Solution Key_CK-12 Calculus Flexbook

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CK-12 Calculus, Solution Key

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Contents www.ck12.org

Contents

1 Functions, Limits, and Continuity, Solution Key 1

1.1 Equations and Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Relations and Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3 Models and Data. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.4 The Calculus. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.5 Finding Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.6 Evaluating Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

1.7 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.8 Infinite Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2 Derivatives, Solution Key 31

2.1 Tangent Lines and Rates of change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.2 The Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.3 Techniques of Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

2.4 Derivatives of Trigonometric Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

2.5 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

2.6 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

2.7 Linearization and Newtons Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

3 Applications of Derivatives, Solution Key 67

3.1 Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 683.2 Extrema and the Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

3.3 The First Derivative Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

3.4 The Second Derivative Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

3.5 Limits at Infinity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

3.6 Analyzing the Graph of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

3.7 Optimization. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

3.8 Approximation Errors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

4 Integration, Solution Key 129

4.1 Indefinite Integrals Calculus. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

4.2 The Initial Value Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

4.3 The Area Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

4.4 Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

4.5 Evaluating Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

4.6 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

4.7 Integration by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

4.8 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

5 Applications of Definite Integrals, Solution Key 164

5.1 Area Between Two Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

5.2 Volumes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

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5.3 The Length of a Plane Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184

5.4 Area of a Surface of a Revolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188

5.5 Applications from Physics, Engineering, and Statistics. . . . . . . . . . . . . . . . . . . . . . . 193

6 Transcendental Functions, Solution Key 198

6.1 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199

6.2 Exponential and Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203

6.3 Differentiation and Integration of Logarithmic and Exponential Functions . . . . . . . . . . . . 2066.4 Exponential Growth and Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

6.5 Derivatives and Integrals Involving Inverse Trigonometric Functions . . . . . . . . . . . . . . . 213

6.6 LHospitals Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217

7 Integration Techniques, Solution Key 219

7.1 Integration by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220

7.2 Integration by Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224

7.3 Integration by Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228

7.4 Trigonometric Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234

7.5 Trigonometric Substitutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239

7.6 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245

7.7 Ordinary Differential Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248

8 Infinite Series, Solution Key 254

8.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255

8.2 Infinite Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259

8.3 Series Without Negative Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261

8.4 Series With Odd or Even Negative Terms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262

8.5 Ratio Test, Root Test, and Summary of Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264

8.6 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267

8.7 Taylor and Maclaurin Series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271

8.8 Calculations with Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278

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www.ck12.org Chapter 1. Functions, Limits, and Continuity, Solution Key

CHAPTER 1 Functions, Limits, andContinuity, Solution Key

Chapter Outline

1.1 EQUATIONS ANDGRAPHS

1.2 RELATIONS ANDFUNCTIONS

1.3 MODELS AND DATA

1.4 THE C ALCULUS

1.5 FINDINGLIMITS

1.6 EVALUATINGLIMITS

1.7 CONTINUITY

1.8 INFINITELIMITS

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1.1. Equations and Graphs www.ck12.org

1.1 Equations and Graphs

1. Letx=1. Then find the correspondingy.

2x3y=52 (1)3y=5

23y=53y=523y=3

y= 1

(1,1)is one solution.

Letx=4. Then find the correspondingy.

2x3y=52 (4)3y=5

83y=53y= 3

y=1

(4,1)is another solution.

To find thexintercept, lety=0 and solve forx.

2x3y=52x3 (0) =5

2x=5

x=5

2

Thexintercept is

52 ,0

.

To find thexintercept, letx=0 and solve fory.

2x3y=52 (0)3y=5

3y=5y= 5

3

Theyintercept is

0,53

.

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The equation gives a linear relationship betweenxandy. Its graph can be sketched through any two solutions. There

is no symmetry.

2. Solve fory:

3x2y=5

y

= 3x2

+5

y=3x25

Ifx=1, theny=3 (1)25=35= 2. One solution is(1,2).Ifx= 1, theny=3 (1)25=35= 2. Another solution is(1,2).To find thexintercepts, lety=0.

y=3x250=3x255=3x2

5

3= x2

5

3= x

Thexintercepts are

5

3,0

.

To find theyintercept, letx=0.

y=3 (0)25y= 5

Theyintercept is(0,5).The graph is a parabola with a= 3,b=0, and c= 5. It is symmetric with respect to x = b2a =0, which is theyaxis.3. Use a graphing calculator. Enter the relationship on the Y=menu. Look at the table of points. There are manysolutions, such as (2,6) and (2,6). The xintercepts are(0,0) , (1,0) ,(1,0). The yintercept is(0,0). Byinspection, the graph is symmetric about the origin.

4. Use a graphing calculator. Enter the relationship on the Y=menu. Look at the table of points. There are manysolutions, such as (2,0) and (1,6). The xintercepts are (0,0) ,(3,0) ,(2,0). The yintercept is (0,0). Byinspection, the graph has no symmetry.

5. The best answer is b. Even though the values of both cars are falling, the value of the BMW is always greater

than that of the Chevy for any value oft.

6. Graph c is the best representation because you would expect a decline as soon as you bought the car and you

would expect that the value would decline more gradually after the initial drop.

7. a. Let represent the length of the pool and let w represent the width of the pool. Then A (w) = l w=(w + 25) w=w2 + 25w.

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1.1. Equations and Graphs www.ck12.org

b.

264=w2 + 25w

0=w2 + 25w + 264

0= (w + 33) (w8)0=w + 33 or 0=w8

33=w or 8=wThe width is 8 yards. The length is 8 + 25=33 yards.

8. The rate of change will be 8,50018,00020082004 =

9,5004

= 2,375. Since the rate of depreciation is constant, the formulafor the changing value of the car is linear. Because at time 0, the value of the car is $18,000, theyintercept of theformula is 18,000. The formula isy= 2,375x + 18,000.9. The formula for the changing value of the car is y=2,375x + 18,000. Whenx=7,y= 2,375 (7) + 18,000=16,625 + 18,000=$1,375.10. A linear model may not be the best function to model depreciation because the graph of the function decreases

as time increases; hence at some point the value will take on negative real number values, an impossible situation

for the value of real goods and products.

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1.2 Relations and Functions

1. Apply the vertical line test to the graph of the relationship shown. The graph gives a function. The domain is the

set of all real numbers, or(,). The range is {2 y 2}.2. A vertical line can cross the graph in more than one place. The graph is not a function.

3. The function is a rational function. Set the denominator=0 and solve for x.

x21=0(x1) (x + 1) =0

x=1 or x= 1

The domain is {x = 1,1}. Graph the function on your graphing calculator.

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1.2. Relations and Functions www.ck12.org

7. This function is the basic function f(x) = 1x

shifted 2 units right and 3 units up and then flipped.

8. f(x) = x2 + 3= (x + 2) + 3. This function is the basic function f(x) = x. Note that there is a

negative sign in front of thex.The new function becomes f(x) =xand the graph of this function is a reflection

of f(x) =

xaround theyaxis. Then the function is shifted 2 units left and 3 units up. It is then flipped upsidedown.

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9. (f o g) (x) = f(g (x)) = 3x+ 2; (go f) (x) =g (f(x)) = 3x + 210. (f o g) (x) = f(g (x)) =

x2

=x; (go f) (x) =g (f(x)) =

x2 =x

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1.3. Models and Data www.ck12.org

1.3 Models and Data

1. a.

b. Based on the plot, I will use linear regression.

c. Use the graphing calculator to find the line of best fit.

Rounding to the nearest ten-thousandth, the line of best fit has the equationy=3.1335x + 0.3296.

d. The equation is a model of the formula for the circumference of a circle: C= d. The slopem is an estimate of. They

interceptb should be 0 but due to measurement errors, it is not 0.

2. a.

b. Use the graphing calculator to find the line of best fit.

Rounding, the line of best fit has the equationy=0.120547x39.0465.d. Letx=700 (number of thousands in 700,000 ). Then

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y=0.120547x39.0465=0.125047 (700)39.0465=45.34

Since we cant have a fractional part of the number of manatees, round up to 46. About 46 manatees will be killed

in the year 2000.

3. a.

b. Use the graphing calculator to find the line of best fit.

Rounding, the line of best fit has the equationy=2.0132x0.2624.The slopemof the line of best fit is close to 2. The model is certainly shows that twice the measurement of the wrist,

x, is the measurement of the neck,y.

c. Letx=52. Then

y=2.0132 (52)0.2624=0.125047 (700)39.0465=104.12

4. a.

b. Based on the sketch, a quadratic model could be used.

c. Use the graphing calculator to find the quadratic model.

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The model isy=0.0194x20.29696x + 37.7636.d. The model might dip and then grow because the first wave of women into the workforce tended to take whatever

jobs they could find without regard to salary.

5. Lety=100. Then solve the equation below forx:

100=0.0194x20.29696x + 37.76360=0.0194x20.29696x62.2364

x=b b24ac

2a

x=0.29696

(0.29696)24 (0.0194) (62.2364)

2 (0.0194)

x=0.29696 4.9177298816

0.0388

x=0.29696 + 2.2175944180

0.0388 or x=

0.296962.21759441800.0388

Because we need a positive solution (for the number of years ), the solution isx= 0.29696+2.21759441800.0388 =64.808.

Based on the model, women will make as much as men 64 years after 1960 or 1960 +64 = 2024. It could be realistic.

6. a.

b. Based on the sketch, use a cubic model.

c.

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The cubic model is y=0.00031x30.01614x2 + 0.257394x + 1.19651273.d. The year 1975 is year 1. The year 2012 is year 37 in the data. Let x=37.

Then

y=0.00031 (37)3

0.01614 (37)2 + 0.257394 (37) + 1.19651273

=4.33

The price of gas could be $4.33 in 2012.

7. Use the graphing calculator to find the linear model.

The best linear model is y=0.051627x + 1.54389.

For the year 2012, let x=37 again.

y=0.051627x + 1.54389

=0.051627 (37) + 1.54389

=$3.45

This models predicts the price per gallon of gas to be $3.45. It is hard to say which model works best.

8. a. Use the graphing calculator to make a scatter plot and find the exponential model.

Rounding, the exponential model is y=1,000 (1.1275)x.b. The earnings triple wheny=$3,000.

Solve the equation 3,000=1,000 (1.1275)x forx.

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3,000

1,000=

1,000 (1.1275)x1,000

3= (1.1275)x

ln (3) =ln (1.1275)x

ln (3) =xln (1.1275)

ln (3)ln (1.1275)

=xln (1.1275)

ln (1.1275)

9.15=x

It will take a little more than 9 years for the earnings to triple.

9. a. y =3,000 (1.1275)x

b. The earnings triple wheny=$9,000. Solve the equation 9,000=3,000 (1.1275)x.

9,000

3,000 =

3,000

(1.1275)x

3,000

3= (1.1275)x

9.15=x

The equation reduced to the same equation as in Exericse 8. The solution is the same. It will take a little more than

9 years for the earnings to triple.

10. a. Use the graphing calculator to make the scatter plot.

b. Use the graphing calculator to find the sine model of the data.

The sine model isy=29.452706sin (0.520555x1.62865) + 51.827236.

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1.4 The Calculus

1. a. The formula for slope of the tangent line for f(x) =x2 atx=3 ism= x232

x3 = x29x3 =

(x3)(x+3)x3 =x + 3.

Px,x2

m=x + 1

P (2.9,8.41) 2.9 + 3=5.9

P (2.95,8.7025) 2.95 + 3=5.95

P (2.975,8.850625) 2.975 + 3=5.975

P (2.995,8.970025) 2.995 + 3=5.995

P (2.999,8.994001) 2.999 + 3=5.999

b. The sequence of slopes are approaching m=6.2. a. Draw tangent lines on the graph of f(x) =x2. The slope of the tangent line is negative forx


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1.4. The Calculus www.ck12.org

The tangent lines have slopes of 0 whenx=0.57 andx= 0.57.c. Using the graph, the tangent line appear to have positive slope forx0.57.d. Using the graph, the tangent line appear to have negative slope for 0.57


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Each width is equal to 12

. Call the rectanglesR1to R4.

The areas are:

R1=1

2 f

3

2

=

1

2 9

4=

9

8

R2=1

2 f(2) = 1

24=2

R3=1

2 f

5

2

=

1

2 25

4 =

25

8

R4= 12 f(3) = 1

29= 9

2

The approximation under the curve isR1+R2+R3+R4= 98+ 2 +

258 +

92 = 10.75.

b. Divide the area under the curve from x=1 tox=3 in eight equal rectangles.

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The areas are:

R1=1

4 f

5

4

=

1

4 25

16=

25

64

R2=1

4 f

3

2

=

1

4 9

4=

9

16

R3=1

4 f

7

4

=

1

4 49

16=

49

64

R4=1

4 f(2) = 1

44=1

R5=1

4 f

9

4 =1

4 81

16=

81

64

R6=1

4 f

5

2

=

1

4 25

4 =

25

16

R7=1

4 f

11

4

=

1

4 121

16 =

121

64

R8=1

4 f(3) = 1

49= 9

4

The approximation of the area under the curve isR1+R2+R3+R4+R5+R6+R7+R8=9.6875.

c.

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The midpoint ofR1is 1+ 32

2 =52

2 = 54

. The midpoint ofR2is32 +2

2 =72

2 = 74

. The midpoint ofR3is 2+ 52

2 =92

2 = 94

. The

midpoint ofR4is52 +3

2 =

112

2 = 11

4.

The areas are:

R1=1

2 f

5

4

=

1

2 25

16=

25

32

R2=1

2 f

7

4

=

1

2 49

16=

49

32

R3=1

2 f9

4 = 1

2 81

16

=81

32

R4=1

2 f

11

4

=

1

2 121

16 =

121

32

The approximation of the area under the curve isR1+R2+R3+R4= 2532+

4932+

8132+

12132 =8.625.

6. a.

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The areas are:

R1=1

4 f

1

4

=

1

4

14

3+ 4

1

4

=

1

4 1

64+ 1

=

63

256

R2=1

4 f

1

2

=

1

4

12

3+ 4

1

2

=

1

41

8+ 2

=

15

32

R3=1

4 f

3

4 =1

4

3

43

+ 43

4 =1

4

37

64+ 3 =

155

256

R4=1

4 f(1) = 1

4

(1)3 + 4 (1)

=1

4 (1 + 4) = 3

4

The approximation of the area under the curve isR1+R2+R3+R4= 63256+

1532+

155256+

34=2.07.

b. The area from x= 1 to x=0 is below thexaxis. We are not finding area under a curve but the are between thecurve and thexaxis. The area fromx = 1 tox =0 is below the xaxis is symmetric to the area under the curvefromx=0 tox=1.

7. The length of the interval is 40=4. Divide 4 by 6 to get that the length of each sub-interval is of length 46 = 23 .

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Call the rectanglesR1to R6.

The areas are:

R1=1

6 f

2

3

=

1

6

2

3=0.136

R2=

1

6 f43 = 16 43 =0.192R3=

1

6 f(2) = 1

6

2=0.236

R4=1

6 f

8

3

=

1

6

8

3=0.272

R3=1

6 f

10

3

=

1

6

10

3 =0.304

R3=1

6 f(4) = 1

62=0.333

The approximation of the area under the curve is R1+R2+R3+R4+R5+R6= 0.136 + 0.192 + 0.236 + 0.272 +0.304 + 0.333=1.473.

8. The average velocity of a falling object from t= ato t=bis given by s(b)s(a)ba =

4.9b24.9a2ba . Simplifying, we get

the average velocity is 4.9(b2a2)

ba = 4.9(ba)(b+a)

ba =4.9 (b + a).

Make a sequence of values ofx that get closer tob=4 and find the average velocity between eachx and 4.

x average velocity =4.9 (4 +x)

3.9 4.9 (4 + 3.9) =4.9 (7.9) =38.71

3.95 4.9 (4 + 3.95) =4.9 (7.95) =38.955

3.99 4.9 (4 + 3.99) =4.9 (7.99) =39.151

3.999 4.9 (4 + 3.999) =4.9 (7.999) =39.1951

3.9999 4.9 (4 + 3.9999) =4.9 (7.9999) =39.19951

The velocity of the ball after 4 seconds is 39.2 m/sec.

19


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limx4

3x33x =180b.

limx4

3x33x

= 180

c.

limx0

3x33x =0d.

The function values are the same as the limits because the function is defined at those values.

4. a. limx3 f(x) =1.5 because f(3) =1.5.

b. limx2 f(x) =0 because f(2) =0.

c. limx1 f(x) =2 because f(1) =2.

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1.5. Finding Limits www.ck12.org

d. limx4 f(x)does not exist because the right-hand limit and the left-hand limit are not the same.

5. a. limx2 f(x) =0 because f(2) =0.

b. limx0 f(x)because the function is not defined at x=0.

c. limx4 f(x)is a number close to 1 but less than 1 because of the horizontal asymptote ofy=1.

d. limx50 f(x)is a number close to 1 but less than 1 because of the horizontal asymptote ofy=1.

6. Use a graphing calculator to make a table of values to find the limit.

The limit exists and limx2x2 + 3

=7

7.

The limit exists and limx1 x+1x21 = 0.5. (Note: you will get an error if you try to find f(1)directly. The limitof values of both sides are approaching 0.5, but the function is not defined at x= 1.8.

The limit exists and limx22x + 5=1.

9.

The limit exists and limx2x2 + 3x

=28.

10.

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The limit does not exist. Note that there is a break in the graph whenx= 1.

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1.6. Evaluating Limits www.ck12.org

1.6 Evaluating Limits

1. limx2x23x + 4

=223 (2) + 4=46 + 4=2

2. limx4 x2

16x4 =limx4 (x4)(x+4)x4 =limx4(x + 4) =4 + 4=8

3. limx4

x2x4 =limx4

x2

x2

x+2 =limx4 1

x+2= 1

4+2= 12+2 =

14

4. The rational function x2x+1 cannot be simplified. Use a graphing calculator to view the graph and you will see that

limx1 x2x+1 does not exist.

5. limx1 10x23x+1 = 10(1)2

3(1)+1 =1023+1 =

122 =6

6.

limx

1

x + 32

x1 =lim

x

1

x + 32

x1

x + 3 + 2

x + 3 + 2=lim

x1(x + 3)2x + 3 + 2x + 34

(x1)x + 3 + 2=lim

x1x + 34

(x1)x + 3 + 2 = limx1 x1(x1)x + 3 + 2=lim

x11

x + 3 + 2

= 11 + 3 + 2

= 1

4 + 2=

1

2 + 2=

1

4

7.

limx5

x225x3125 = limx5

(x5) (x + 5)(x5) (x2 + 5x + 25)

=limx5

x + 5

x2 + 5x + 25=

5 + 5

52 + 5 (5) + 25=

10

75=

2

15

8. limx1(g o f(x)) =limx1 g (f(x)) =limx1 g

1x+1

The function f(x) = 1

x+1 is not defined atx=1, which meansg

1x+1

is not defined there. Thus, limx1 g (f(x))

does not exist.

9. Use the Squeeze Theorem.

limx5

(5x11) limx5

f(x) limx5

x24x + 9

5 (5)11 limx5

f(x) 524 (5) + 914 lim

x5f(x) 14

Thus, limx5 f(x) =14, too.

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10. The function sin

1x

is bounded below by x4 and bounded above by x4.

First, use the Squeeze Theorem:

limx0

x4 limx0

sin

1

x

lim

x0x4

04

limx0 sin1x 040 lim

x0sin

1

x

0

Thus, limx0 sin

1x

=0. Then limx0x4 sin

1x

=limx0x4 limx0 sin

1x

=00=0.

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1.7. Continuity www.ck12.org

1.7 Continuity

1.

The function has a break atx= 1. It is continuous everywhere except for x= 1.2.

The function has a break atx= 2. It is continuous everywhere except for x= 2.3.

limx0+

x

1 + x1 = limx0+

x1 + x1

1 +

x + 1

1 + x + 1

= limx0+

x

1 +

x + 1

1 +

x1

= limx0+

x

1 +

x + 1

x

= limx0+

1 +

x + 1

= 1 +

10 + 1

=1 + 1=2

4.

limx2

x38|x2|(x2)= limx2

x38(x2) (x2)

= limx2

(x2)x2 + 2x + 4(x2) (x2)

= limx2

x2 + 2x + 4

x + 2

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x y= x38|x2|(x2)

1.99 1194.08

1.999 11,994

1.9999 119,994

limx2

x38|x2|(x2)=

5. Forx>1,|x1|=x1. Then limx1+ 2x|x1|(x1) =limx1+ 2x(x1)

(x1) =limx1+2x=2.

6. Forx>2, |x + 2|=x + 2. Then

limx2+ |x + 2

|+x + 2

|x + 2|x2 = limx2+(x + 2) +x + 2

(x + 2)x2

7. f(2) = (2)3 + 2 (2)2 (2) + 1= 8 + 8 + 2 + 1=3f(3) = (3)3 + 2 (3)2 (3) + 1= 27 + 18 + 3 + 1= 5By the Intermediate Value Theorem, there is an xvaluec with f(c) =0.8. f(9) =

9 3

91= 0.08

f(10) =

10 3

101=0.008By the Intermediate Value Theorem, there is an xvaluec with f(c) =0.

9. The valuex =a is considered a maximum because it is a high point in the graph and the graph turns back down.Because it is not the highest point in the interval,x=ais called a relative maximum. The value x=cis a maximumof the interval and is the absolute maximum because it is the highest maximum The value x=bis a minimum of theinterval and is, in fact, an absolute minimum of the interval because it is the lowest value of the interval. The value

x=dis neither a maximum nor a minimum.

10. [Note: I think that this question should be replaced with an easier question - see comments on pdf file.]

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1.8. Infinite Limits www.ck12.org

1.8 Infinite Limits

1.

x f(x) = (x + 2)2

(x2)213.01 1248.76

3.001 12498.75

3.0001 124998.75

limx

3+

(x + 2)2

(x2)2

1= +

2.

limx

(x + 2)2

(x2)21 = limxx2 + 4x + 4

x24x + 41

= limx

x2 + 4x + 4

x24x + 3

= limx

x2

x2+ 4x

x2+ 4

x2

x2

x2 4x

x2+ 3

x2

= limx1 + 4

x+ 4

x2

1 4x

+ 3x2

=1 + 0 + 0

10 + 0=1

3.

x f(x) = (x + 2)2

(x2)21

1.01 455.281.001 4505.251.0001 45,005.25

limx1+

(x + 2)2

(x2)21 =

4. limx 2x1x+1 =limx2xx 1

xxx

+ 1x

=limx2 1

x

1+ 1x

= 201+0 = 2

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5. limx x5+3x4+1

x31 =limxx5

x5 3x4

x5 + 1

x5

x3

x5+ 1

x5

=limx1+ 3

x4+ 1

x51

x2+ 1

x5

The limit of the denominator is 0, so this is an indeterminate form. We can argue the limit in this way: the numerator

approaches 1 asx goes to . The denominator approaches 0 and is a positive quantity because 1x2> 1

x5. The ratio

1+ 3x4

+ 1x5

1

x2+ 1

x5

is a positive quantity that increases without bound because 1 divided by a very small positive number is a

large positive number.

Thus, limx x5+3x4+1x31 = .

6.

limx

3x42x2 + 3x + 12x42x2 = limx

3x4

x4 2x2

x4 + 3x

x4+ 1

x4

2x4

x4 2x2

x4

= limx

3 2x2

+ 3x3

+ 1x4

2 + 2x2

=30 + 0 + 0

2 + 0 =

3

2

7.

limx

2x3x + 3x52x3 + 2x3 = limx

2x3

x5 x

x5+ 3

x5

x5

x5 2x3

x5 + 2x

x5 3

x5

= limx

2x2 1

x4+ 3

x5

1 2x2

+ 2x4 3

x5

= 00 + 010 + 00 =0

8. Zero: Set numerator=0.

(x + 4)2 =0

x + 4=0

x= 4

Vertical asymptotes: Set denominator=0.

(x4)21=0x28x + 161=0

x28x + 15=0(x5) (x3) =0

x=5 or x=3

The vertical asymptotes arex=5 orx=3.

End behavior:

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1.8. Infinite Limits www.ck12.org

limx

(x + 4)2

(x4)21 = limxx2 + 8x + 16

x28x + 15

= limx

x2

x2+ 8x

x2+ 16

x2

x2

x2 8x

x2+ 15

x2

= limx 1 +

8

x+16

x2

1 8x

+ 15x2

= 1 + 0 + 01 + 0 + 0

= 1

limx

(x + 4)2

(x4)21 = limx

x2 + 8x + 16

x28x + 15

= limx

x2

x2+ 8x

x2+ 16

x2

x2

x2 8x

x2+ 15

x2

= limx

1 + 8x

+ 16x2

1

8

x

+ 15

x2

=1 + 0 + 0

1 + 0 + 0=1

9. Zero:

There are no vertical asymptotes.

limx

3x3x2 + 2x + 2 = lim

x3x3x2 + 2x + 2 =

10. 2x28

x+2

= 2(x24)

x+2

= 2(x2)(x+2)x+2

=2 (x

2)

Zero:

2 (x2) =0x2=0

x=2

There are no vertical asymptotes. There is a discontinuity at x= 2.

limx

2x28x + 2

= limx

[2 (x

2)] =

limx

2x28x + 2

= limx [2 (x2)] =

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www.ck12.org Chapter 2. Derivatives, Solution Key

CHAPTER 2 Derivatives, Solution KeyChapter Outline

2.1 TANGENTL INES ANDRATES OF CHANGE

2.2 THE D ERIVATIVE

2.3 TECHNIQUES OF DIFFERENTIATION

2.4 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS

2.5 THE C HAINR UL E

2.6 IMPLICIT DIFFERENTIATION

2.7 LINEARIZATION ANDNEWTONS METHOD

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2.1. Tangent Lines and Rates of change www.ck12.org

2.1 Tangent Lines and Rates of change

1. a. Lety= f(x). Forx0=3, f(x) = 12x2 = 12(3)

2 = 92 and forx0=4, f(4) = 12x2 = 12(4)

2 = 162 =8.

The average rate of change ofy with respect to x over[3,4]is msec= f(x1)f(x0)

x1x0 = f(3)f(1)

31 = 8 92

43 =72

1 = 72

.

b. First, find f (x).

f (x0) = limh0

f(x0+ h) f(x0)h

=limh0

12(x0+ h)

2 12x0

h

=limh0

12x20+ 2hx0+ h

2 12x0

h

=limh02hx0+ h

2

h

=limh0

h (2x0+ h)

h

=limh0

(2x0+ h)

=2x0

The instantaneous rate of change ofy= f(x)with respect tox at x0=3 is f (3) =2x0=2 (3) =6.

c. The slope of the tangent line at x1=4 is f (4) =2x0=2 (4) =8.

d. The slope of the secant line between x0=3 andx1=4 is the same as the average rate of change ofy with respect

tox over the interval[3,4]. Thenmsec= f(x1)f(x0)

x1x0 = 8 92

43 =72

1 = 72

.

e.

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2. a. Lety= f(x). Forx0=2, f(2) = 12

and forx0=3, f(3) = 13

.

The average rate of change ofy with respect to x over[2,3]is msec= f(x1)f(x0)

x1x0 =13 1232 =

26 36

1 = 16 .b. First, find f (x).

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3. The slope of the tangent line at a point x is f (x).

f (x0) = limh0

f(x0+ h) f(x0)h

=limh0

(x0+ h)2 + 1x20+ 1

h

=limh0

x20+ 2x20h + h

2 + 1x201h

=limh0

2x20h + h2

h

=limh0

h2x20+ hh

=limh0

2x20+ h

=2x20

The slope of the tangent line at x0=6 is f (6) =2x20=2 (6) =12.

4. a. Lety= f(x). Forx0=1, f(1) = 1

1=1 and forx0=4, f(3) =

13

=

3

3 .

The average rate of change ofy with respect tox over [1,3] is msec= f(x1)f(x0)

x1x0 = f(3)f(1)

31 =

33 1

2 =

3

3 1

12

=

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3

6 12 .b.

f (x0) = limh0

f(x0+ h) f(x0)h

=limh

0

1x0+h

1x0

h

=limh0

1x0+h

1x0

h

x0+ h

x0

x0+ h

x0

=limh0

(

x0+h)(

x0)(

x0+h) (

x0+h)(

x0)

x0

h

x0+ h

x0

=limh0

x0

x0+ h

h

x0+ h

x0

=limh0

x0

x0+ h

h

x0+ h

x0

x0+

x0+ h

x0+

x0+ h=lim

h0x0 (x0+ h)

h

x0+ h

x0

x0+

x0+ h

=limh0

hh

x0+ h

x0

x0+

x0+ h

=limh0

1x0+ h

x0

x0+

x0+ h

= 1x0+ 0

x0

x0+

x0+ 0

= 1

x0

x0

x0+

x0= 1

x0

x0

2

x0

= 12

x03

c. f (1) = 12

13 = 12

5. a. h (35) =4.9 (3.5)2 =6002.5 meters

b.

v= h (35)h (0)350

=4.9 (35)24.9 (0)2

35

=4.9 (35)2

35

=4.9 (35)

=171.5 m/sec

c. Seth (t) =200 and solve for the numbers of minutestthat pass for the rocket to travel 200 meters.

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200=4.9t2

200

4.9 = t2

40.82= t2

6.39= t2

Next, find the average velocity:

v=200h (0)

6.390=

2004.9 (0)26.39

= 200

6.39

=31.3 m/sec

d. Writeh (t) = f(t)do avoid confusingh (t)withh in the formula for the instantaneous rate of change.

f (t0) = limh0

f(t0+ h) f(t0)h

=limh0

4.9 (t0+ h)24.9t20

h

=lim

h0

4.9

t20+ 2ht0+ h

2

4.9t20

h

=limh0

4.9t20+ 9.8hx0+ 4.9h24.9t20

h

=limh0

9.8ht0+ 4.9h

h

=limh0

4.9h (2x0+ 1)

h

=limh0

4.9 (2t0+ 1)

=9.8t0

Rewrite the formula with h (t): h (t) =9.8t0. Thenh (35) =9.8 (35) =343 m/sec.6. a.

v= (2)(0)

20=

9.9 (2)39.9 (0)320

=79.2

2

=39.6 nanometers/nanosecond

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b.

(t0) = lim

h0(t0+ h)(t0)

h

=limh0

9.9 (t0+ h)39.9t30

h

=limh0

9.9t30+ 3t

20 h + 3t0h

2 + h39.9t30

h

=limh0

9.9t30+ 29.7t20 h + 29.7t0h

2 + 9.9h39.9t30h

=limh0

29.7t20 h + 29.7t0h2 + 9.9h3

h

=limh0

9.9h

3t20+ 3t0h + h2

h

=limh0

9.9

3t20+ 3t0h + h

=29.7t20

The instantaneous velocity att0=2 is (2) =29.7 (2)2 =118.8 nanometers/nanosecond.

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2.2 The Derivative

1.

f (x) = limh0

f(x + h) f(x)h

=limh

0

6 (x + h)26x2h

=limh0

6x2 + 2xh + h2

6x2h

=limh0

6x2 + 12xh + 6h26x2h

=limh0

12xh + 6h2

h

=limh0

6h (2x + h)

h

=limh0

6 (2x + h)

=6

(2x

)=12x

Whenx0=3, y= f(3) =6 (3)2 =54. The slope of the tangent line is m= f (3) =12 (3) =36. Then the equation

of the tangent line atx=3 is

yy0=m (xx0)y54=36 (x3)y54=36x108

y=36x54

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2.2. The Derivative www.ck12.org

2.

f (x) = limh0

f(x + h) f(x)h

=limh0

x + h + 2 x + 2h

=limh0

x + h + 2 x + 2

h

x + h + 2 +

x + 2

x + h + 2 +

x + 2

=lim

h0(x + h + 2) (x + 2)

h

x + h + 2 +

x + 2

=limh0

x + h + 2x + 2h

x + h + 2 +

x + 2

=limh0

h

hx + h + 2 + x + 2=lim

h01

x + h + 2 +

x + 2

= 1

x + 2 +

x + 2

= 1

2

x + 2

When x0= 8, y= f(8) = 8 + 2= 10. The slope of the tangent line is m = f (8) = 12

8 + 2= 1

2

10. Then

the equation of the tangent line atx=3 is

yy0= m (xx0)y

10=

1

2

10

(x8)

y

10= 1

2

10x 4

10

y= 1

2

10x 4

10+

10

3.

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f (x) = limh0

f(x + h) f(x)h

=limh0

3 (x + h)32 3x32h

=limh0

3x3 + 3x2h + 3xh2 + h3

23x3 + 2h

=limh0

3x3 + 9x2h + 9xh2 + 3h33x3h

=limh0

9x2h + 9xh2 + 3h3

h

=limh03h3x2 + 3xh + h2

h

=limh0

3

3x2 + 3xh + h2

=9x2

Whenx0= 1, y= f(1) =3 (1)3

2= 5. The slope of the tangent line is m= f (1) =9 (1)2

=9. Thenthe equation of the tangent line atx=3 is

yy0=m (xx0)y

(

5) =9 (x

(

1))

y + 5=9x + 6

y=9x + 4

4.

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f (x) = limh0

f(x + h) f(x)h

=limh0

1x+h+2 1x+2

h

=limh0

1(x+h+2)

(x+2)(x+2) 1(x+2) (x+h+2)(x+h+2)

h

=limh0

(x+2)(x+h+2)(x+h+2)(x+2)

h

=limh0

x+2xh2(x+h+2)(x+2)

h

=limh0

h(x+h+2)(x+2)

h

=limh0

h(x + h + 2) (x + 2) h

=limh0

1(x + h + 2) (x + 2)

= 1(x + 2) (x + 2)

= 1(x + 2)2

Whenx0= 1, y= f(1) = 1(x+2) = 11+2= 1. The slope of the tangent line ism= f (1) = 1(1+2)2 = 1

12=

1. Then the equation of the tangent line atx= 1 is

yy0=m (xx0)y1= 1 (x (1))y1= x1

y= x

5.

f (x) = limh0

f(x + h) f(x)h

=limh0

a (x + h)2b ax2bh

=limh0

ax2 + 2x2h + h2bax

2 + b

h

=limh0

ax2 + 2axh + ah2ax2h

=limh0

2axh + ah2

h

=limh0

ah

2x + ah2

h

=limh0

a

2x + ah2

=2ax

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When x0= b, y= f(b) = ab2 b. The slope of the tangent line is m= f (b) =2ab. Then the equation of the

tangent line atx= 1 is

yy0=m (xx0)y ab2b =2ab (xb)

yab2 + b=2abx2ab2y=2abxab2b

6. To find the derivative of f(x) =x13 , we will use the relationship below. It shows that the difference of two numbers

as a difference of cubes and then factor the difference of two cubes.

ab=

a13

3

b13

3=

a13 b 13

a

13

2+ a

13 b

13 +

b

13

2

=

a13 b 13

a

23 + a

13 b

13 + b

23

f (x) = lim

h0f(x + h) f(x)

h

=limh0

(x + h)13 (x) 13h

=limh0

(x + h) 13 (x) 13

h

(x + h) 23 + (x + h) 13 x 13+x 23

(x + h)

23 + (x + h)

13 x

13+x

23

=lim

h0x + hx

h

(x + h)23 + (x + h)

13 x

13+x

23

=lim

h0h

h

(x + h)23 + (x + h)

13 x

13+x

23

=lim

h01

h(x + h)23 + (x + h)

13 x

13+x

23

= 1

x23+x

13x

13+x

23

= 1

3x23

whenx0=1, y= f(1) = (1)13 =1. The slope of the tangent line is m= f (1) = 1

3(1)23

= 13

.

Then the equation of the tangent line atx=1 is

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yy0=m (xx0)y1= 1

3(x1)

y1= 13

x 13

y= 13

x 13

+ 1

y=1

3x +

2

3

7. Use Exercise 5. For f(x) =5x22, a=5 andb= 2 and f (x) =2ax=25x=10x. Thus, dydx

x=1

=10 (1) =

10.

8. The function f(x) = 3

xis continuous atx=0 because it satisfies these three conditions:

1. f(0) = 30=0

2. limx0+ 3

x=0 and limx0 3

x=0. Thus, limx0 3

x=0.3. limx0 3

x= f(0) =0.

f(x)is not differentiable at x=0 because f (x) = 13x

23

is not defined forx=0.

9. The function is continuous atx=1 because it satisfies these three conditions:

1. f(1) =12 + 1=22. limx1 f(x) =limx1

x2 + 1

=1 + 1=2 and limx1+ f(x) =limx1+2x=2. Thus, limx1 f(x) =2.

3. limx1 f(x) = f(1) =2.

Forx =1, f(x) = x2 + 1, which gives a =1 and b =1. By Exercise 5, f (x) = 2abxfor x 1. Thus, f (x) = 2xand is defined atx=1.

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10. Using the property that f(x +y) = f(x) +f(y) + 3xy,

f(x) = f(x + 0) = f(x) +f(0) + 3x (0)

f(x) = f(x + 0) = f(x) +f(0) + 0

f(x) = f(x) +f(0)

f(x) f(x) = f(0)0= f(0)

Because fis differentiable, then we can find for all x:

f(x) = lim

h0f(x + h)

f(x)

h

=limh0

f(x) +f(h) + 3xh f(x)h

=limh0

f(h) + 3xh

h

=limh0

f(h)

h+lim

h03xh

h

=4 +limh0

3x

=4 + 3x

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2.3. Techniques of Differentiation www.ck12.org

2.3 Techniques of Differentiation

1.

y=5x7

dy

dx=5

d

dx

x7

=5

7x71

=35x6

2.

y=1

2

x32x2 + 1

dy

dx=

1

2

d

dx

x32x2 + 1

=1

2

d

dx

x3

+ d

dx

2x2+ ddx

[1]

=1

2

3x3122x21+ 0

=1

2 3x24x

=3

2x22x

3.

y=

2x3 12

x2 + 2x +

2

dy

dx=

d

dx

2x3 12

x2 + 2x +

2=

d

dx

2x3

+

d

dx

1

2x3

+ d

dx[2x] +

d

dx

2

=

2d

dx

x3 1

2

d

dx

1

2x2

+ 2d

dx[x] +

d

dx

2

=

2

3x31 1

2

2x21

+ 2 [1] + 0

=3

2x2 22

x + 2

=3

2x2

2x + 2

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8.

y=

x + 1

x

y=

x12

+

1

x 12

y=x12 +x

12

dy

dx=

d

dx

x

12 +x

12

=

d

dx

x

12

+

d

dx

x

12

=

1

2x

12 22

+

1

2xx

12 22

=1

2x

12 1

2x

32

= 1

2x12

1

2x32

= 1

2

x 1

2 2

x3

9.

y= 3

x + 3

dy

dx=

d

dx

3x + 3

=(

x + 3) ddx

[3]3 ddx

[

x + 3]

(

x + 3)2

=(

x + 3)03 ddx

x

12+ 3

(

x + 3)2

=03 1

2x

12 22+ 0

(

x + 3)2

=32x

12

(

x + 3)2

= 3

2x12(

x + 3)2

= 3

2

x (

x + 3)2

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10.

y=4x + 1

x29dy

dx=

d

dx 4x + 1

x2

9

=

x29 ddx[4x + 1] (4x + 1) ddx x29

(x29)2

=

x29 [4] (4x + 1) [2x]

(x29)2

=

4x236 8x2 + 2x

(x29)2

=4x2368x22x

(x29)2

=4x22x36

(x29)2

11.

F= GmM

r2

dF

dr=

d

dr

G

mM

r2

= ddr

GmMr2

= 2GmMr21= 2GmMr3

=2GmM

r3

12.

d

d

0+

3

30

=

(30) dd0+

30+3 dd[30]

(30)2

=(30)

0+ 3

2 0+30

(30)2

=(30)

0+ 3

2

(30)2

=0+ 3

2

30

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13.

y= 2

x3

=2x3

dy

dx=

d

dx 2x3

=2 (3)x

3

1

= 6x4d2y

dx2=

d

dx

6x4= 6 (4)x41=24x5

d3y

dx3=

d

dx

24x5

=24 (5)x51=

120x6

= 120x6

d3y

dx3

x=1

= 12016

= 120

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2.4 Derivatives of Trigonometric Functions

1.

y=x sinx + 2

y=xd

dx[sinx] + sinx

d

dx[x] +

d

dx[2]

=x cosx + sinx1 + 0=x cosx + sinx

2.

y=x2 cosx

x tanx

1

y=x2 ddx

[cosx] + cosxd

dx

x2x d

dx[tanx] + tanx

d

dx[x]

+ d

dx[1]

=x2 (sinx) + cosx (2x) x sec2x + tanx (1)0= x2 sinx + 2x cosxx sec2x tanx

3.

y=sin2x

=sinx sinx

y=sinx ddx

[sinx] + sinx ddx

[sinx]

=sinx cosx + sinx cosx

=2sinx cosx

4.

y=sinx1sinx + 1

y=(sinx + 1) d

dx(sinx1) (sinx1) d

dx(sinx + 1)

(sinx + 1)2

=(sinx + 1) (cosx) (sinx1) (cosx)

(sinx + 1)2

=sinx cosx + cosx (sinx cosx cosx)

(sinx + 1)2

=sinx cosx + cosx sinx cosx + cosx

(sinx + 1)2

= 2cosx

(sinx + 1)2

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2.4. Derivatives of Trigonometric Functions www.ck12.org

5.

y=cosx + sinx

cosx sinxy=

(cosx sinx) ddx

(cosx + sinx) (cosx + sinx) ddx

(cosx sinx)(cosx sinx)2

=(cosx sinx) ( sinx + cosx) [(cosx + sinx) ( sinxcosx)]

(cosx sinx)2

=cosx sinx + cos2x + sin2xcosx sinx cosx sinx cos2x sin2x cosx sinx

(cosx sinx)2

=2cossinx + 12cosx sinxcos2x sin2x

(cosx sinx)2

=2cosx sinx + 1 + 2cosx sinx + cos2x + sin2x

(cosx sinx)2

=2cosx sinx + 1 + 2cosx sinx + 1

(cosx sin)2

=

2

(cosx sinx)2

= 2

cos2x2cosx sinx + sin2x=

2

12cosx sinx6.

y=

x

tanx+ 2

= x

12

tanx

+ 2

y=(tanx) d

dx

x

12

x12

ddx

(tanx)

(tanx)2 + 0

=(tanx)

12x

12

x12

sec2x

tan2x

=(tanx)

12x

12

tan2x

x12

sec2x

tan2x

= 1

2

x tanx

x

cos2x sin2xcos2x

= cotx

2

x x

sin2x

= cotx

2

x x csc2x

7.

y=cscx sinx +x

=1 +x

y=0 + 1=1

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8.

y=secx

cscx

= sinx

cosx

=tanx

y=sec2x

9.

y=cscx

y= cscx cotxy=cotx

d

dx(cscx) (cscx) d

dx(cotx)

=cotx (cscx cotx) + cscxcsc2xy

6

=cot2

6

csc

6

csc3

6

=32 ( (2)) (2)3

=3 (2) (8)=6 + 8

=14

10.

d

dx[cosx] = lim

h0cos (x + h) cosx

h

=lim

h0

cosx cos h sinx sin hcosxh

=limh0

cosx cos hcosx

h sinx sin h

h

=limh0

cosx (cos h1)

h sinx sin h

h

=cosx limh0

(cos h1)

h

sinx lim

h0

sin h

h

=cosx0 sinx1= sinx

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4.

f(x) =sin3x

f (x) =3sin31x ddx

(sinx)

=3sin2x (cosx)=3sin2x cosx

5.

f(x) =sinx3

f (x) =cosx3 d

dx

x3

=cosx33x2

=3x2 cosx3

6.

f(x) =sin3x3

f (x) =3sin31x3 d

dx

sin

x3

=3sin2x3cosx33x2

=9x2 sin2x3

cosx3

7.

f(x) =tan

4x5

f (x) =sec2

4x5 d

dx

4x5

=sec2

4x5

45x4=20x4 sec2

4x5

8.

f(x) =

4x sin2 (2x)

=

4x sin2

(2x) 12

f (x) =1

2

4x sin2 (2x) 12 22 d

dx

4x sin2 (2x)

=1

2

4x sin2 (2x) 12 [42sin (2x) cos (2x)2]

=44sin (2x) cos (2x)

2

4x sin2 (2x)

=22sin (2x) cos (2x)

4x sin2 (2x)

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2.5. The Chain Rule www.ck12.org

9.

f(x) = sinx

cos (3x2)

f (x) =cos (3x2) d

dx[sinx] sinx d

dx[cos (3x2)]

cos2 (3x2)=

cos (3x2)cosx sinx [sin (3x2)] (3)cos2 (3x2)

=cos (3x2)cosx + 3sinx [sin (3x2)]

cos2 (3x2)

10.

f(x) = (5x + 8)3x3 + 7x

13f (x) = (5x + 8)3 d

dx

x3 + 7x

13+x3 + 7x

13 ddx

(5x + 8)3

= (5x + 8)3

13

x3 + 7x

12

3x2 + 7

+

x3 + 7x

13

3 (5x + 8)2 (5)

=13 (5x + 8)3

x3

+ 7x12

3x2

+ 7

+ 15x

3

+ 7x13

(5x + 8)2

11.

f(x) =

x32x5

3

f (x) =3

x32x5

2 d

dx

x32x5

=3

x32x5

2(2x5) ddx

[x3] (x3) ddx

[2x5](2x5)2

=3

x32x5

2

(2x5) (1) (x3) (2)(2x5)2

=3 (x3)2(2x5)6 (x3)3

(2x5)24

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2.6 Implicit Differentiation

1.

x2 +y2 =500

d

dx

x2 +y2

=

d

dx[500]

d

dx

x2

+ d

dx

y2

= d

dx[500]

2x + 2ydy

dx=0

2ydy

dx = 2xdy

dx=2x

2y

dy

dx= x

y

2.

x2y + 3xy2=1d

dx

x2y + 3xy2 = d

dx[1]

d

dx

x2y

+

d

dx[3xy] d

dx[2] =

d

dx[1]

ydy

dx

x2

+x2 d

dx[y] +y

dy

dx[3x] + 3x

d

dx[y]0=0

y (2x) +x2dy

dx+y (3) + 3x

dy

dx=0

2xy +x2dy

dx+ 3y + 3x

dy

dx=0

x2dy

dx+ 3x

dy

dx= 2xy3y

x2 + 3x

dydx

= 2xy3ydy

dx=2xy3y

x2 + 3xdy

dx=y (2x + 3)

x (x + 3)

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2.6. Implicit Differentiation www.ck12.org

3.

1

x

+1

y

=1

2x1 +y1 =

1

2d

dx

x1 +y1

=

d

dx

1

2

d

dx

x1

+

d

dx

y1

=0

x2y2 dydx

=0

y2 dydx

=x2

dy

dx = x

2

y2dy

dx= y

2

x2

4.

x y=

3

x12 y 12 =

3

d

dx

x

12 y 12

=

d

dx

3

d

dx

x

12

d

dx

y

12

=0

1

2x

12

1

2y

12

dy

dx =0

12

y12

dy

dx= 1

2x

12

dy

dx=12x

12

12y

12

dy

dx=

y12

x12

dy

dx=

yx

=

y

x

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5.

sin

25xy2

=x

d

dx

sin

25xy2

=

d

dx[x]

cos

25xy2

y2 (25) + 25x

2y

dy

dx

=1

cos

25xy2

25y2 + 50xydy

dx

=1

25y2 cos

25xy2

+ 50xy cos

25xy2 dy

dx=1

50xy cos25xy2 dydx =

1

25y2 cos25xy2dy

dx=

125y2 cos25xy250xy cos (25xy2)

6.

tan3x2y2 =tan

4

d

dx

tan3

x2y2 = dy

dx

tan

4

3tan2

x2

y2

sec2

x2

y2

2x2ydy

dx

=0

3tan2x2y2sec2 x2y2(2x)3tan2 x2y2sec2 x2y22y dy

dx

=0

3tan2 x2y2sec2 x2y22y dydx

= 3tan2 x2y2sec2 x2y2(2x)

dy

dx=3tan2 x2y2sec2 x2y2(2x)3tan2 (x2y2) sec2 (x2y2) (2y)

dy

dx=

x

y

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7.

x2yy2x= 1d

dx

x2yy2x = d

dx[1]

d

dxx2y

d

dxy2x =0

ydy

dx

x2

+x2 d

dx[y]

x

dy

dx

y2

+y2 d

dx[x]

=0

y (2x) +x2

dy

dx

x2 (2y)

dy

dx

y2 (1) =0

2xy +x2

dy

dx

2x2y

dy

dx

y2 =0

x2

dy

dx

2x2y

dy

dx

=y22xy

x22x2y

dy

dx=y22xy

dy

dx=

y22xyx22x2y

m= y22xyx22x2y

=1212

=1

1 =1

8.

sin (xy) =y

d

dx[sin (xy)] =

d

dx[y]

cos (xy)

y

d

dx[x]x d

dx[y]

=

dy

dx

cos (xy)

y (1)x dydx

=

dy

dx

y cos (xy)x cosx dydx

=dy

dx

y cos (xy) =dy

dxx cosx dy

dx

y cos (xy) =dy

dx(1x cosx)

y cos (xy)

1x cosx =dy

dx

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m= y cos (xy)

1x cosx=

1cos (1)1cos

=1 (1)

1 += 1

1 +

9.

x3y3 =5

d

dx

x3y3

=

d

dx[5]

y3

d

dx x3

+x3

d

dx y3

=0

y3

3x2

+x3

3y2 dydx

=0

3x2y3 + 3x3y2dy

dx=0

3x3y2dy

dx= 3x2y3

dy

dx=3x2y3

3x3y2

dy

dx= y

x

d2y

dx2=

d

dx

y

x

= xddx

[y]y ddx

[x]

x2

= xdydxy (1)x2

= xdydxy

x2

= x

dydx

y

x2

= xy

x

yx2

= yyx2

= 2yx2

=2y

x2

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10.

y2 =kx

2ydy

dx=k

dy

dx=

k

2y

Then slopem= k2y0

at(x0,y0). Note that y20=kx0.

The equation of the tangent line is

yy0=m (xx0)yy0= k

2y0(xx0)

2y0y2y20=k(xx0)2y0y

2kx0=kx

kx0

2y0y=kx kx0+ 2kx02y0y=kx + kx0

2y0y=k(x +x0)

y0y=1

2k(x +x0)

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2.7 Linearization and Newtons Method

1. First, find f (x).

f (x) = d

dx

x2 + 1

x

=x (2x) x2 + 11

x2

=2x2x21

x2

=x21

x2

f (a) = f (1) = 12

112 =0 and f(1) = 12

+11 =2.

Then the linearization of f(x)at a=1 is:

f(x) f(a) +f (a) (xa) f(1) +f (1) (x1) 2 + 0 2

2.

f (x) =sec2x

f () =sec2 =1f() =tan =0

Then

f(x) f(a) +f (a) (xa)

f() +f () (x

)

0 + 1 (x)x

3.

f(x) = (1 +x)n

f (x) =n (1 +x)n1

Then

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2.7. Linearization and Newtons Method www.ck12.org

f(x) f(a) +f (a) (xa) (1 + a)n + n (1 + a)n1 (xa) (1 + a)n + n (1 + a)n1xn (1 + a)n1 a

Sincea is much less than 1, we can leta=0. Then

f(x) 1n + n (1)n1x 1 + nx

4. a. (1 +x)n 1 + nxtells us that(1x)4 1 + 4 (x) =14x.b.

1x= (1x) 12 1 + 12(x) =1 1

2x.

c. 51 +x

5 (1 +x) 12 =51 12x =5 52xd.

3

1 3

(x1)2

=

1 3

(x1) 2

3

=

1 +

3

(1x) 2

3

=

1 + 3 (1x)1 2

3

1 + 3 (1x)1 2

3 (1 + 3 (1 +x)) 23 = (4 + 3x) 23 =

4

1 +

3

4x

23

=423

1 +

3

4x

23

=423

1 +

2

3 3

4x

=4

23

1 +

1

2x

e. (1.003)99 = (1 + 0.003)99 1 + 99 (0.003) =1.2975. Graph the function first.

f(x) =x3 + 33

f (x) =3x2

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Looking at the graph, set x1= 1.5. By Newtons Method,

xn+1=xn f(xn)f (xn)

=xn x3 + 3

3x2

Then

x2= 1.5 (1.5)3 + 3

3 (1.5)2 = 1.444

x3= 1.44 (1.44)3 + 3

3 (1.44)2 = 1.442

Thus,x 1.442.6. Graph the function first.

f(x) = x + 31 +xf (x) = 1 +3

2(1 +x) 12 (1)

= 1 + 321 +x

First zero: Looking at the graph, setx1=1.1. By Newtons Method,

xn+1=xn f(xn)f (xn)

=xn x + 31 +x

1 + 321+x

Thenx2=1.1 1.1 + 31 + 1.1

1 + 321+1.1

=1.1404

x3=1.1404 1.1404 + 31 + 1.1404

1 + 321+1.1404

=1.146

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2.7. Linearization and Newtons Method www.ck12.org

Thus,x 1.146.Second zero:

Looking at the graph, set x1=8. By Newtons Method,

xn+1=xn

f(xn)

f (xn)=xn x + 3

1 +x1 + 3

21+x

Thenx2=8 8 + 31 + 8

1 + 321+8

=7.9201

x3=7.9201 7.9201 + 31 + 7.9201

1 + 321+7.9201

=7.854

Thus,x 7.854.

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www.ck12.org Chapter 3. Applications of Derivatives, Solution Key

CHAPTER 3 Applications of Derivatives,Solution Key

Chapter Outline

3.1 RELATED RATES

3.2 EXTREMA AND THE MEA NVALUET HEOREM

3.3 THE F IRST DERIVATIVE T EST

3.4 THE S ECONDDERIVATIVE T EST

3.5 LIMITS AT INFINITY

3.6 ANALYZING THE GRAPH OF AF UNCTION

3.7 OPTIMIZATION

3.8 APPROXIMATION ERRORS

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3.1. Related Rates www.ck12.org

3.1 Related Rates

1. a. Answers will vary.

b. Answers will vary.2.

4x2 + 16y2 =32

4

2x

dx

dt

+ 16

2y

dy

dt

=0

8xdx

dt+ 32y

dy

dt=0

Substitute(2,1)and

dx

dt =3 into the last equation and solve for

dy

dx .

8xdx

dt+ 32y

dy

dt=0

16 (1) (3) + 32dy

dt=0

48 + 32dy

dt=0

32dy

dt= 48

dy

dt=48

32 =

3 ft

2 sec

3. Draw a diagram of the situation. The runner is 23(60) =40 ft from first base. The players rate is dx

dt = 18ft

sec .

The variabley represents the distance between the runner and home plate. The variablex represents the distance

traveled by the runner. The rate at which the distance between the runner and home plate is changing is dydt

. The

runner is 23(60) = 40 ft from first base. The players rate of change is

dxdt

= 18ftsec . The diagram shows that a righttriangle is formed with x, the side of the diamond, and y. Use the Pythagorean Theorem to solve for y.

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602 + 402 =y2

3600 + 1600=y2

5200=y25200=y

Now, differentiate 602 +x2 =y2 with respect to timetand substitute the known values to find dydt

.

602 +x2 =y2

0 + 2xdx

dt=2y

dy

dt

2xdx

dt=2y

dy

dt

2 (40)18 ft

sec =2

5200

dydt

1440 ft

sec =25200 dydt

1440 ft

2

5200

sec=

dy

dt

720 ft5200sec

=dy

dt

9.98 ft

sec dy

dt

4. Draw a diagram of the situation. The balloon was 300 ft from the ground. The balloons rate of change wasdxdt

= 20ftsec .

The variabley represents the distance between Mr. Smiths place and the balloons place. The variable x represents

the height of the balloon. The rate at which the distance between Mr. Smiths place and the balloons place was

changing is dydt

.The diagram shows that a right triangle is formed with x, the height of the balloon, and y. Use thePythagorean Theorem to solve fory.

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3002 + 1002 =y2

90,000 + 10,000=y2

100,000=y2

100,000=y

Now, differentiate 1002 +x2 =y2 with respect to timetand substitute the known values to find dydt

.

1002 +x2 =y2

0 + 2xdx

dt=2y

dy

dt

2xdx

dt=2y

dy

dt

2 (300)20 ft

sec =2

100,000

dydt

12,000 ft

sec =2

100,000 dy

dt12,000 ft

2

100,000

sec=

dy

dt

6,000 ft100,000sec

=dy

dt

18.97 ft

sec dy

dt

5. Draw a diagram of the situation. Let x represent the distance traveled by the first train. The rate of change of the

first train was dxdt

= 65mihr

. Letyrepresent the distance traveled by the second train. The rate of change of the second

train was dy

dt

= 75mi

hr . At 3 PM, the distancey =130 mi and the distance x =120 mi. Let s represent the distance

between the two trains.

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Use the Pythagorean Theorem to solve fors.

1202 + 1302 =s2

14,400 + 16,900=s2

31,300=s2

31,300=s

Now, differentiatey2 +x2 =s2 with respect to timetand substitute the known values to find dsdt

.

y2 +x2 =s2

2ydy

dt+ 2x

dx

dt=2s

ds

dt

ydy

dt+x

dx

dt=s

ds

dt

(130)75 mi

hr + (120)

65 mi

hr = 31,300 ds

dt9,750 mi

hr +

7,800 mi

hr =

31,300

ds

dt17,750 mi

31,300 hr=

ds

dt

99.20 mi

hr ds

dt

6. Draw a diagram of the situation. Let x represent the distance on the ground between the bottom of the ladder and

the wall. Let y represent the height of the ladder against the wall. The rate of change of the ladder is dydt

= 6ftsec.The distance between the bottom of the ladder and the wall is 17 ft.

Use the Pythagorean Theorem to solve forx wheny=8.

x2 + 82 =172

x2 + 64=289

x2 =225

x=15

Now, differentiate 172 +x2 =y2 with respect to timetand substitute the known values to find dxdt

.

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x2 +y2 =172

2xdx

dt+ 2y

dx

dt=0

xdx

dt+y

dx

dt=0

(15)dx

dt+ (8)

6 ftsec

=0

(15)dx

dt 48 ft

sec =0

(15)dx

dt=

48 ft

secdx

dt=

48 ft

(15) sec

dx

dt 16 ft

5 sec

7. A= l wwhere represents the length of the rectangle, w represents the width, and A represents the area of therectangle. Then dl

dt= 6 ftmin and

dwdt

= 2 ftmin . Differentiate the equationA=lwwith respect to timet.

A=lw

dA

dt=

dl

dtw + l

dw

dt

= 6 ft

min(15) + (25)

2 ft

min

=90 ft

min +50 ft

min

=140 ft

min

8. When d pdt

= 10week, find dx

dt.

9. Lets=length of one side of the cube. Then volume V=s3.

V=s3

dV

dt=3s2

ds

dtdV

dt=3 (6 in.)2

1 in.

4 min

=27 in.3

4 min

10. a. A= r2

Solve forrwhenA=36 in.2

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36 = r2

36=r2

6=r

A= r2

dAdt

=2rdrdt

24 in.

min =2(6)

dr

dt24 in.

(12) min=

dr

dt

2 in.

min=

dr

dt

b.

C=2r

dC

dt=2

dr

dt

=2

2 in.

min

=4 in.

min

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3.2. Extrema and the Mean Value Theorem www.ck12.org

3.2 Extrema and the Mean Value Theorem

1. The absolute maximum is at x=7. The absolute minimum is at x=4. There is a relative maximum at x=2. Theextreme values of f are f(7) =7 and f(4) =

1.

2. The absolute maximum is at x=7. The absolute minimum is atx=9. There is a relative minimum at x=3. Theextreme values of f are f(7) =9 and f(9) =0.

3. The absolute minimum is at x =0. There is no maximum because the function is not continuous on the closedinterval. The extreme value of f is f(0) =1.

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4.

f(x) = x26x + 4f (x) = 2x6

Find the critical values of f.

Solve f (x) =0.

2x6=02x=6

x= 3f(3) = (3)26 (3) + 4= 9 + 18 + 4=13

Check the endpoints:

x= 4f(4) = 426 (4) + 4=12

x=1

f(1) = 126 (1) + 4= 3

Compare function values to find the maximum and minimum. The absolute maximum is atx = 3 because f(3)is the greatest value. The absolute minimum is at x= 1 because f(1) is the smallest value. The extrema aref(13) =13 and f(1) = 3.

5.

f(x) =x3x4f (x) =3x24x3

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Solve f (x) =0.

3x24x3 =0x2 (34x) =0

x2 =0 or 34x=0x=0 or 4x= 3x=0 or x=

3

4

Find the function values: f(0) =0 and f

34

=

34

3 34

4= 2764 81256= 108256 81256= 27256 0.1055.

Find function values of the endpoints:

0 is one endpoint and was already checked.

f(2) =816= 8

The absolute maximum atx= 34 . The absolute minimum atx=2. The extrema are f

34

0.1055 and f(2) = 8.

6.

f(x) = x2 + 4x2

= x2 + 4x2

f (x) = 2x8x3 = 2x 8x3

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Solve f (x) =0.

2x 8x3

=0

2x4

8=0

x44=0x2

4

x2 +

4

=0x22x2 + 2 =0

x22=0x

2=0 or x +

2=0

x= 2 or x= 2x2 + 2=0

x2 = 2

There are no real number solutions forx2 + 2=0.

Since the variable is in the denominator of one term of f (x), set that denominator equal to 0.

x3 =0

x=0

f (x)is undefined for x=0. This is another critical value.

Find the function values of the critical values in the interval [2,0]and of the endpoints.

f

2

= 2 +42

= 2 + 2=0

f(0)is undefined.

Find function values of the endpoints:

0 is one endpoint and f(0)was already found to be undefined.

f(2) = 4 +44

= 3.

The absolute minimum is at x=

2. The extrema is f

2

= 0. There is no absolute maximum as f(x)

approachesas x approaches 0.

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7. f(x) =3x212xis continuous and differentiable because it is a polynomial.Solve 3x312x=0:

3x312x=03xx24 =0

3x=0 orx24=0x=0 orx=2 orx= 2

On[2,0] ,f(2) = f(0) =0. On[0,2] ,f(0) = f(2) =0

f (x) =9x212

Set f (x) =0 and solve for the critical values.

9x212=033x2

4 =03x24=0

x= 2

3=

2

3

3 or x= 2

3= 2

3

3

By Rolles Theorem, there is at least one critical value in(2,0). That value is c = 2

33

. There is at least one

critical value in(0,2). That value is c= 2

33

.

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8.

f(x) =x2 2x1 =x

22 (x1)1

f (x) =2x + 2 (x1)2 =2x 2(x1)2

On the interval[1,0] ,f(1) =1 2

2 = 2 and f(0) =2. By Rolles Theorem, there is a critical value in (1,0).Solve

f (x) =0

2x 2(x1)2 =

0

2x (x1)22=02xx22x + 12=0

xx22x + 11=0x32x2 +x1=0[I cannot continue to solve the problem algebraically as written.]

9. f(x)is continuous on[1,2].

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

= x(x+2)x2

= xx2x2

= 2x2

has the interval(1,2)in its domain.

There is a numberc such that f(2) f(1) = (21)f (c).

f(2) =4

2

=2

f(1) =3

1=3

f(2) f(1) = (21)f (c)23=1f (c)1= f (c)

Solve forc.

2

c2

=

1

2= c22=c2

2=c

The value ofc is

2.

10. On[0,r] ,f(0) =0. Also, f(r) =0 becauseris a root of f. Note that f (x) =3x2 + 2a1x + a2is the derivativeof f(x). Then by Rolles Theorem, f (x) = 3x2 + 2a1x + a2 has a root in the interval (0,r). Thus, f (x) = 3x2 +2a1x + a2has a positive root that is less than rbecause there is a root in (0, r).

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3.3. The First Derivative Test www.ck12.org

3.3 The First Derivative Test

1. The function is increasing on(0,3). The function is decreasing on(3,6). The function is constant on(6,+).

2. The function is increasing on(,0)and(3,7). The function is decreasing on(0,3).3. a. Draw tangent lines to the graph to help you solve these problems.

f (3)is positive. f (3)>0.b. f (1)is negative. f (1)0.

4.

f(x) =x2 1x

=2xx1

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

x2

Find the critical values.

2x + 1

x2=0

2x3 + 1

x2 =0

2x3 + 1=0

2x3 = 1x3 = 1

2

x= 30.5= 3

0.5= 0.79

f 3

0.5

=1.89

f (x)is undefined for x=0.

Set up the intervals and make a table. Find test points to substitute into the derivative and check the sign of thederivative.

Interval, 3

0.5

3

0.5,0

(0,+)

Test pointx=c c= 1 c= 0.1 c=1f (c) 2 + 1= 2 0.2 + 1

(0.1)2 =99.8 2 + 1=3

sign of fx f (x)0 f (x)>0

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By the First Derivative Test, fchanges from negative to positive atx = 3

0.5. The critical value ofx = 3

0.5 is a

local minimum. The function decreases on, 3

0.5

and on(0,+). The function increases on

3

0.5,0

.

5.

f(x) =x215

f (x) =5x214 (2x)


5x214 (2x) =0

x214 =0 or 2x=0x21=0 orx=0x=1 orx= 1 or x=0f(1) =0

f(1) =0f(0) = 1

Interval (,1) (1,0)Test pointx=c c= 2 c= 0.5f (x) =5

x214 (2x) 5(2)214 (2 (2)) 5(0.5)214 (2 (0.5))

f (c) =5 (3)4 (4) =5 (0.251)4 (1)Sign of f (x) f (x)


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Interval (0,1) (1,+)

Test pointx=c c=0.5 c=2

f (x) =5x214 (2x) 5(0.5)214 (2 (0.5)) 52214 (22)

f (c) =5 (0.25

1)4 (1) =5 (4

1)4 (4)

Sign of f (x) f (x)>0

By the First Derivative Test, there is an absolute minimum at x=0. The function is decreasing on(,1)and on(1,0). The function is increasing on(0,1)and on(1,+).

6.

f(x) =x214

f (x) =4x213 (2x)


4x21

3(2x) =0

x213 =0 or 2x=0x21=0 orx=0

x=1 orx= 1 orx=0

f(1) =0

f(1) =0f(0) =1

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Interval (,1) (1,0)Test pointx=c c= 2 c= 0.5f (x) =4

x213 (2x) 4(2)213 (2 (2)) 4(0.5)213 (2 (0.5))

f (c) =4 (3)3 (4) =4 (1.25)3 (1)Sign of f (x) f (x)


89/286


Interval (,2) (2,+)Test pointx=c c= 3 c=0f (x) = 2x4 2 (3)4 2 (0)4f (c) =64=2 =04= 4Sign of f (x) f (x)>0 f (x)


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Interval (,3) (3,1) (1,+)Test pointx=c c= 4 c=0 c=3f (x) =3x2 + 3x9 3 (4)2 + 3 (4)9 3 (0)2 + 3 (0)9 3 (3)2 + 3 (3)9f (c) =48 + 12

9=51 =

9 =27 + 9

9=27

Sign of f (x) f (x)>0 f (x)0

The function is increasing on(,3)and on(3,+). The function is decreasing on(3,1).b. There is a relative maximum at x= 3 with f(3) =28. There is a relative minimum at x=1 with f(1) = 4.c.

9.

f(x) =x23 (x5) =x 53 5x 23

f (x) =5

3x

23 10

3x

13


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Solution Key_CK-12 Calculus Flexbook