1-7 Inverse Relations and Functions

Graph each function using a graphing calculator, and apply the horizontal line test to determine whether its inverse function exists. Write yes or no.

2. f (x) = x2 – 16x + 64

SOLUTION:

The graph of f (x) = x2 – 16x + 64 below shows that it is possible to find a horizontal line that intersects the graph of

f(x) more than once. Therefore, you can conclude that an inverse function does not exist.

4. f (x) = 3x − 8

SOLUTION:

It appears from the portion of the graph of f (x) = 3x − 8 shown below that there is no horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does exist.

6. f (x) = 4

SOLUTION: The graph of f (x) = 4 below shows that it is possible to find a horizontal line that intersects the graph of f (x) more than once. That horizontal line is the same as the graph of f (x) itself. Therefore, you can conclude that an inverse function does not exist.

8. f (x) = −4x2 + 8

SOLUTION:

The graph of f (x) = −4x2 + 8 below shows that it is possible to find a horizontal line that intersects the graph of f (x)

more than once. Therefore, you can conclude that an inverse function does not exist.

10. f (x) =

SOLUTION:

It appears from the portion of the graph of f (x) = shown below that there is no horizontal line that intersects

the graph of f (x) more than once. Therefore, you can conclude that an inverse function does exist.

12. f (x) = x3

SOLUTION:

It appears from the portion of the graph of f (x) = x3 shown below that there is no horizontal line that intersects the

graph of f (x) more than once. Therefore, you can conclude that an inverse function does exist.

Determine whether each function has an inverse function. If it does, find the inverse function and state any restrictions on its domain.

14. f (x) = 4x5 – 8x

4

SOLUTION:

The graph of f (x) = 4x5 – 8x

4 below shows that it is possible to find a horizontal line that intersects the graph of f (x)


16. f (x) =

SOLUTION:

It appears from the portion of the graph of f (x) = shown below that there is no horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does exist.

The function f has domain [–8, ) and range [0, ).

f−1

(x) = x2 − 8

From the graph of y = x2 – 8 below, you can see that the inverse relation has domain (– , ) and range [8, ).

By restricting the domain of the inverse relation to [0, ), the domain and range of f are equal to the range and

domain of f –1, respectively. Therefore, f

−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |

SOLUTION: The graph of f (x) = | x – 6 | below shows that it is possible to find a horizontal line that intersects the graph of f (x) more than once. Therefore, you can conclude that an inverse function does not exist.

20. g(x) =

SOLUTION:

It appears from the portion of the graph of g(x) = shown below that there is no horizontal line that intersects

the graph of g(x) more than once. Therefore, you can conclude that an inverse function does exist.

The function g has domain (– , 0) (0, ) and range [– , 1) (1, ).

g−1(x) =

From the graph y = below, you can see that the inverse relation has domain [– , 1) (1, ) and range (–

, 0) (0, ).

The domain and range of g are equal to the range and domain of g –1, respectively. Therefore, no further

restrictions are necessary. g−1

(x) = for x ≠ 1.

22. g(x) =

SOLUTION:

It appears from the portion of the graph of g(x) = shown below that there is no horizontal line that

intersects the graph of g(x) more than once. Therefore, you can conclude that an inverse function does exist.

The function g has domain (–3, ) and range (0, ).

g−1(x) = −3 +

From the graph y = −3 + below, you can see that the inverse relation has domain (– , 0) (0, ) and range

(–3, ).

By restricting the domain of the inverse relation to (0, ), the domain and range of g are equal to the range and

domain of g –1

, respectively. Therefore, g−1

(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:

It appears from the portion of the graph of h(x) = shown below that there is no horizontal line that intersects

the graph of h(x) more than once. Therefore, you can conclude that an inverse function does exist.

The function h has domain and range .

h−1

(x) =

From the graph y = below, you can see that the inverse relation has domain and range

.

The domain and range of h are equal to the range and domain of h –1, respectively. Therefore, no further

restrictions are necessary. h−1

(x) = for x ≠ .

26. SPEED The speed of an object in kilometers per hour y is y = 1.6x, where x is the speed of the object in miles per hour. a. Find an equation for the inverse of the function. What does each variable represent? b. Graph each equation on the same coordinate plane.

SOLUTION: a.

In the original equation y is speed in km/h and x = speed in mi/h. Then for the inverse y = speed in mi/h, x = speed in km/h b.

Show algebraically that f and g are inverse functions.

28. f (x) = 4x + 9

g(x) =

SOLUTION: f(x) and g(x) are inverses if f [g(x)] = g[f (x)] = x. Find each composition to show that f (x) and g(x) are inverses.

30. f (x) = + 8; x ≥ 0

g(x) =


Note that the domain restriction is needed because .

32. f (x) =

g(x) = − 8; x ≥ 0


Note that the domain restriction is needed to avoid the square root of a negative.

34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =


Use the graph of each function to graph its inverse function.

38.

SOLUTION: Graph the line y = x and reflect the points. Then connect the points with a smooth curve that resembles the original graph.

40.


42.


44. JOBS Jamie sells shoes at a department store after school. Her base salary each week is $140, and she earns a 10% commission on each pair of shoes that she sells. Her total earnings f (x) for a week in which she sold x dollars worth of shoes is f (x) = 140 + 0.1x.

a. Explain why the inverse function f −1(x) exists. Then find f −1

(x).

b. What do f −1(x) and x represent in the inverse function?

c. What restrictions, if any, should be placed on the domains of f (x) and f −1

(x)? Explain.

d. Find Jamie’s total sales last week if her earnings for that week were $220.

SOLUTION: a. The graph of the function is linear, so it passes the horizontal line test. Therefore, it is a one-to-one function and it has an inverse.

Thus, the inverse function is f −1

(x) = 10x − 1400.

b. In the original equation x represents her sales. Thus in the inverse function, x represents Jamie’s earnings for a

week, and f −1(x) represents her sales.

c. x ≥ 0; Jamie cannot have negative sales. d.

Determine whether each function has an inverse function.

46.

SOLUTION: The graph does not pass the Horizontal Line Test.

48.


Determine if f −1 exists. If so, complete a table for f −1.

50.

SOLUTION: No output value corresponds with more than one input value, so an inverse exists.

52.

SOLUTION: An output value corresponds with more than one input value, so an inverse does not exist.

54. TEMPERATURE The formula f (x) = x + 32 is used to convert x degrees Celsius to degrees Fahrenheit. To

convert x degrees Fahrenheit to Kelvin, the formula k(x) = (x + 459.67) is used.

a. Find f −1. What does this function represent?

b. Show that f and f −1 are inverse functions. Graph each function on the same graphing calculator screen.

c. Find [k f ](x).What does this function represent? d. If the temperature is 60°C, what would the temperature be in Kelvin?

SOLUTION: a.

f −1 represents the formula used to convert degrees Fahrenheit to degrees Celsius.

b.

c.

[k f ](x) = x + 273.15; represents the formula used to convert degrees Celsius to degrees Kelvin.d.

Restrict the domain of each function so that the resulting function is one-to-one. Then determine the inverse of the function.

56.

SOLUTION: The domain needs to be restricted so that the graph passes the Horizontal Line Test. There is more than one way to

accomplish this. Sample answer: x ≤ −9

We are selecting the negative side of the domain, so we will replace the absolute value symbols with the opposite of the expression.

f −1 (x) = x – 11

Since the range of the restricted function f is y ≤ –2, we must restrict the inverse relation so that the domain is x ≤

2. Therefore, f −1 (x) = x – 11.

58.


accomplish this. Sample answer: x ≥ −5

We are selecting the positive side of the domain, so we will replace the absolute value symbols with the expression.

f−1

(x) = x – 1

Since the range of the restricted function f is y ≥ –4, we must restrict the inverse relation so that the domain is x ≥ –

4. Therefore, f −1 (x) = x – 1, x ≥ –4.

State the domain and range of f and f −1, if f −1 exists.

60. f (x) = x2 + 9

SOLUTION:

f −1 does not exist

62. f (x) =

SOLUTION: Graph f .

There appears to be asymptotes at x = 3 and y = 4. f: D = x | x 3, x , R = y | y 4, x

The domain and range restrictions of f correspond with the range and domain restrictions of f−1

. Therefore, no further restrictions are needed.

f −1: D = x | x 4, x , R = y | y 3, x

64. FLOWERS Bonny needs to purchase 75 flower stems for banquet decorations. She can choose between lilies and hydrangea, which cost $5.00 per stem and $3.50 per stem, respectively. a. Write a function for the total cost of the flowers. b. Find the inverse of the cost function. What does each variable represent? c. Find the domain of the cost function and its inverse. d. If the total cost for the flowers was $307.50, how many lilies did Bonny purchase?

SOLUTION: a. Let x represent the number of stems of hydrangea. b.

c(x) = 3.5x + 5(75 − x).

c −1(x) = 250 − ; x represents the total cost and c

−1(x) represents the number of stems of hydrangea

c. Bonny is not purchasing more than 75 stems. Therefore, the domain of c(x) is x | 0 ≤ x ≤ 75, x ∈ . The range of c(x) is from c(0) to c(75).

The domain of c −1

(x) is equal to the range of c(x), or x | 262.5 ≤ x ≤ 375, x ∈

d.

Remember that x is the number of hydrangea. The number of lilies is 75 − 45 or 30.

Use f (x) = 8x – 4 and g (x) = 2x + 6 to find each of the following.

69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:

Now, find the inverse.

73. (f · g)−1

(x)

SOLUTION:

Now, set this equal to y find the inverse.

The domain of the inverse function equals the range of the original function. The function (f · g)(x) has a range of [–1.25, ∞]. Therefore, the domain of (f · g)-1(x) is x ≥ –1.25.

Use f (x) = x2 + 1 with domain [0, ∞) and g (x) = to find each of the following.

75. [f −1 g−1](x)

SOLUTION: First, find the inverse of f (x) and g(x).

Use the inverses to simplify the original expression.

The domain of a composition is all the x-values in the domain of the inner function whose range values are in the domain of the outer function.

The domain of g−1(x) equals the range of g(x), which is [0, ∞).

Over this domain, the range of g−1(x) is [4, ∞).

The domain of f −1(x) is x – 1 ≥ 0 or [1, ∞) .

All the range values of g-1(x) are contained in the domain of f −1(x).

Therefore, [f −1 g−1](x) = for x ≥ 0.

77. [f g]−1

(x)

SOLUTION: First, find [f g](x).

Now, find the inverse of the composition.

The domain of [f g]

–1(x) equals the range of [f g](x).

First, find the domain of g(x) which is x – 4 ≥ 0, or x ≥ 4. Then the range of g(x) is y ≥ 0. All members of the range are acceptable domain values for f (x), which is [0, ∞).

For x ≥ 4 the range of [f g](x) = x – 3 is y ≥ 1. Thus, the domain of [f g]–1

(x) is x ≥ 1.

Therefore, [f g]–1

(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)

Start by finding the inverse of g(x).

Next, find the product.

The domain of the product is the intersection of the domains of f (x) and g–1(x).

The domain of f (x) is [0, ∞). The domain of g–1(x) equals the range of g(x). Since the range of

is y ≥ 0, the domain of g–1(x) is x ≥ 0. The intersection of is [0, ∞) and x ≥ 0 is x ≥ 0. So, the domain of (f · g−1

) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.

81. COPIES Karen’s Copies charges users $0.40 for every minute or part of a minute to use their computer scanner. Suppose you use the scanner for x minutes, where x is any real number greater than 0. a. Sketch the graph of the function, C(x), that gives the cost of using the scanner for x minutes. b. What are the domain and range of C(x)? c. Sketch the graph of the inverse of C(x). d. What are the domain and range of the inverse? e . What real-world situation is modeled by the inverse?

SOLUTION: a. For every integer value of time (x) the function should have a point at 0.4x. For non integer values of x, the function will be rounded up to the next integer and then multiplied by 0.4. This will look like a step function with y-values at multiples of 0.4, with closed circles at the end of each step.

We can use C(x) = 0.4[[(x + 1)]] to represent our cost. However, if x is an integer, we cannot add one. For example: C(0.7) = 0.4([[0.7]] + 1] = 0.4(0 + 1) = 0.4 C(1.0) = 0.4([[1]]) = 0.4(1) = 0.4 C(1.01) = 0.4([[1.01]] + 1) = 0.4(1 + 1) = 0.8 Therefore, we need to use a piecewise function.

b. The minutes are rounded up, so the domain will consist of whole numbers. D=x | x , R= y | y positive multiples of 0.4 c. To graph the inverse, interchange the axes.

d. D= x| x positive multiples of 0.4 R = y | y e . The inverse gives the number of possible minutes spent using the scanner that costs x dollars.

eSolutions Manual - Powered by Cognero Page 1

1-7 Inverse Relations and Functions


2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.









2. f (x) = x2 – 16x + 64

SOLUTION:



4. f (x) = 3x − 8

SOLUTION:


6. f (x) = 4


8. f (x) = −4x2 + 8

SOLUTION:



10. f (x) =

SOLUTION:



12. f (x) = x3

SOLUTION:




14. f (x) = 4x5 – 8x

4

SOLUTION:




16. f (x) =

SOLUTION:



f−1

(x) = x2 − 8




−1(x) = x

2 − 8 for x ≥ 0.

18. f (x) = | x – 6 |


20. g(x) =

SOLUTION:




g−1(x) =


, 0) (0, ).



(x) = for x ≠ 1.

22. g(x) =

SOLUTION:




g−1(x) = −3 +


(–3, ).


domain of g –1


(x) = −3 + for x > 0.

24. h(x) =

SOLUTION:




h−1

(x) =


.



(x) = for x ≠ .


SOLUTION: a.



28. f (x) = 4x + 9

g(x) =


30. f (x) = + 8; x ≥ 0

g(x) =



32. f (x) =

g(x) = − 8; x ≥ 0



34. g(x) = + 5

f (x) = x2 – 10x + 33; x ≥ 5



36. f (x) =

g(x) =



38.


40.


42.




(x).



(x)? Explain.




(x) = 10x − 1400.





46.


48.



50.


52.







SOLUTION: a.


b.

c.



56.




f −1 (x) = x – 11


2. Therefore, f −1 (x) = x – 11.

58.




f−1

(x) = x – 1


4. Therefore, f −1 (x) = x – 1, x ≥ –4.


60. f (x) = x2 + 9

SOLUTION:


62. f (x) =

SOLUTION: Graph f .




f −1: D = x | x 4, x , R = y | y 3, x



c(x) = 3.5x + 5(75 − x).






d.



69. [f –−1 g−1](x)

SOLUTION:

71. [f g]−1

(x)

SOLUTION:


73. (f · g)−1

(x)

SOLUTION:




75. [f −1 g−1](x)









77. [f g]−1

(x)



The domain of [f g]




(x) is x ≥ 1.


(x) = x + 3 for x ≥ 1.

79. (f · g−1) (x)

SOLUTION:

(f · g−1) (x) = f (x) · g−1

(x)






) (x) is x ≥ 0.

Therefore, (f · g

−1) (x) = x

4 + 5x

2 + 4 for x ≥ 0.








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1-7 Inverse Relations and Functions