Calculus Concept Collection - Chapter 7

Inverse Functions

Answers

1. 1 1( )

3

xf x

2. 3x

3. 3 1x

4. Function has an inverse. 5. Function does not have an inverse. 6. Function does not have an inverse.

7. 1

2

x

8. 1

2

x

9. ( ) 2( 5) 0f x x on 5.x

10. ( ) sinf x x , which is negative on the interval in question, so ( )f x is monotonically

decreasing.

11. 1 7 1( )

3 6

xf x

x

; domain 2x , range

7

3y

12. 1( ) 4 2f x x ; domain 4x , range 2y

13. 1 5( )

11 13

xf x

x

; domain 13

11x , range

5

11y

14. 1 1/7( ) ( 2)f x x ; domain: all real, range: all real

15.

2

1 1( )

1

xf x

x

; domain 1x and 1x ; range 0y and 1y

Exponential and Logarithmic Functions

Answers

1. 3x

2. 3x ln

3. 8x

4. 5/2x e

5. 0.757x

6. 2x ln and 5x ln

7. 2x

8. 23x e

9. /5

12 1

10yx

10. 3

ln2 3

yx

11. 3

7x

12. 73 1

2

ex

13. log5 log5

362log 6 log5log

5

x

14. 137

6x

15. 1

2x ; 2log(2 7) log3 log( 1) 2 9 7 3x x x x , or

1, 4

2x , but 4x is

extraneous (is not a solution to the original equation).

Derivatives of Exponential and Logarithmic Functions

Answers

1. 2

'ln 7 (2 5)

yx

, 5

2x

2. 2

1'

ln10 logy

x x

, 0x

3. 2

5'

ln10 ( 4) log ( 4)y

x x

, 4x

4. 2

2 32 5

6(ln )log log

ln 2 ln 5

xy x x ; then

12ln'

ln 2 ln 5

xy

x

5. cos

' cotsin

xy x

x , x n ( n an integer)

6. 2

3

15cos 5 sin5' 15tan5

cos 5

x xy x

x

, 5

2x n

( n an odd integer)

7. 1

'( 1)

yx x

, 0x

8. cos(ln ) cot(ln )

'sin(ln )

x xy

x x x , x n ( n an integer)

9. 66 xy e

10. 3 22 3 2 6(9 4 ) x xy x x e

11. 2 36' 2 )( xxy x e ; extrema at 2

2 4(0,0), ( , )

3 9e; 2 3'' (9 12 2) xy x x e , and

for 0x , " 0y means a local minimum; and for 2

3x , " 0y means a maximum.

12. 2

4

( )x xy

e e

13. cot cscx x xy e e e , xe n or ln( )x n with n a positive integer.

14. a. log( ) ln( )

ln10x x

x xy

e e so that

1 1 ln'

ln10

xx xy e

x

:

b. Extrema are located where 1 1 ln

' 0ln10

xx xy e

x

, i.e., where ln 1 0x x .

A calculator can be used to show 1.7632x .

15. 22

21

' ln 1xy ex

x

Integrals of Exponential and Logarithmic Functions

Answers

1. 4 4x xe dx e C

2. 2 3 2 32 x xe dx e C

3. 3 2

3 2 55

3ln 5

xx dx C

4. (7 2) 7 2x xe dx e x C

5. 6

5 (1 3 )(1 3 )

18

xx x e

e e dx C

6. 4

6 (1 5 ) 1 (1 5 )5 (1 5 )(1 5 )

ln 5 2 5

x xx x x dx C

7. xe C

8. /22 xe C

9. 21ln | 4 6 7 |

2x x C

10. ln | |x xe e C

11. ln 2

12. 21

ln13

13. ln ln x

14. Use ln , dx

u x dux

:

/2 /2

1 0

4cos(ln )4cos( ) 4[sin( / 2) sin(0)] 4

e xdx u du

x

.

15. 2 2 2

ln 5 ln 2

(ln (ln

3

4ln 4) 2) 2)(ln

x xdx d

xx dx

x x xx xx

; let ln 2u x , dx

dux

.

2 2

ln 5 3ln[ ln 2 ]

ln 2(ln

1 3

4ln 4)

xdx du

uxu x

xd C

xx u

, ln 2x

Exponential Growth and Decay

Answers

1. 2 Parts: a. 298.1 million b. 2067

2. Hint: use 11

ktA Ce

3. TBS

4. 14.4 C

5. 324 seconds, or about five and a half more minutes. 6. 7 grams 7. 2 Parts:

a. 210 W/m

b. 63 dB 8. 30 dB 9. 93 dB

10. If 3P is the population 3 years ago, then 5

350000 (1 0.05)P ; 3 39,176P people.

11.

14/1

20 252

T

where T is the half-life in days; 43.5T days.

12. 5$10000 (1 0.065) $7146V

13. 44 10R I ;

7.9 77.9 10 7.943 10R I ; The larger earthquake has an intensity

7943 times the smaller earthquake.

14. 8t days

15.

( / )1

lim lim 1 lim 1/

nt n r rt

n n n

rA P P

n n r

;

Let n

xr

, then

( / )1 1 1

lim 1 lim 1 lim 1/

rt rtn r rt x x

rt

n x xP P P Pe

n r x x

.

Inverse Trigonometric Functions

Answers

1. 1sin 0 0

2. arccos( 1)

3. 1tan ( 1)4

4. 1 5cot ( 3)

6

5. csc( 2)6

arc

6. 1 2 3sec

3 6

7. 2

arcsin( )2 4

8. 1 3

sin(arccos )4 4

9. 3 7

tan(arccos )4 3

10. 1 3 3 3sin(tan )

2 13

11. 1 15 15 161sec(csc )

8 161

12. 1 2 5cos(sin )

3 3

13. 1 3 3 13sin(tan )

2 13

14. 1 15 17sec(cot )

8 15

15. 1 5 12tan(cos )

13 5

Derivatives of Inverse Trigonometric Functions

Answers

1. 4

2

1x x

2. 2 1 2

1

(1 )(tan )x x

3. 1 2

1

(cos ) 1x x

4. 4

8

4

1

x

x

e

e

5. 4 2

2 ln

1 (ln )

x x x

x x

6. 2

1

2

36 tan

1

xx x

x

7. 1

2 (1 2 )x x

8. 2

6

6

1 4

t

t

9. 4

2

1

t

t

10. 2

2 2

21 6

cot (1 3 )1 (1 3 )

tt

t

11. 2

2 3/2(1 )

t

t

12. 2

4 2

( 1)

1

y

y y

13. 1 1 4 2

22

1 2(sin 2 tan ) 0 6 3 0

11

dx x x x

dx xx

; Use of Descarte’s rule

indicates there is 1 positive root, 1 negative root, and 2 complex roots. The two real roots

are 3 2 3 0.68, 3 2 3 0.68x , with ( 0.68) 0.45f (a maximum) and

(0.68) 0.45f (a minimum).

Figure 7.7.2

14. 2

1

1 x

15.

arcsin arcsin2

arcsin 22 2

3/22 2 2

21

11 2 10

1 1 1

x x

x

e xex

d e x xx x

dx x x x

Now, 2 21 0

2x x x ; but 0x is a constraint to satisfy the equation.

Checking the derivative: 0dy

dx for

2

2x and 0

dy

dx for

2

2x .

Therefore 2

0.6452

f

is a maximum for the function.

Figure 7.7.3

Integrals Involving Inverse Trigonometric Functions

Answers

1. 1sin3

xC

2. / 6

3. 2 11ln( 1) 3tan

2x x C

4. ln 2

5. 1 2 2

6. 1sec x C

7. 2 15 4ln(9 1) tan 3

18 3x x C

8.

3/2 21

0

tan 22 6

x

9. 2

17 7tan

14 7

xeC

10. 13tan 4x C TBS

11.

2

1 1

0

2sin sin 0.73

3 3

x

12. 2 1

2 2 2

2 24 2sin

24 4 4

x x xdx dx dx x C

x x x

13. 5

2 2 1 2

4 4 2

3 3 3 3 33 tan ( )

2 2 21 1 1

x x dudx x dx x x x C

x x u

14. 1

2 2 2

5 5 5 1 9 5 2sin

9 9 2 2 981 4 121

9

x xx

xx

e edx dx du e C

e ue

15.

2 1

2 2 2

2 5 2 4 9 9 2 2( 2ln( 4 6) tan

2 24 6 4 6 ( 2) 2

x x xdx dx dx x x C

x x x x x

Hyperbolic Functions

Answers

1. 0

2. 1

3. 0

4. 4

3

5. 5

4

6. 15

2 sinh 2 6sinh 12sinh(1.25) 19.22 2 6

dL h

h

meters

7. ln(5)x

8. ln(2)x

9. 3sinh cosh 1 3tanh 1 sechx x x x ;

Square both sides and expand: 2 2 2 2(3tanh 1) sech 1 tanh 9tanh 6tanh 1x x x x x ;

2tanh (5tanh 3) 0 tanh 0x x x or 3

tanh5

x ; tanh 0 0x x , but does not work!

3

tanh ln(2)5

x x , which works!

Solution is: ln(2)x .

10. sinh tan2

x xe ex y

means 2 2cosh 1 sinh 1 tan sec

2

x xe ex x y y

sinh cosh tan secxx x e y y means ln(sec tan )x y y .

11. 1 1 4

sechcosh 5 / 4 5

xx

12. 5

coth4

x

13.

27 25

cosh 124 24

x

14. cosh 2 11.5x , using 2 2cosh 2 1cosh cosh 2 2cosh 1

2

xx x x

15. 2csch coth 1 8 2 2x x ; choose negative because coth 0x

Derivatives and Integrals of Hyperbolic Functions

Answers

1. 2 312 cosh 4x x

2. 2 2

2

2 sinh sinh cosh cosh

2sinh

x x x x x

x

3. 8sinh 4 cosh 4x x

4. 3cosh3 2sinh( 7)x x

5. 6cosh3 sinh3x x

6. 3 2 3 2 2sech sech tanh sech sech (1 sech ) sech [2sech 1]x x x x x x x x

7. 2

2

6 2csch

1( 1) xx

8. cosh 6

12

xC

9. 1 sinh 2

2 2

xx C

10. 3 3

3 3

3 3

1 4 44 tanh3 4 4 ln ln 2cosh3

3 3 3

x xx x

x x

e e dux dx dx e e C x

ue e

4 4

4 tanh3 ln cosh3 ln 2 ln cosh33 3

x dx x C x C

11. 1

tanh4

x C

12. cothx x C

13. sinh 2 2x x C

14. 21

ln2 2

xx C

15. 2

2

1 csch3cosh3 csch 3x cosh3 coth3 csch3x

3sinh 3

xx dx x dx x dx C

x

Inverse Hyperbolic Functions

Answers

1. 22 2ln 1 ln 1

22

11

21sinh ln 1

2 2 2

x x x x x xe e xx x

x x x

2. 1sinh 0 0

3. 1cosh 0

does not exist, because the domain of 1cosh x

is 1x .

4. 1tanh 0 0

5. 1sinh 2 ln 2 5

6. 1cosh 1 0

7. 1 1 1tanh ln 3

2 2

8. 1 1

coth2 1

ex

e

9. 1 1 2

csch3 2 3( 1)

ex

e

10. 2cosh1 1

4 8

ex

e

11. 2sinh cosh 0 2tanh 1x x x , or 1 1 ln 3tanh

2 2x

.

12. 2sinh 53tanh 5sech 5 3 5 3sinh 5 5 1 sinh

cosh cosh

xx x x x

x x

The above is simplified to (8sinh 15)sinh 0x x , so that

1 15sinh ln 4 1.39

8x

, or 1sinh (0) 0x ;

13. 5

2 tanh3

xy

is a one-to-one function. The inverse is 13tanh ( 2) 5y x where

1x .

14. 2sinh( )y x is symmetric about the y-axis, line 0x , and is therefore not one-to-one. It

can be made one-to-one by restricting the domain to 0x . In this case, the inverse

function is 1sinhy x , where 0x .

15. 2sinh ( 4) 3y x is symmetric about the line 4x , and is therefore not one-to-one. It

can be made one-to-one by restricting the domain to 4x . In this case, the inverse

function is 14 sinh 3y x , where the domain is restricted to 3x .

Derivatives and Integrals Involving Inverse Hyperbolic

Functions

Answers

1. 1

2

8sinh 8

1 64

dx

dx x

2. 1 2

2

14cosh 7

49 1

d xx

dx x

for 7 1x

3. 1 2 1 2 1

2

6(tanh 3 ) (tanh 3 ) tanh 3

1 9

d xx x x x

dx x

for 3 1x

4. 1

2

csc cotcosh (csc ) csc

csc 1

d x xx x

dx x

5. 1

2

sintanh (cos )

1 cos

d xx csxx

dx x

for cos 1x

6. 2 1 2 1

2

3cosh 3 2cosh 3

9 1

x xde x e x

dx x

for 3 1x

7. 1 1 1 2

sech ( 3 )( 3)( 2)3 1 (3 ) 2 3

d xx

dx x xx x x

for 2 3x

8. Using the u-substitution b

u xa

results in:

1

2 2 2 2 2 2cosh

1 11

Adx A dx A a du A du A bx C

a a b b b ab x a u ubx

a

9. 1

2

44csch 3

1 9dx x C

x x

10.

2

1

2

cosh sinh sinh sinhcosh ln 1

3 3 3sinh 9

x x x xdx C C

x

11.

22

1

42

11

2 1 1 2tanh ( ) ln12 2 44 12

xx x

dx C Cx x

with 2 2x .

12. 1

2 2

3 3 3 2csch ( 1)

2 2( 1) 2 4 6 ( 1) 2[( 1) 2]dx dx x C

x x x x x

13. 1

2

1 1 4cosh

4 516 25dx x C

x

14. 1

2

1 1tanh

2 24

xx

x

edx e C

e

15. 1

2 2

cosh cosh 2 2tanh sinh 2

2 2sinh 4sinh 2 2 (sinh 2)

x xdx dx x C

x x x