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Odd - Even Calculus Solutions

Constant Rule: Power Rule Scalar Rule

0 d

cdx

1 ( )

The Sum and Difference Rules

n nd d dyx nx cf x c

dx dx dx

Trigonometric Rule:

( ) ( ) ( ) ( ) sin( ) cos( )

( ) ( ) ( ) ( )

d df x g x f x g x x x

dx dx

dyf x g x f x g x

dx

cos( ) sin( ) d

x xdx

distance•Rate=

time

in position•Average Velocity= = =

in time

=f (c)=slope of the tangent line at the point c, f(c)

final initial

final initial

s sChange s

Change t t t

dy

x cdx

RECALL:

RATE OF CHANGE:






1

2( )

1

12(1)1 2

f x x

risef

run

3( )

1(1) 3

1

3

f x x

risef

run

1

2( )

112(1)

1 2

f x x

risef

run

1( )

1(1) 1

1

f x x

risef

run

( ) 2 ( ) 0f x f x

7 7 1 8

7

8

1( ) 7 7

7( )

y x f x x xx

f xx

8 8 1 9

8

9

1( ) 8 8

8( )

y x f x x xx

f xx

1

5 5

1 41

5 5

5 4

( )

1 1 1( )

5 5 5

f x x x

f x x xx

1

4 4

1 31

4 4

34

( )

1 1 1( )

4 4 4

f x x x

f x x xx

( ) 1

( ) (1) 1 0 1

f x x

dy d d dyx

dx dx dx dx

( ) 3 1

(3 ) (1) 3 0 3

f x x

dy d d dyx

dx dx dx dx

#1

a)

b)

#2)

a)

b)

#3#24, Find the derivative of the function:

#3) 8 0y y

#4)

#5) 6 5( ) 6y x f x x

#6) 8 7( ) 8y x f x x

#7)

#8)

#9)

#10)

#11)

#12)






2

2

2 1

( ) 2 3 6

( 2 ) (3 ) (6)

2 2 3 0 4 3

f t t t

dy d d dt t

dt dt dt dt

dyt t

dt

2

2

2 1

( ) 2 3

( ) (2 ) (3)

2 2 0 2 2

f t t t

dy d d dt t

dt dt dt dt

dyt t

dt

2 3

2 2

( ) 4

( ) 2 4 3 2 12

g x x x

g x x x x x

3

2

8

0 3

y x

y x

3

3 1 2

( ) 2 4

( ) 3 2 0 3 2

s t t t

s t t t

3 2

3 1 2 1 1 1 2

( ) 2 3

( ) 2 3 2 3 6 2 3

f x x x x

f x x x x x x

sin cos2

cos ( sin )2

cos sin2

y

y

y

( ) cos

( ) sin

g t t

g t t

2 1cos

2

12 sin

2

y x x

y x x

5 sin

0 cos cos

y x

y x x

1

1 1 1

2

2

13sin 3sin

3sin 3cos

13cos 3cos

y x x xx

y x x x x

y x x xx

3

3

3 1

4

5 52cos 2cos

(2 ) 8

5 15( 3) 2sin 2sin

8 8

y x x xx

y x x xx

#13)

#14)

#15)

#16)

#17)

#18)

#19)

#20)

#21)

#22)

#23)

#24)






2

2

2 1

3

3

5Original

2

5Rewrite:

2

5Differentiate: ( 2)

2

5Simplify: 5

yx

y x

y x

y xx

2

2

2 1

3

3

2Original

3

2Rewrite:

3


3

4 4Simplify:

3 3

yx

y x

y x

y xx

3

3

3 1

4

4

3Original

2

3Rewrite:

8


8

9 9Simplify:

8 8

yx

y x

y x

y xx

2

2

2 1

3

3

Original 3

Rewrite: 9

Differentiate: ( 2)9

2 2Simplify:

9 9

yx

y x

y x

y xx

11 12 12 2

11

2

3

2

3

Original

Rewrite:

1Differentiate:

2

1 1Simplify:

2 2

xy

x

xy x x

x

y x

y xx

3

3

3 1

2

4Original

Rewrite: 4

Differentiate: 4 3

Simplify: 12

yx

y x

y x

y x

2

2

3

3

3( ) 3

Point: (1,3)

6 ( ) 6

(1) 6

f x xx

f x xx

f

2 2

3 1 3( )

5 5

3 3 3 25 5

5 95 5 9 3

25

f tt t

f

#25)

#26)

#27)

#28)

#29)

#30)

#31#38, Find the slope of the graph of the function

at the given point. Use the derivative feature of a

graphing utility to confirm your results.

#31)

#32)






3

2 2

1 7( )

2 5

1Point: 0,

2

7 21( ) 0 3

5 5

(0) 0

f x x

f x x x

f

3

2 2

( ) 3 6

Point: 2,18

( ) 3 2 6

(2) 6 4 24

f x x

f x x x

f

2

2

( ) (2 1)

Point: 2,18

( ) 4 4 1

( ) 8 4

(2) 8 2 4 20

f x x

f x x x

f x x

f

2

2 2

( ) 3(5 )

Point: 5,0

( ) 3(25 10 ) 75 30 3

( ) 30 6

(5) 30 30 0

f x x

f x x x x x

f x x

f

( ) 4sin

Point: 0,0

( ) 4cos 1

(0) 4 1 3

f

f

f

( ) 2 3cos

Point , 1

( ) 3sin

( ) 3sin( ) 0

g t t

g t t

g

2 2

3

3

4

3

( ) 5 3

6( ) 2 6 2

2 6( )

f x x x

f x x x xx

xf x

x

2 2

3

4 3

3

( ) 3 3

6( ) 2 3

2 3 6( )

f x x x x

f x xx

x xf x

x

2 2 3

3

4

4

4( ) 4

12( ) 2 12 2

g t t t tt

g t t t tt

2

2

3

1( )

2( ) 1

f x x x xx

f xx

#33)

#34)

#35)

#36)

#37)

#38)

#39#52, Find the derivative of the function

#39)

#40)

#41)

#42)






3 2 3 2

2 2 2 2

2

2 1 3

3

3 3

3 4 3 4( )

( ) 3 4

( ) 1 4 ( 2) 1 8

8 8( ) 1

x x x xf x

x x x x

f x x x

f x x x

xf x

x x

21

2

2 3 1( ) 2 3

1( ) 2

x xh x x x

x

h xx

2 3

2

( 1)

3 1

y x x x x

y x

2

2 3

2

3 (6 5 )

18 15

36 45

y x x x

y x x

y x x

11

3 32

1 21 111

3 32 2

3 2

3 3 32 2 2

3 32 2

3 62 7

( ) 6 6

1 1 1( ) 6 2

2 3 2

1 2 4( )

2 2 2

4 4( )

2 2

f x x x x x

f x x x x x

x xf x

x x x x x x

x x x xf x

x x x

4 2

5 3

4 2 1 11 1

5 3 5 3

1 1

5 3

( )

4 2 4 2( )

5 3 5 3

4 2( )

5 3

h s s s

h s s s s s

h s

s s

2 1

3 3

1 2

3 3

1 2

3 3

( ) 4

2 1( ) 0

3 3

2 1( )

3 3

f t t t

f t t t

f t

t t

1 1

3 5 3 5

1 11 1

3 5

2 4

3 52 4

3 5

4 2 4 2

5 3 5 3

2 4 4 2 2 4

3 5 5 3 3 5

4 2

5 3

22

15

5 34 2

15 22

( )

1 1( )

3 5

1 1 1 1( )

3 53 5

5 3 5 3( )

15 15 15

5 3( )

15

5 3( )

15

f x x x x x

f x x x

f x x x

x x

x x x xf x

x x x x x

x xf x

x

x xf x

x

#43)

#44)

#45)

#46)

#47)

#48)

#49)

#50)






( ) 6 5cos

6( ) 5sin

2

f x x x

f x xx

1

3

3

11

3

4

3

2( ) 3cos 2 3cos

2( ) 3sin

3

2( ) 3sin

3

f x x x xx

f x x x

f x x

x

4 2

3

3 2 Point: (1,0)

4 6

4 6 2 ( ) 21

Tangential equation:

0 2( 1)

2 2

y x x

y x x

y m slopex

y x

y x

3

2

Point: (-1,-2)

3 1

3 1 4 ( ) 41


( 2) 4( 1)

4 2

y x x

y x

y m slopex

y x

y x

3

4

34

31

47

4

2=2 Point: (1,2)

3 3

22

3 3( )

1 2 2


32 ( 1)

2

3 7

2 2

y xx

y x

x

y m slopex

y x

y x

2 3 2

2

( +2 )( 1) 3 2

Point: (1,6)

3 6 2

3 6 2 11 ( ) 111


6 11( 1)

11 5

y x x x x x x

y x x

y m slopex

y x

y x

4 2

3

3

2

8 2

4 16

Horizontal tangent line at which has 0

4 16 0

4 ( 4) 0 4 ( 2)( 2) 0

4 0 0 (0) 2

2 0 2 (2) 14

2 0 2 ( 2) 14

( ) has horizontal tangent line at p

y x x

y x x

y

x x

x x x x x

x x f

x x f

x x f

f x

oints:

(0,2) ; (2, -14) ; (-2, -14)

#51)

#52)

#53#56, Find an equation to the tangent line to the

graph of f(x) at the given point

#53)

#54)

#55)

#56)

#57#62, determine the points at which the graph of

the function has a horizontal tangent line

57)






3

2

2

3 1

tangent line at which has y =0

3 1 0 for all x

( ) has no point whose tangent line is

horizontal

y x x

y x

Horizontal

but x

f x

2

3

3

1

2

Horizontal tangent line at which has 0

2but 0 for all x (- ,0) (0, )

( ) has no point whose tangent line is

horizontal

yx

yx

y

x

f x

2 1

2


2 0 0

(0) 1

( ) has horizontal tangent line at: (0,1)

y x

y x

Horizontal

x x

f

f x

sin , 0 2

1 cos


1 cos 0 cos 1

0 2 ,

( )

( ) has horizontal tangent line at: ( , )

y x x x

y x

Horizontal

x x

x x

f

f x

3 2cos , 0 2

3 2sin


33 2sin 0 sin

2

20 2 , ,

3 3

3 2 2 31 1

3 3 3 3

( ) has horizontal tangent line at points:

,3

y x x x

y x

Horizontal

x x

x x

f f

f x

3 2 2 31 ; , 1

3 3 3

2

2

2

2 2

2

: ( )

: 4 9

( ) 2

The line is tangent

( )We have two facts

2 4

4 9

2 4

(2 4) 4 9

2 4 4 9

9 3

2

10

Function f x x kx

Line y x

f x x k

f x y

x k

x kx x

k x

x x x x

x x x x

x x

k

k

#58)

#59)

#60)

#61)

#62)

#63#66, find k such that the line is tangent to the

graph of the function

#63)






2

2

: ( )

: 4 7

( ) 2

The line is tangent


2 4

4 7

2

4(2) 7 4

3

Function f x k x

Line y x

f x x

f x y

x

k x x

x

k

k

2

22

2

2

: ( )

3: 3

4

( )

The line is tangent

( )

We have two facts 3( )

4

34 33, x 0

4 344

3, x 0 4 3

4

4 33 12

4 3

kFunction f x

x

Line y x

kf x

x

f x y

f x

kkx

xxx

kk x

x

k xx

x

k x

2

2

2

3 12

3

4

6 12 0

6 ( 2) 0

2

0 (eliminated because x 0)

2 3

x x

xk

x x

x x

x

x

As x k

2

: ( )

: 4

( )2

The line is tangent to ( )


( ) ( )

4, 0•2 4

1, 022

2 4 4

4 2 4 4

But the line

Function f x k x

Line y x

kf x

x

f x

f x y

f x m slope

k x x xx x

kx

k xx

x x x

As x k

4 is only tangent

to ( ) , 0

4

y x

f x k x k

k

#64)

#65)

#66)

#67)

a) AB

The average rate of change is the secant slope. The slope

which seems to be a vertical line has greatest slope.

b)

The average rate of change of the function between A

and B is greater than the instantaneous rate of change at

B because it has greater slope.

c)

First, draw a secant line CD, then draw a line that is

tangent to the curve CD and parallel to secant CD, then

mark the tangent point.






( ) 0

rate of change is decreasing

( )

has greatest slope as x 0

f x

f x x

f x

( ) ( ) 6

g(x) has the same shape as f(x),

but g(x) is f(x) shifted up 6 units

( ) ( )

g x f x

g x f x

( ) 5 ( )

g(x) has opposite sign to f(x) at a particular

x=c

Its magnitude of instantaneous slope is 5 times f(x)'s.

( ) 5 ( ) ( ) 5 ( )

g x f x

g x f x g x f x

2

1

2

2

1

2

1 2

1

2

1

2

1

6 5

2

2 6

2 2 6

34 6

2

32 3

2

32 6 3

2

9 3 9(3 / 2) Point: ,

4 2 4

7 3 7(3 / 2) Point: ,

4 2 4

9 3Tangent line to y : 3

4 2

Tangent li

y x

y x x

y x

y x

y y x x

x x

y

y

f

f

y x

2

1

2

7 3ne to y : 3

4 2

93

4

113

4

t

t

y x

y x

y x

#68)

Figure

#69)

#70)

73)Need graph

Figure:






1 2

2

1 2

11

2 2

2

tangent-1

tangent-2

Intersection point:

1y , x 0

x 1 1

intersects at (1,1), ( 1, 1)

(1,1)

y 1y 11

1

11

2

These two lines are perpendicular

(-1

y xx

x

y y

x

yyx

x

y x

y x

1

2

tangent-1

tangent-2

,-1) SAME STEPS AS ABOVE

y 11

11

2

These two lines are also perpendicular

x

yx

y x

y x

( ) 3 sin 2

( ) 3 cos

Horizontal tangent line indicates f (x)=0

3 cos 0 cos 3

but 1 cos 1

cos -3

f(x) does not have horizontal tangent line

f x x x

f x x

x x

x

There is no x value such that x

5 3

4 2

4 2

4 2

2

2

2 2

( ) 3 5

( ) 5 9 5

( ) 3

( ) 5 9 5 3

5 9 2 0

,

5 9 2 0

4 9 9 4.5.2

2 2.5

9 81 40

10

41 9 0

10

9 410

10

: however, 0

We can't find su

f x x x x

f x x x

f a slopex a

f a a a

a a

Let z a

z z

b b acz

a

z

z

a z z

a

5 3

ch that ( ) 3

( ) 3 5 does not have a tangent line

with a slope of 3.

f a

f x x x x

#74) Show that 1

2

1

y x

yx

have tangent lines that are

perpendicular to each other at their point of

intersection.

#75)

#76)






0 0

0 0

( ) ( ) ( )( )

4

( )

1( ) , x 0

2

1, x 0

42

4

1 1( )

42 4

1 (4)(4) 2

4 8

Tangent point: (4,2)

1Tangent line: 2 ( 4)

4

11

4t

y y f x f x f xf x

x x x x x

f x x

f xx

x

xx

x

f x

ff

y x

y x

0 0

0 0

2

2

2

( ) ( ) ( )( )

5

2( ) , x 0

2( ) , x 0

2

2, x 0

5

2 2, x 0

(5 )

0 (eliminated due to x 0)

5

2

5 8

2 25

5

8 2

25

y y f x f x f xf x

x x x x x

f xx

f xx

x

x x

x x x

x

x

f

f

5 4

5 2 55

2

5 4Tangent point: ,

2 5

4 8 5Tangent line:

5 25 2

8 8

25 5t

f

y x

y x

#77#78, find an equation of the tangent line to the

graph of the function f(x) through the point (x0,y0) not

on the graph. Given: 0

0

( )y y

f xx x

#77)

78)






( ) ( ), ( ) ( )

because

( ) ( )

If f x g x then f x g x

FALSE

f x g x D

( ) ( ) , ( ) ( )If f x g x C then f x g x

TRUE

2If y= , 2

FALSE

0 ( is a constant)

dythen

dx

dy

dx

1 , then

x dyIf y

dx

TRUE

( ) 3 ( ), ( ) 3 ( )If g x f x then g x f x

TRUE

-1

1

1 1 ( ) , ( )

( )

n n

n

f f x then f xx nx

FALSE

nf x

x

sec

sec

( ) 2 7 [1,2]

•

f(1)=9

f(2)=11

(2) (1) 11 92

2 1 1

21

22

Linear function has the average

rate of change equals to the function's

slope.

ant

ant

f t t

Average rate of change m

f fm

dy

tdt

dy

tdt

2

sec

sec

( ) 3 [2, 2.1]

( ) 2

•

f(2)=1

f(2.1)=1.41

(2.1) (2) 1.41 1 414.1

2 1 2.1 2 10

42

4.22.1

In a small interval of t, the average rate

change is c

ant

ant

f t t

f t t


f fm

dy

tdt

dy

tdt

lose to instantaneous rate of

change.

#83)

#84)

#85)

#86)

#87)

#88)

#89#92, find the average rate of change of the

function over the given interval. Compare this average

rate of change with the instantaneous rates of change

at the endpoints of the interval.

#89)

#90)






2

secant

sec

1( ) , [1, 2]

1( ) , [1, 2]

•

f(1)= 1

1f(2)=

2

11

(2) (1) 12

2 1 1 2

11

1

2 4

1 11

4 2

In an increasing close interval,

the average rate

ant

f xx

f xx


f fm

dy

xdx

dy

xdx

of change is

between two instantaneous rates of

change.

0 0

2

1362 ft, 0

( ) 16 1362 ft

( ) 32 m/s

s v

s t t

dsv t t

dt

(1) 16 1362 1346

(2) 64 1362 1298

(2) (1) 1298 134648 ft/s

2 1 1

s

s

s sv

( ) 32 ft/s

(1) 32 ft/s

(2) 64 ft/s

dsv t t

dt

v

v

2

When the object reaches the ground, s(t)=0

16 1362 0

9.23 s

since 0, we choose t 9.23 s

t

t

t

( ) 32 ft/s

v(9.23)= 295.36 ft/s

v t t

sec

( ) sin 0,6

( ) cos( )

(0) 0

1sin

6 6 2

sec

1(0)36 2

06 6

(0) 1

3

6 2

3 31 (0) ( )

2 6

f x x

f x x

f

f

Average rate ant slope

f f

m

f

f

f Average rate f

#91)

#92)

#93#94, Vertical Motion, use the position function 2

0 0( ) 16s t t v t s for free-falling object.

#93)

a)

b) Determine the average velocity on the interval [1,2]

c) Find the instantaneous velocities when t=1 and t=2

d) Find the time required for the coin to reach

ground level

e) Find the velocity of the coin at impact






2

0 0

0 0

2

2

2

( ) 16

220 feet, 22 ft/s

( ) 16 22 220

( ) 32 22

Velocity after 3 seconds thrown

(3) 118 ft/s

Velocity after falling 108 feet

220 108 112 ft

112 16 22 220

f

s t t v t s

s v

s t t t

dsv t t

dt

v

s

t t

b bt

4

2

2 s

v(2)= 32 22 86 ft/s

ac

a

t

t

2

0 0

0

0

2

( ) 4.9

120 m/s

0

( ) 4.9 120 m

( ) 9.8 120 m/s

(5) 71 m/s

v(10)=22 m/s

s t t v t s

v

s

s t t t

dsv t t

dt

v

2

0 0

0

2

0

2

0

0

( ) 4.9

0

(6.8) 4.9

0 4.9(6.8)

227 m

s t t v t s

v

s t s

s

s

#94)

#95)

#96)

#97)

#98)

#99 and #100)

#102) Convert






3 3

2 2

2

cm

3 cm

(4) 48 cm

V s

dVs

ds

V

2 2 m

2 m

(4) 8 m

A s

dAs

ds

V

2

0 0

0 0

0 0

2 2

0 0

0

0

0 0

0

( ) , [t , t ]2

(t ) (t )

t (t )

t t

2 2

2

( )

( )

Average velocity is equivalent to

instantaneous velocity at

ats t C t t

s t s tsv

t t t

a t a t

v att

v at

s t at

s t at

t t

2

2

1,008,000( ) 6.3 , [350,351]

The change in anual cost:

(351) (350) 5083.09 50851.91

351 350 1

The instantaneous rate of change:

1,008,000C (Q)= 6.3

1,008,000(350) 6.3 1.93

350

(350)

C Q QQ

C Cm

Q

C

C m

X miles 1gallon 1 year

1 galon $1.55 15,000 miles

2

1

1.55 150001.55 15000 $$

1.55 15000 23250 $( ) = , 0

23250, 0

x year year

x

C x xx x year

dCx

dx x

X 10 15 20 25 30 35 40

C 2325 1550 1162 930 775 664 581

C’(x) -233 -103 -58 -37 -26 -19 -15

The driver who gets 35 miles/g will benefit more

because he spends less than $664/ year, and his rate of

change is greater.

#103)

#104)

#105)

#106)

#107) ( )N f p

a) (1.479)f is the instantaneous rate of change of the

number of gallons of unleaded gasoline sold at the price

$1.479.

b) (1.479)f <0 because as the price increases, the

number of gallons decreases.






( )a

dTk T T

dt

2( ) , point (0,1)

Tangent line: y x 1, point (1,0)

(0) 1

1

( ) 2 (1) 2

(1) (tangent slope)

2 1

( ) and the tangent line share (1,0)

(1,0)

(1) 0

1

f x ax bx c

f c

c

f x ax b f a b

f m

a b

f x

f a b c

a b

2

2 1 2

1 3

1 1

( ) 2 3 1

a b a

a b b

c c

f x x x

2 2

2 2

1( ) , point ( , )

1( )

1 1( ) ( )

According to the graph:

1 2

2 2

f x a bx

f a ba

f x f ax a

h hm h

a a

2

2

Tangent line: ( 2)2

2

2 22

2

2

1 2As

(2 )

2

2 2

2 (2 )

(2 )

11 1

12

2

12

2

hy x

hy x h

hab h b h ha

bh

a

b ha a a

ha

a a ah

a a

a

ba a

h

A bh

#108)

#109)

#110)

#110 continues






3

2

2

3 2

3 3 2

( ) 9

3 9

(1, 9) Tangent line:

9 (3 9)( 1)

( )

3 3 9

9 3 3 9

0 (0) 0

3 3 81

2 2 8

3 81There are two tangent points: (0,0); ,

2 8

(0,0), (

f x x x

y x

At

y x x

f x y

y x x x

x x x x x

x f

x f

f

tan

2

tangent

tangent

0) 9

9

3 81 3 3 3 9, , 3 9

2 8 2 2 2 4

81 9 3 9 27

8 4 2 4 4

9

9 27

4 4

genty x

f

y x y x

y x

y x

2

2 2

2 2

tangent

) ( )

( ) 2

Tangent line: y-a=2x 2

( ) and y share the tangent point

( ) 2

( )

have two tangent points

( , ), ( , )

( , )

( ) 2

2 ( )

2

a f x x

f x x

y x a

f x

f x y x x a

x a f a a

We

a a a a

a a

f a a

y a a x a

y ax a

tangent

tangent

tangent

( , )

( ) 2

2 ( )

2

20,

2

a a

f a a

y a a x a

y ax a

y a x aa

y a x a

2

2

2 2

2

2

2

)

(a,0)

( )

( ) 2

Tangent line: 2 ( ) 2 2

( ) and y share the tangent point

( ) 2 2

2 (2 ) 4

0 (0) 0

have two tangent points

(0,0), (2 , 4 )

(0,0)

(0) 0

0

(2 , 4 )

b

f x x

f x x

y x x a y x xa

f x

f x y x x xa

x a f a a

x f

We

a a

f

y

a a

2

2

tangent

tangent

2

tangent

(2 ) 4

4 4 ( 2 )

4 4

0,

4 4

f a a

y a a x a

y ax a

ya

y ax a

#111)

#112)

#112 continues:






3

2

3

2

2 2

3

2 2

32

2 2

, 2( )

, 2

(2) (2) 8

( ) (2) 8lim lim

2 2

( ) (2) 8lim lim

2 2

( 8)lim lim ( 2 4) 12

2

According to the defi

x x

x x

x x

ax xf x

x b x

f a a

f x f x b a

x x

f x f ax a

x x

a xa x x a

x

0

2

2

nition:

( ) ( )lim ( ), 2

( ) 2 (2) 4

1'(2) 12 12 4

3

( ) is differentiable everywhere as:

8lim exists

2

8 4

48 4

3

1

3( ) is differentiable (- , ) with

h

x

f x h f xf x x

h

f x x f

f a a a

f x

x b a

x

b a

b a

a

f x

b

4

3

cos , 0 ( )

, x 0

( ) is differentiable ( . ) as

(0 ) (0 )

0, ( ) sin

(0) 0

x 0, ( )

(0)

0

x xf x

ax b

f x

f f

x f x x

f

f x a

f a

a

0 0

0

(0) (0)

( ) (0) cos( )lim lim 0

0

cos( ) 11 so that lim 0

0

1

x x

x

f a b b

f x f x b

x x

xb

x

a

b

1

2

( ) sin( )

( ) sin

sin , 0

sin , 2( )

sin , 0

sin( ) , 0

(0 ) cos(0) 1

'(0 ) cos(0) 1

(0)

( ) cos( ) 1

( ) cos( ) 1

( )

(2 )

f x xf x

f x x

x x

x xf x

x x

x x

f

f

f DNE

f

f

f DNE

f

cos(2 ) 1

(2 ) cos(2 ) 1

(2 )

( ) is differentiable n , (n+1) at which

n is an integer.

: 0, ( ) (0, )

f

f DNE

f x

Ex n f x is differentiable

#113)

#114)

#114 continues:

#115)

ap-sat tutorial 2.2.pdf · page 1 odd - even calculus solutions > @ constant rule: ... f x f x(...

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