Calculus Summer Pkt

8/10/2019 Calculus Summer Pkt

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Coral Springs Charter

SUMMER REVIEW PACKET

For students in entering CALCULUS AP

Name: __________________________________________________________

1. This packet is to be handed in to your Calculus teacher on the first day of the school year.

. All !ork must be sho!n in the packet "# on separate paper attached to the packet.$. Completion of this packet is !orth one%half of a ma&or test grade and !ill be counted in your first

marking period grade.

1



Summer Review Packet for Students Entering Calculus (all levels)

Complex Fractions

Simplif each of the follo!ing"

1.

'

a− a

'+ a

.

−(

x +

'+1)

x +

$.

( −1

x − $

'+1'

x − $

(.

x

x +1−1

x x

x +1+1

x

'.

1− x

$ x − (

x +$

$ x − (

F#nctions

$et f * x+ = x +1 and g * x+ = x −1 " Fin% each"

,. f *+ = ____________ -. g *−$+ = _____________ . f *t +1+ = __________

/. f g *−+ = __________ 1). g f *m + + = ___________ 11. f * x + h+ − f * x+

h= ______

$et f ( x) = sin x Fin% each exactl"

1. f π

2

= ___________ 1$. f

2π

3

= ______________



$et f * x+ = x 0 g * x+ = x +'0 and h* x+ = x −1 " Fin% each"

1(. h f *−+ = _______ 1'. f g * x −1+ = _______ 1,. g h* x$+ = _______

Fin% f * x + h+ − f * x+

h for the gi&en f#nction f.

1-. f * x+ = / x + $ 1. f * x+ = '− x

Intercepts an% Points of Intersection

To find the %intercepts0 let y 2 ) in your e3uation and sol4e.

To find the y%intercepts0 let 2 ) in your e3uation and sol4e

Fin% the x an% intercepts for each"

1/. y = 2 x − 5 ). y = x2 + x − 2

1. y = x 16 − x2 . y

2 = x3 − 4 x

$



Use s#'stit#tion or elimination metho% to sol&e the sstem of e(#ations"

Fin% the point)s* of intersection of the graphs for the gi&en e(#ations"

$. x + y = 8

4 x − y = 7(.

x2 + y = 6

x + y = 4'.

x2 − 4 y

2 − 20 x − 64 y − 172 = 0

16 x2 + 4 y2 − 320 x + 64 y +1600 = 0

Inter&al +otation

,. Complete the table !ith the appropriate notation or graph.

Sol#tion Inter&al +otation ,raph

−2 < x ≤ 4

−1, 7)

-

5ol4e each e3uation. 5tate your ans!er in 6"T7 inter4al notation and graphically.

-. 2 x − 1 ≥ 0 . −4 ≤ 2 x − 3 < 4 /. x

2−

x

3> 5

.omain an% Range

Fin% the %omain an% range of each f#nction" Write o#r ans!er in I+TERVA$ notation"

$). f ( x) = x2 − 5 $1. f ( x) = − x + 3 $. f ( x) = 3sin x $$. f ( x) = 2

x −1

(



In&erses

Fin% the in&erse for each f#nction"

/0" f ( x) = 2 x + 1 /1" f ( x) = x2

3

Also0 recall that to P#"89 one function is an in4erse of another function0 you need to sho! that: f (g( x)) = g( f ( x)) = x

Example2

If2 f ( x) = x − 9

4and g( x) = 4 x + 9 sho! f(x) and g(x) are in&erses of each other"

f (g( x)) = 4 x − 9

4

+ 9 g( f ( x)) =

4 x + 9( )− 9

4

= x − 9 + 9 = 4 x + 9 − 9

4

= x = 4 x

4

= x

f (g( x)) = g( f ( x)) = x therefore they are inverses

of each other.

Pro&e f and g are in&erses of each other"

/3" f ( x) = x

3

2g( x) = 2 x3 /4" f ( x) = 9 − x

2 , x ≥ 0 g( x) = 9 − x

'



E(#ation of a line

Slope intercept form2 y = mx + b Vertical line2 2 c *slope is undefined+

Point5slope form2 y − y1 = m( x − x1

) 6ori7ontal line2 y 2 c *slope is )+

$. se slope%intercept form to find the e3uation of the line ha4ing a slope of $ and a y%intercept of '.

$/. ;etermine the e3uation of a line passing through the point *'0 %$+ !ith an undefined slope.

(). ;etermine the e3uation of a line passing through the point *%(0 + !ith a slope of ).

(1. se point%slope form to find the e3uation of the line passing through the point *)0 '+ !ith a slope of <$.

(. Find the e3uation of a line passing through the point *0 + and parallel to the line y = 5

6 x − 1 .

($. Find the e3uation of a line perpendicular to the y% ais passing through the point *(0 -+.

((. Find the e3uation of a line passing through the points *%$0 ,+ and *10 +.

('. Find the e3uation of a line !ith an %intercept *0 )+ and a y%intercept *)0 $+.

,



2

-2

(-1,0)

(0,-1)

(0,1)

(1,0)

Ra%ian an% .egree Meas#re

(,. Con4ert to degrees: a.5π

6 b.

4π

5c. .,$ radians

(-. Con4ert to radians: a. 45o b. −17

o c. $- o

Unit Circle

(. a.) sin180o

b.) cos270

o

c.) sin(−90o) d .) sinπ

e.) cos360

o f .) cos(−π )

,raphing Trig F#nctions

,raph t!o complete perio%s of the f#nction"

(/. f ( x) = 5sin x '). f ( x) = cos x − 3

Trigonometric E(#ations2

5ol4e each of the e3uations for 0 ≤ x < 2π . =solate the 4ariable0 sketch a reference triangle0 find all the

solutions !ithin the gi4en domain0 0 ≤ x < 2π . #emember to double the domain !hen sol4ing for a double

angle. se trig identities0 if needed0 to re!rite the trig functions. *5ee formula sheet at the end of the packet.+

'1. sin x = − 1

2'. 2cos x = 3

'$. cos2 x = 1

2'(. sin2

x = 1

2

-



Minor Axis

Major Axis

ab

c FOCUS (h + c, k)FOCUS (h - c, k)

CENTER (h, k)

( x − h)2

a2 + ( y − k )2

b2 = 1

''. sin2 x = − 3

2',. 2cos2

x − 1 − cos x = 0

'-. 4cos2 x − 3 = 0 '. sin2 x + cos 2 x − cos x = 0

In&erse Trigonometric F#nctions2

For each of the follo!ing8 express the &al#e for 9: in ra%ians"

'/. y = arcsin − $

,). y = arccos −1( ) ,1. y = arctan*−1+

For each of the follo!ing gi&e the &al#e !itho#t a calc#lator"

,. tan arccos

$

,$. sec sin

−11

1$

,(. sin arctan1

'

,'. sin sin−1 -

.

Circles an% Ellipses

r 2 = ( x − h)2 + ( y− k )2



4

2

-2

-4

-5 5

4

2

-2

-4

-5 5

4

2

-2

-4

-5 5

4

2

-2

-4

-5 5

For a circle centered at the origin0 the e3uation is x2 + y

2 = r 2

0 !here r is the radius of the circle.

For an ellipse centered at the origin0 the e3uation is x

2

a2 +

y2

b2 = 1 0 !here a is the distance from the center to the

ellipse along the %ais and ' is the distance from the center to the ellipse along the y%ais. =f the largernumber is under the y2 term0 the ellipse is elongated along the y%ais. For our purposes in Calculus0 you !ill not

need to locate the foci.

,raph the circles an% ellipses 'elo!2

,,. x2 + y

2 = 16 ,-. x2 + y

2 = 5

,. x

2

1+

y2

9= 1 ,/.

x2

16+

y2

4= 1

-)%/ *#epresented by net set of problems+

/



Vertical Asmptotes

;etermine the 4ertical asymptotes for the function. 5et the denominator e3ual to >ero to find the %4alue for

!hich the function is undefined. That !ill be the 4ertical asymptote.

/). f ( x) = 1

x2

/1. f ( x) = x

2

x2 − 4

/. f ( x) = 2 + x

x2 (1 − x)

6ori7ontal Asmptotes

;etermine the hori>ontal asymptotes using the three cases belo!.

Case I. ;egree of the numerator is less than the degree of the denominator. The asymptote is y 2 ).

Case II" ;egree of the numerator is the same as the degree of the denominator. The asymptote is the ratio ofthe lead coefficients.

Case III. ;egree of the numerator is greater than the degree of the denominator. There is no hori>ontalasymptote. The function increases !ithout bound. *=f the degree of the numerator is eactly 1 more than the

degree of the denominator0 then there eists a slant asymptote0 !hich is determined by long di4ision.+

.etermine all 6ori7ontal Asmptotes"

/$. f ( x) =

x2 − 2 x + 1

x

3

+ x − 7/(. f ( x)

=

5 x3 − 2 x

2 + 8

4 x − 3 x 3 + 5/'. f ( x)

=

4 x5

x2 − 7

?iscellaneous @no!ledge

/,. #ationali>e the denominator: *a+ *b+ *c+

1)



/-. 5ol4e for x *do not use a calculator+:

*a+ '* x 1+ 2 ' *b+ *c+ log x 2 $ *d+ log$ x 2 log$( % ( log$'

/. 5implify: *a+ log' log* x % 1+ % log* x % 1+ *b+ log(/ % log$

//. Factor completely: *a+ x, % 1, x( *b+ ( x$ % x % ' x ') *c+ x$ - *d+ x( %1

1)). Find all real solutions to: *a+ x, % 1, x( 2 ) *b+ ( x$ % x % ' x ') 2 ) *c+ x$ - 2 )

1)1. Find the remainders on di4ision of

*a+ x' % ( x( x$ % - x 1 by x *b+ x' % x( x$ x % x ( by x$ 1

1). Find the domain of the function

1)$. Find the domain and range of the functions:

11



cos2 x = cos2 x − sin

2 x

= 1− 2sin2 x

= 2cos2 x − 1

Form#la Sheet

#eciprocal =dentities: csc x = 1

sin xsec x =

1

cos xcot x =

1

tan x

Buotient =dentities: tan x = sin x

cos x cot x = cos x

sin x

Pythagorean =dentities: sin2 x + cos2

x = 1 tan2 x + 1 = sec2

x 1 + cot2 x = csc2

x

;ouble Angle =dentities: sin2 x = 2sin x cos x

tan2 x = 2tan x

1 − tan2 x

ogarithms: y = loga x is e3ui4alent to x = a y

Product property: logb mn = logb m + logb n

Buotient property: logb

m

n= log

b m − logb n

Po!er property: logb m p = p logb m

Property of e3uality: =f logb m = logb n 0 then m 2 n

Change of base formula: loga n =

logb n

logb a

5lope%intercept form: y = mx + b

Point%slope form: y − y1 = m( x − x1)

5tandard form: A 6y C 2 )

1

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Calculus Summer Pkt