparticular types, or from particular sections of the textbook. Certain sections of thetextbook remain examination material even though no problems are included in theWeBWorK assignments.

H.1.6 May I assume that the distribution of topics on quizzes and finalexaminations will parallel the distribution of topics in the WeBWorKassignments?

No! While the order of topics on WeBWorK assignments should conform to the orderof the lectures, there are some topics on the syllabus that will not appear in WeBWorKquestions. Use WeBWorK for the areas it covers, and supplement it by working prob-lems from your textbook. Also, remember that WeBWorK — which checks answer

104Uniform E-mail Address


only — cannot ascertain whether you are using a correct method for solving problems.But, if you write out a solution to an odd-numbered textbook problem, you can compareit with the solution in the Solutions Manual; and, if in doubt, you can show your workto a Teaching Assistant at one of the many office hours that they hold through the week.

H.1.7 WeBWorK provides for different kinds of “Display Mode”. Whichshould I use?

“Display mode” is the mode that you enter when you first view a problem; and, later,when you submit your answer. You may wish to experiment with the different formats.The default is images mode, which should look similar to the version that you printout (cf. next question). The lowest quality is text mode, which is essentially the waythe author of the problem entered his data into the system; this mode is related tothe TEX and LATEX systems that mathematicians use in typesetting their documents;the notes that you are reading here were prepared using LATEX; it contains formattinginstructions in a “markup” language, and is difficult for inexperienced readers. Themode called jsMath requires the presence of certain TEX fonts on your computer, andshould be avoided unless you are using your own computer, and are prepared to followthe instructions to install the new fonts. This requires a major investment of time, andis not recommended unless there are other reasons why you need to work with TEX orLATEX.

H.1.8 WeBWorK provides for printing assignments in “Portable DocumentFormat” (.pdf), “PostScript” (.ps) forms. Which should I use?

Most newer home computers have already been loaded with the Acrobat Reader for .pdffiles; if the Reader has not been installed on your computer105, you will find instructionsfor downloading this (free) software in §1.5.5 of these notes. If you are not happy with.pdf files, and wish to print and view PostScript files, you may require such (free) softwareas Ghostscript and Ghostview, available at

http://www.cs.wisc.edu/∼ghost/gsview/index.html

Most computers available to you on campus should be capable of printing in eitherof .pdf and PostScript formats.

H.1.9 What is the relation between WeBWorK and WebCT?

There is none. WebCT is the proprietary system of Web Course Tools that has beenimplemented by McGill University. You may access the web page for this course, and

105At the time these notes were written, the latest version of the Reader was 7.0.8, but recent, earlierversions should also work properly.


WeBWorK through your WebCT account106, and WebCT will link you to the appropri-ate server for WeBWorK. If you follow this route to WeBWorK, you will still have tolog in when you reach the WeBWorK site. At the present time we will be using WebCTprimarily for the posting of grades, and as a convenient repository for links to notes andannouncements in the course. We are not planning to use the potential WebCT sitesthat exist for the tutorial sections: use only the site for the lecture section in which youare registered.

H.1.10 Which browser should I use for WeBWorK?

We recommend that you use Internet Explorer, Netscape, or Mozilla. While otherbrowsers may give satisfactory results, your instructors and tutors do not have timeto correct errors in your WeBWorK records that could be attributed to idiosyncraciesin another browser. Information about browsers supported by WebCT may be obtainedat

http://www.mcgill.ca/webct/

H.1.11 What do I have to do on WeBWorK?

After you sign on to WeBWorK, and click on “Begin Problem Sets”, you will see alist of Assignments, each with a due date. Since there is no limit to the number ofattempts at problems on P0 or the other “Practice” assignments, you may play withthese assignments to learn how to use the WeBWorK software.

You may print out a copy of your assignment by clicking on “Get hard copy”. This isyour version of the assignment, and it will differ from the assignments of other studentsin the course. You should spend some time working on the assignment away from thecomputer. When you are ready to submit your solutions, sign on again, and select thesame assignment. This time click on “Do problem set”. You can expect to become morecomfortable with the system as you attempt several problems; but, in the beginning,there are likely to be situations where you cannot understand what the system findswrong with some of your answers. It is useful to click on the Preview Answers button tosee how the system interprets an answer that you have typed in. As the problems maybecome more difficult, you may have to refer to the “Help” page, and also to the “Listof functions” which appears on the page listing the problems. Don’t submit an answeruntil you are happy with the interpretation that the Preview Answers button shows thatthe system will be taking of your answer.

106http://webct.mcgill.ca


H.1.12 How can I learn how to use WeBWorK?

As soon as your instructor announces that the WeBWorK accounts are ready, sign onand try assignment P0, which does not count. The system is self-instructive, so we willnot burden you with a long list of instructions.

You will need to learn how to enter algebraic expressions into WeBWorK as it is codedto read what you type in a way that may different from what you expect. For example,the symbol ^ is used for writing exponents (powers). If you type 2^3, WeBWorK willinterpret this as 23 = 8. However, if you type 2^3+x, WeBWorKwill interpret it as23 +x, i.e. as 8+x; if you wish to write 23+x, you have to type 2^(3+x). You may obtainmore information from the List of Available Functions, available online, or at

http://webwork.math.rochester.edu/webwork_system_html

/docs/docs/pglanguage/availablefunctions.html

H.1.13 Where should I go if I have difficulties with WeBWorK ?

If you have difficulties signing on to WeBWorK, or with the viewing or printing func-tions on WeBWorK, or with the specific problems on your version of an assignment,you may send an e-mail distress message directly from WeBWorK by clicking on theFeedback button. You may also report the problem to your instructor and/or your

tutor, but the fastest way of resolving your difficulty is usually the Feedback . Pleasegive as much information as you can. (All of the instructors and tutors are able to viewfrom within WeBWorK the answers that you have submitted to questions.)

If your problem is mathematical, and you need help in solving a problem, you shouldconsult one of the tutors at their office hours; you may go to any tutor’s office hours, notonly to the hours of the tutor of the section in which you are registered.

H.1.14 Can the WeBWorK system ever break down or degrade?

Like all computer systems, WeBWorK can experience technical problems. The systemsmanager is continually monitoring its performance. If you experience a difficulty whenonline, please click on the Feedback button and report it. If that option is not availableto you, please communicate with either instructor by e-mail.

If you leave your WeBWorK assignment until the hours close to the due time onthe due date, you should not be surprised if the system is slow to respond. This isnot a malfunction, but is simply a reflection of the fact that other students have alsobeen procrastinating! To benefit from the speed that the system can deliver under normalconditions, do not delay your WeBWorK until the last possible day! If a systems failureinterferes with the due date of an assignment, arrangements could be made to change that


date, and an e-mail message could be broadcast to all users (to the e-mail addresses onrecord), or a message could be posted on WeBCT or the WeBWorK sign-on screen.107

H.1.15 How many attempts may I make to solve a particular problem onWeBWorK?

Practice Assignments P1 — P7 are intended to prepare you for Assignments A1 — A7,and permit unlimited numbers of attempts; your grades on these “Practice” do notcount in your term mark. For the problems on assignments A1 — A7 you will normallybe permitted about 5 tries: read the instructions at the head of the assignment.

H.1.16 Will all WeBWorK assignments have the same length? the samevalue?

The numbers of problems on the various assignments may not be the same, and theindividual problems may vary in difficulty. Assignments A1 — A7 will count equally inthe computation of your grade.

H.1.17 Is WeBWorK a good indicator of examination performance?

A low grade on WeBWorK has often been followed by a low grade on the examination.A high grade on WeBWorK does not necessarily indicate a likely high grade on the

examination.To summarize: WeBWorK alone is not enough to prepare this course; but students

who don’t do WeBWorK appear to have a poor likelihood of success in MATH 140:that is one reason why we have made the WeBWorK assignments compulsory.

107But slowness of the system just before the due time will not normally be considered a systemsfailure.


I Contents of the DVD disks for

Larson/Hostetler/Edwards

These excellent disks were produced to accompany the textbook, Calculus of a SingleVariable: Early Transcendental Functions, 3rd Edition[22] (called LHE in the charts be-low). The correspondence shown to sections of [11] are only approximate. (NOTE THATTHIS BOOK DOES NOT FOLLOW STEWART’S CONVENTIONS FOR INVERSESECANT/COSECANT!)

DVD LHE Stewart# Section Subject Minutes Section1 P Chapter P: Preparation for Calculus1 P.1 Graphs and Models 451 P.2 Linear Models and Rates of Change 27 A101 P.3 Functions and Their Graphs 48 1.11 P.4 Fitting Models to Data 21 1.21 P.5 Inverse Functions 48 1.61 P.6 Exponential and Logarithmic Functions 30 1.5

DVD LHE Stewart# Section Subject Minutes Section1 1 Chapter 1: Limits and Their Properties1 1.1 A Preview of Calculus 11 2.11 1.2 Finding Limits Graphically and Numeri-

cally25 2.2, 2.4

1 1.3 Evaluating Limits Analytically 28 2.31 1.4 Continuity and One-Sided Limits 22 2.51 1.5 Infinite Limits 18 2.6

DVD LHE Stewart# Section Subject Minutes Section1 2 Chapter 2: Differentiation1 2.1 The Derivative and the Tangent Line Prob-

lem68 2.1

1 2.2 Basic Differentiation Rules and Rates ofChange

34 2.3

1 2.3 The Product and Quotient Rules andHigher Order Derivatives

25 3.2, 3.7


DVD LHE Stewart# Section Subject Minutes Section2 2 Chapter 2 (continued): Differentiation2 2.4 The Chain Rule 44 3.52 2.5 Implicit Differentiation 50 3.62 2.6 Derivatives of Inverse Functions 17 3.5, 3.8, 3.92 2.7 Related Rates 34 3.102 2.8 Newton’s Method 26 4.9

DVD LHE Stewart# Section Subject Minutes Section2 3 Chapter 3: Applications of Differentiation2 3.1 Extrema on an Interval 41 4.12 3.2 Rolle’s Theorem and the Mean Value The-

orem15 4.2

2 3.3 Increasing and Decreasing Functions andthe First Derivative Test

19 4.3

2 3.4 Concavity and the Second Derivative Test 24 4.32 3.5 Limits at Infinity 23 2.62 3.6 A Summary of Curve Sketching 43 4.52 3.7 Optimization Problems 37 4.72 3.8 Differentials 51 3.11

DVD LHE Stewart# Section Subject Minutes Section3 4 Chapter 4: Integration3 4.1 Antiderivatives and Indefinite Integration 40 4.10

DVD LHE Stewart# Section Subject Minutes Section4 7 Chapter 7: Integration by Parts Trigono-

metric Substitution Partial FractionsL’Hopital’s Rule

4 7.7 Indeterminate Forms and L’Hopital’s Rule 22 4.4

(The coverage extends to part of the material for Math 141 as well.)


J References

J.1 Stewart Calculus Series

[1] J. Stewart, Single Variable Calculus (Early Transcendentals), Fifth Edition. Thom-son * Brooks/Cole (2003). ISBN 0-534-39330-6.

[2] J. Stewart, Calculus (Early Transcendentals), Fifth Edition. Thomson *Brooks/Cole (2003). ISBN 0-534-39321-7.

[3] D. Anderson, J. A. Cole, D. Drucker, Student Solutions Manual for Stewart’s SingleVariable Calculus (Early Transcendentals), Fifth Edition. Thomson * Brooks/Cole(2003). ISBN 0-534-39333-0.

[4] J. Stewart, Single Variable Calculus (Early Transcendentals), Fifth Edition. Thom-son * Brooks/Cole (2003); bundled with Student Solutions Manual for Stew-art’s Single Variable Calculus (Early Transcendentals), Fifth Edition. Thomson* Brooks/Cole (2003). ISBN 0-17-6425411.

[5] J. Stewart, Single Variable Essential Calculus (Early Transcendentals). Thomson* Brooks/Cole (2006). Thomson * Brooks/Cole (2003). ISBN 0-495-10957-6.

[6] J. Stewart, Calculus (Early Transcendentals), Fifth Edition. Thomson *Brooks/Cole (2003); bundled with Student Solutions Manual for Stewart’s SingleVariable Calculus (Early Transcendentals), Fifth Edition. Thomson * Brooks/Cole(2003). ISBN 0-534-10307-3.

[7] R. St. Andre, Study Guide for Stewart’s Single Variable Calculus (Early Transcen-dentals), Fifth Edition. Thomson * Brooks/Cole (2003). ISBN 0-534-39331-4.

[8] Video Outline for Stewart’s Calculus (Early Transcendentals), Fifth Edition.Thomson * Brooks/Cole (2003). ISBN 0-534-39325-X. 17 VCR tapes.

[9] Interactive Video Skillbuilder CD for Stewart’s Calculus: Early Transcendentals,5th Edition. Thomson * Brooks/Cole (2003). ISBN 0-534-39326-8.

[10] H. Keynes, J. Stewart, D. Clegg, Tools for Enriching Calculus, CD to accompany[1] and [2]. Thomson * Brooks/Cole (2003). ISBN 0-534-39731-X.

[11] J. Stewart, Single Variable Calculus (Early Transcendentals), Fourth Edition.Brooks/Cole (1999). ISBN 0-534-35563-3.

[12] J. Stewart, Calculus (Early Transcendentals), Fourth Edition. Brooks/Cole (1999).ISBN 0-534-36298-2.


[13] D. Anderson, J. A. Cole, D. Drucker, Student Solutions Manual for Stewart’sSingle Variable Calculus (Early Transcendentals), Fourth Edition. Brooks/Cole(1999). ISBN 0-534-36301-6.

[14] J. Stewart, L. Redlin, S. Watson, Precalculus: Mathematics for Calculus, FifthEdition. Thomson * Brooks/Cole. (2006). ISBN: 0-534-49277-0.

[15] J. Stewart, Trigonometry for Calculus. Thomson * Brooks/Cole. ISBN: 0-17-641227-1.

J.2 Other Calculus Textbooks

J.2.1 R. A. Adams

[16] R. A. Adams, Calculus, Single Variable, Fifth Edition. Addison, Wesley, Longman,Toronto (2003). ISBN 0-201-79805-0.

[17] R. A. Adams, Calculus of Several Variables, Fifth Edition. Addison, Wesley, Long-man, Toronto (2003). ISBN 0-201-79802-6.

[18] R. A. Adams, Calculus: A Complete Course, Fifth Edition. Addison, Wesley, Long-man, Toronto (2003). ISBN 0-201-79131-5.

[19] R. A. Adams, Student Solution Manual for Adams’, Calculus: A Complete Course,Fifth Edition. Addison, Wesley, Longman, Toronto (2003). ISBN 0-201-79803-4.

[20] R. A. Adams, Calculus: A Complete Course, with Solution Manual, Fifth Edition.Addison, Wesley, Longman, Toronto (2003). ISBN 0-131-30565-4.

[21] R. A. Adams, Calculus: A Complete Course Manual, Sixth Edition. Addison,Wesley, Longman, Toronto (2006). ISBN 0-321-27000-2.

J.2.2 Larson, Hostetler, et al.

[22] Calculus Instructional DVD Program, for use with (inter alia) Lar-son/Hostetler/Edwards, Calculus of a Single Variable: Early Transcendental Func-tions, Third Edition [23]. Houghton Mifflin (2003). ISBN 0-618-25097-2.

[23] R. Larson, R. P. Hostetler, B. H. Edwards, D. E. Heyd, Calculus, Early Transcen-dental Functions, Third Edition. Houghton Mifflin Company, Boston (2003). ISBN0-618-22307-X.


J.2.3 Edwards and Penney

[24] C. H. Edwards, Jr., and D. E. Penney, Single Variable Calculus, Early Transcen-dentals, Sixth Edition. Prentice Hall, Englewood Cliffs, NJ (2002). ISBN 0-13-041407-7.

[25] C. H. Edwards, Jr., and D. E. Penney, Calculus with Analytic Geometry, EarlyTranscendentals Version, Fifth Edition. Prentice Hall, Englewood Cliffs, NJ (1997).ISBN 0-13-793076-3.

[26] C. H. Edwards, Jr., and D. E. Penney, Student Solutions Manual for Calculus withAnalytic Geometry, Early Transcendentals Version, Fifth Edition. Prentice Hall,Englewood Cliffs, NJ (1997). ISBN 0-13-079875-4.

[27] C. H. Edwards, Jr., and D. E. Penney, Single Variable Calculus with AnalyticGeometry, Early Transcendentals Version, Fifth Edition. Prentice Hall, EnglewoodCliffs, NJ (1997). ISBN 0-13-793092-5.

[28] C. H. Edwards, Jr., and D. E. Penney, Student Solutions Manual for Single VariableCalculus with Analytic Geometry, Early Transcendentals Version, Fifth Edition.Prentice Hall, Englewood Cliffs, NJ (1997). ISBN 0-13-095247-1.

J.2.4 Others, not “Early Transcendentals”

[29] G. H. Hardy, A Course of Pure Mathematics, 10th edition. Cambridge UniversityPress (1967).

[30] H. S. Hall, S. R. Knight, Elementary Trigonometry, Fourth Edition. Macmillanand Company, London (1905).

[31] S. L. Salas, E. Hille, G. J. Etgen, Calculus, One and Several Variables, 9th Edition.John Wiley & Sons, Inc. (2003). ISBN 0471-23120-7.

J.3 Other References

[32] D. Ebersole, D. Schattschneider, A. Sevilla, K. Somers, A Companion to Calculus.Brooks/Cole (1995). ISBN 0-534-26592-8.

[33] McGill Undergraduate Programs Calendar 2006/2007. Also accessible athttp://www.coursecalendar.mcgill.ca/ugcal2006-07

Documents

McGILL UNIVERSITY FACULTY OF SCIENCE DEPARTMENT OF MATHEMATICS AND STATISTICS MATH · PDF fileMcGILL UNIVERSITY FACULTY OF SCIENCE DEPARTMENT OF MATHEMATICS AND STATISTICS MATH 140