1-6 Function Operations and Composition of...

Find (f + g )(x), (f – g )(x), (f ⋅ g )(x), and (x) for each f (x) and

g (x). State the domain of each new function.

1. f (x) = x2 + 4

g(x) =

SOLUTION:

D = [0, ) for all of the functions except , for which

.

2. f (x) = 8 – x3 g(x) = x – 3

SOLUTION:

D = (− , ) for all of the functions except , for which

.

3. f (x) = x2 + 5x + 6 g(x) = x + 2

SOLUTION:


.

4. f (x) = x – 9 g(x) = x + 5

SOLUTION:


.

5. f (x) = x2 + x g(x) = 9x

SOLUTION:


. Even though there appears to be no restriction in

the simplified function, there is in the original.

6. f (x) = x – 7 g(x) = x + 7

SOLUTION:


.

7. f (x) =

g(x) = x3 + x

SOLUTION:

for all functions. Even though there appears to be

no restriction in the simplified function (f · g)(x), there is in the original.

8. f (x) =

g(x) =

SOLUTION:

for all of these functions. Even though there

appears to be no restriction in the simplified function (f · g)(x) or

, there is in the originals.

9. f (x) =

g(x) = 4

SOLUTION:

D = (0, ) for all of these functions. Even though there appears to be no restriction in the simplified function (f · g)(x), there is in the original.

10. f (x) =

g(x) = x4

SOLUTION:


appears to be no restriction in the simplified function (f · g)(x), there is in the original.

11. f (x) =

g(x) = − 3

SOLUTION:

For (f + g)(x), (f − g)(x), and (f · g)(x), D = [−5, ). For ,

12. f (x) =

g(x) =

SOLUTION:

For (f + g)(x), (f − g)(x), and (f · g)(x), D = [4, ). For , D =

(4, ).

13. BUDGETING Suppose a budget in dollars for one person for one month is approximated by f (x) = 25x + 350 and g(x) = 15x + 200, where f is the cost of rent and groceries, g is the cost of gas and all other expenses, and x = 1 represents the total cost at the end of the firstweek. a. Find (f + g)(x) and the relevant domain. b. What (f + g)(x) represent? c. Find (f + g)(4). What does this value represent?

SOLUTION: a.

We are dealing with time, so the relevant domain is x ≥ 0. b. (f + g)(x) all factors that influence the budget. c.

The budget for 4 weeks.

14. PHYSICS Two different forces act on the movement of an object being pushed across a floor: the force of the person pushing the object and the force of friction. If W is work in joules, F is force in newtons,

and d is displacement of the object in meters, Wp(d) = Fp d describes

the work of the person and Wf (d) = Ff d describes the work created by

friction. The increase in kinetic energy necessary to move the object is

the difference between the work done by the person Wp and the work

done by friction Wf .

a. Find (Wp – Wf )(d).

b. Determine the net work expended when a person pushes a box 50 meters with a force of 95 newtons and friction exerts a force of 55 newtons.

SOLUTION: a.

b. Substitute d = 50, Fp = 95, and Ff = 55 into the expression d(Fp −

Ff ).

The net work expended is 200 newtons per meter.

For each pair of functions, find [ f o g ](x), [g o f ](x), and [ f o g ](6).

15. f (x) = 2x – 3 g(x) = 4x – 8

SOLUTION:

16. f (x) = –2x2 – 5x + 1 g(x) = –5x + 6

SOLUTION:

17. f (x) = 8 – x2

g(x) = x2 + x + 1

SOLUTION:

18. f (x) = x2 – 16

g(x) = x2 + 7x + 11

SOLUTION:

19. f (x) = 3 – x2

g(x) = x3 + 1

SOLUTION:

20. f (x) = 2 + x4

g(x) = −x2

SOLUTION:

Find f o g .

21. f (x) =

g(x) = x2 – 4

SOLUTION:

To find f ◦ g, you must first be able to find g(x) = x2 – 4, which can be done for all real numbers. Then you must be able to evaluate f (x) =

for each of these g(x)-values, which can only be done when g(x)

≠ –1. This means that we must exclude from the domain those values

for which x2 – 4 = –1.

Therefore, the domain of f ◦ g is {x| x ≠ ± , x R}. Now find [f ◦ g](x).

Therefore, for x ≠ ± .

22. f (x) =

g(x) = x2 + 6

SOLUTION:

To find f ◦ g, you must first be able to find g(x) = x2 + 6, which can be

done for all real numbers. Then you must be able to evaluate f (x) =

for each of these g(x)-values, which can only be done when g

(x) ≠ 3. This means that we must exclude from the domain those values

for which x2 + 6 = 3.

Since is not a real number, there are no x-values in the domain of

g such that x2 + 6 = 3. This means that we do not need to restrict the

domain of f ◦ g; the domain of f ◦ g is all real numbers. Now find [f ◦ g](x).

Therefore, .

23. f (x) =

g(x) = x2 – 4

SOLUTION:

To find f ◦ g, you must first be able to find g(x) = x2 – 4, which can be


for each of these g(x)-values, which can only be done when g(x) ≥ –4. This means that we must exclude from the domain those

values for which x2 – 4 < –4.

Since x2 will never be less than 0, we do not need to restrict the domain

of f ◦ g. The domain of f ◦ g is all real numbers. Now find [f ◦ g](x).

Therefore, .

24. f (x) = x2 − 9

g(x) =

SOLUTION:

To find f ◦ g, you must first be able to find g(x) = , which can

only be done for x ≥ –3. Then you must be able to evaluate f (x) = x2 −

9 for each of these g(x)-values, which can be done for all real numbers.

Therefore, the domain of f ◦ g is {x| x ≥ −3, x R}. Now find [f ◦ g](x).

Therefore for x≥ −3.

25. f (x) =

g(x) =

SOLUTION:


only be done for x ≤ 6. Then you must be able to evaluate f (x) = for

each of these g(x)-values, which can only be done when g(x) ≠ 0. This means that we must also exclude from the domain those values for

which = 0.

Therefore, x ≠ 6. Combining these restrictions, the domain of [f o g](x),is {x| x < 6, x R}. Now find [f ◦ g](x).

Therefore, for x < 6.

26. f (x) =

g(x) =

SOLUTION:

To find f ◦ g, you must first be able to find g(x) = , which can only be done for x ≥ –8. Then you must be able to evaluate f (x) =


≠ 0. This means that we must also exclude from the domain those

values for which = 0.

Therefore, x ≠ –8. Combining these restrictions, the domain of [f o g](x), is {x| x > −8, x R}. Now find [f ◦ g](x).

Therefore, for x > −8.

27. f (x) =

g(x) = x2 + 4x – 1

SOLUTION:

To find f ◦ g, you must first be able to find g(x) = x2 + 4x – 1, which

can be done for all real numbers. Then you must be able to evaluate f

(x) = for each of these g(x)-values, which can only be done when g(x) ≥ –5. This means that we must also exclude from the

domain those values for which x2 + 4x – 1 < –5.

Since (x + 2)2 will never be less than 0, we do not need to restrict the

domain of f ◦ g. The domain of f ◦ g is all real numbers. Now find [f ◦ g](x).

Therefore, .

28. f (x) =

g(x) = x2 + 8

SOLUTION:



for each of these g(x)-values, which can only be done when g(x) ≥ 2. This means that we must also exclude from the domain those

values for which x2 + 8 < 2.

Since x2 will never be less than –6, we do not need to restrict the


Therefore, .

29. RELATIVITY In the theory of relativity, m(v) = , where c

is the speed of light, 300 million meters per second, and m is the mass ofa 100-kilogram object at speed v in meters per second. a. Are there any restrictions on the domain of the function? Explain their meaning. b. Find m(10), m(10,000), and m(1,000,000). c. Describe the behavior of m(v) as v approaches c. d. Decompose the function into two separate functions.

SOLUTION: a. {v | 0 ≤ v < c, v ∈R}; The speed of the object v cannot be equal to

the speed of light c. Otherwise, we would obtain , which is

undefined. Also, the speed v cannot be greater than c. Otherwise, we would have to find the square root of a negative number, which is an imaginary number. Lastly, the speed cannot be less than zero because speed cannot be negative. b.

c. Test values closer to c.

As v approaches c, m(v) approaches ∞.

d. Sample answer: Let and

Find two functions f and g such that h(x) = [f o g ](x). Neither function may be the identity function f (x) = x.

30. h(x) = + 7

SOLUTION:

Sample answer: f (x) = + 7; g(x) = 4x + 2

31. h(x) = – 8

SOLUTION:

Sample answer: f (x) = − 8; g(x) = x + 5

32. h(x) = | 4x + 8 | − 9

SOLUTION:

Sample answer: f (x) = | x | − 9; g(x) = 4x + 8

33. h(x) = [[−3(x – 9)]]

SOLUTION: Sample answer: f (x) = [[−3x]]; g(x) = x – 9

34. h(x) =

SOLUTION:

Sample answer: f (x) = ; g(x) =

35.

SOLUTION:

Sample answer: f (x) = x3; g(x) = + 4

36. h(x) =

SOLUTION:

Sample answer: f (x) = ; g(x) = x + 2

37. h(x) =

SOLUTION:

Sample answer: f (x) = ; g(x) = x – 5

38. h(x) =

SOLUTION:


39. h(x) =

SOLUTION:

Sample answer: f (x) = ; g(x) = x – 1

40. QUANTUM MECHANICS The wavelength λ of a particle with mass m kilograms moving at v meters per second is represented by λ =

, where h is a constant equal to 6.626 · 10–34

.

a. Find a function to represent the wavelength of a 25-kilogram object as a function of its speed. b. Are there any restrictions on the domain of the function? Explain their meaning. c. If the object is traveling 8 meters per second, find the wavelength in terms of h. d. Decompose the function into two separate functions.

SOLUTION: a.

b. The value of v must be greater than 0. The speed of the object cannot be equal to zero or negative. If it is, then wavelength is negative or the function is undefined. c.

d. Sample answer: Notice that the function involves division and multiplication. So one way to write f(v) as a composition of two functions is to write one function involving multiplication and one

involving division. a(v) = , b(v) = 25v, and f (v) = a[b(v)] = .

41. JOBS A salesperson for an insurance agency is paid an annual salary plus a bonus of 4% of sales made over $300,000. Let f (x) = x – 300,000and h(x) = 0.04x. a. If x is greater than $300,000, is the bonus represented by f [h(x)] or by h[f (x)]? Explain your reasoning. b. Determine the amount of bonus for one year with sales of $450,000.

SOLUTION: a. h[f (x)]; The bonus is found after subtracting 300,000 from the total sales. b.

Therefore, the amount of bonus for one year with sales of $450,000 is $6,000. [You need a sentence like this to get the units on the answer.]

42. TRAVEL An airplane flying 0.5 mile above a landing strip at 275 milesper hour passes a control tower at time t = 0 hours.

a. Find the distance d between the airplane and the control tower as a function of the horizontal distance h from the control tower to the plane.b. Find d as a function of time t. c. Find d o h. What does this function represent? d. If the plane continued to fly the same distance from the ground, how far would the plane be from the control tower after 10 minutes?

SOLUTION: a. The plane is consistently 0.5 miles above ground, so we can use the Pythagorean theorem and h to find d. b. h(t) = 275t

c. [d o h](t) = ; d o h represents the distance of the plane from the control tower as a function of time.

d. Time is 10 minutes, or hour.


43.

SOLUTION:

Let g(x) represent the more complicated part of the function, . Therefore, we have f (x) = x + c, where c is some expression that will

convert g(x) to .

Focusing strictly on the x, we can simplify f (x) to f (x) = x − . Now,

we need to find a way to convert to x. To do this, solve for c.

Finally, replace c with x.

Sample answer: f (x) = x − ; g(x) =

Check your solution.

44.

SOLUTION:

Let g(x) represent the more complicated part of the function, .Therefore, we have f (x) = x + c, where c is some expression that will

convert g(x) to .

Focusing strictly on the x, we can simplify f (x) to f (x) = x + . Now,


Finally, replace c with x and simplify .

Sample answer: f (x) = x + , g(x) = .


45.

SOLUTION:

Let g(x) represent the more complicated part of the function, x2.

Therefore, we have f (x) = + c where c is some expression that

will convert g(x) to 5x. Focusing strictly on the x, we can simplify f (x) to f (x) = x + 5c. Now,

we need to find a way to convert x2 to . To do this, take the square

root. Replace c with .

Sample answer: f (x) = + 5 ; g(x) = x2


46. h(x) = + 9x

SOLUTION:

Let g(x) represent the more complicated part of the function, −7x.

Therefore, we have f (x) = + c, where c is some expression that will convert g(x) to 9x. Focusing strictly on the x, we can simplify f (x) to f (x) = x + 9c. Now,

we need to find a way to convert −7x to x. To do this, multiply by .

Replace c with .

Sample answer: ; g(x) = −7x


47.

SOLUTION: Let g(x) be some function, such as g(x) = x + 4. Analyze the first term

of h(x), . The first term of f (x) will need to be in the form of a

fraction in order to convert g(x) to . For g(x) to equal the

numerator x, 4 would have to be subtracted from g(x), or g(x) − 4. Thus, the numerator of the first term of f (x) is x − 4 since g(x) − 4 = x or (x + 4) − 4 = x. The denominator of the first term of f (x) must result in g(x), or x + 4, converting to 2x − 1. To arrive at 2x, we must multiply g(x) by 2 or 2 · (x + 4) = 2x + 8. If we subtract 9 from this expression, we arrive at 2x + 8 − 9 or 2x − 1. Thus, we multiplied g(x) by 2 and then subtracted 9 or 2g(x) − 9. So, the denominator of the first term of f

(x) is 2x − 9. The first term of f (x) is .

Now analyze the second term of h(x), . The second term of f

(x) will need to be in the form of a fraction inside a radical in order to

convert g(x) to . Since the numerator does not have an x term, the

numerator of the fraction inside the radical for f (x) is 4. For g(x) to equal the denominator x, 4 would have to be subtracted from g(x), or g

(x) − 4. Thus, the denominator of the fraction inside the radical for f (x) is x − 4 since g(x) − 4 = x or (x + 4) − 4 = x. Thus, the second term of f

(x) is .

Sample answer: f (x) = + ; g(x) = x + 4


48. h(x) = +




numerator x2 − 4, we have to square g(x), (x + 2)2 or x2 + 4x + 4. We

now have the first term x2, but have to eliminate the 4x. If we multiply

g(x) by −4, we get −4(x + 2) or −4x − 8. Adding the two expressions

together, we arrive at (x2 + 4x + 4) + (−4x − 8) or x2 − 4. Therefore,

the numerator of the fraction of the first term of f (x) is x2 − 4x. For g

(x) to equal the denominator x, 2 would have to be subtracted from g

(x), or g(x) − 2. Thus, the denominator of the fraction for the first term of f (x) is x − 2 since g(x) − 2 = x or (x + 2) − 2 = x. So, the first term

of f (x) is .

Now analyze the second term of h(x), . The second term of f (x)

will need to be in the form of a fraction in order to convert g(x) to

. The numerator of the second term of f (x) must result in g(x),

or x + 2, converting to 3x − 5. To arrive at 3x, we must multiply g(x) by3 or 3 · (x + 2) = 3x + 6. If we subtract 11 from this expression, we arrive at 3x + 6 − 11 or 3x − 5. Thus, we multiplied g(x) by 3 and then subtracted 11 or 3g(x) − 11. So, the numerator of the second term of f(x) is 3x − 5. The denominator of the second term of f (x) must result in g(x), or x + 2, converting to 5x. To arrive at 5x, we must multiply g(x) by 5 or 5 · (x + 2) = 5x + 10. If we subtract 10 from this expression, wearrive at 5x + 10 − 10 or 5x. Thus, we multiplied g(x) by 5 and then subtracted 10 or 5g(x) − 10. So, the denominator of the second term of

f(x) is 5x − 10. The second term of f (x) is .



Use the given information to find f (0.5), f (−6), and f (x + 1). Round to the nearest tenth if necessary.

49. f (x) – g(x) = x2 + x – 6, g(x) = x + 4

SOLUTION:

50. f (x) + g(x) = + – , g(x) = 2x

SOLUTION:

51. g(x) – f (x) + = 9x2 + 4x, g(x) =

SOLUTION:

52. g(x) = f (x) – 18x2 + , g(x) =

SOLUTION:

Find [f o g o h](x).53. f (x) = x + 8

g(x) = x2 – 6

h(x) = + 3

SOLUTION:

The domain of h(x) = + 3 is {x| x ≥ 0, x R}. To evaluate g o h ,

you must be able to evaluate g(x) = x2 – 6 for each of these h(x)-

values, which is possible for all real h(x)-values. Therefore the domain

of g o h is {x| x ≥ 0, x R}. Now find .

To find f o g o h, you must be able to evaluate f (x) = x + 8, for each of these g[h(x)]-values, which can be done for all real g[h(x)]-values.

Therefore, the domain of f ◦ g ◦ h is {x| x ≥ 0, x R}. Now find [f o g o h](x).

Therefore, for x ≥ 0.

54. f (x) = x2 – 2 g(x) = 5x + 12

h(x) =

SOLUTION:

The domain of h(x) = is {x| x ≥ 0, x R}. To evaluate g o h , you must be able to evaluate g(x) = 5x + 12 for each of these h(x)-values, which is possible for all real h(x)-values. Therefore the domain


To find f o g o h, you must be able to evaluate f (x) = x2 – 2, for each ofthese g[h(x)]-values, which can be done for all real g[h(x)]-values.



55. f (x) =

g(x) = x2 – 3

h(x) =

SOLUTION:

The domain of h(x) = is {x| x 0, x R}. To evaluate g o h , you

must be able to evaluate g(x) = x2 – 3 for each of these h(x)-values,

which is possible for all real h(x)-values. Therefore the domain of g o his {x| x 0, x R}. Now find .

To find f o g o h, you must be able to evaluate use f (x) = , for each of these g[h(x)]-values, which can only be done for g[h(x)] ≥ –5.

That is when ≥ –5

Notice that x2 will never be less than or equal to a negative number, so

the composition of f with g[h(x)] add no additional restriction to the

domain of f ◦ g ◦ h. Therefore, the domain of f ◦ g ◦ h is {x| x 0, xR}. Now find [f o g o h](x).

Therefore, = for x 0.

56. f (x) =

g(x) = x2 – 4x + 1

h(x) = x + 2

SOLUTION:

The domain of h(x) = x + 2 is all real numbers. To evaluate g o h, you

must be able to evaluate g(x) = x2 – 4x + 1 for each of these h(x)-


of g o h is all real numbers. Now find .

To find f o g o h, you must be able to evaluate use f (x) = , for each

of these g[h(x)]-values, which can only be done for g[h(x)] ≠ 0. That is

when ≠ 0. This means that we must exclude from the domain

those values for which = 0.

Therefore, the domain of f ◦ g ◦ h is {x| x ≠ , x R}. Now find [f o g o h](x).

Therefore, = for x ≠ .

57. If f (x) = x + 2, find g(x) such that:a. (f + g)(x) = x2 + x + 6

b. (x) =

SOLUTION: a.

b.

58. If f (x) = , find g(x) such that:a. [f o g](x) = |6 x| b. [g o f ](x) = 200x + 25

SOLUTION: a.

b.

Let y = 2 . Find x in terms of y .

Substitute for x and 2 .

Therefore, g(x) = 50x2 + 25.

59. If f (x) = 4x2, find g(x) such that:a. (f · g)(x) = x

b.

SOLUTION: a.

b.

60. INTEREST An investment account earns interest compounded quarterly. If x dollars are invested in an account, the investment I(x) after one quarter is the initial investment plus accrued interest or I(x) = 1.016x. a. Find [I o I](x), [I o I o I](x), and [I o I o I o I](x). b. What do the compositions represent? c. What is the accounts annual percentage yield?

SOLUTION: a.

b. The compositions represent the compounded interest for 6 months, 9 months, and 1 year.

c. The annual percentage yield is the percentage growth after one year,in this case about 6.6%.

Use the graphs of f (x) and g (x) to find each function value.

61. (f + g)(2)

SOLUTION:

62. (f – g)(−6)

SOLUTION:

63. (f · g)(4)

SOLUTION:

64. (−2)

SOLUTION:

65. [f o g](−4)

SOLUTION:

66. [g o f ](6)

SOLUTION:

67. CHEMISTRY The average velocity v(m) of gas molecules at 30°C in

meters per second can be represented by ,

where m is the molar mass of the gas in kilograms per mole. a. Are there any restrictions on the domain of the function? Explain their meaning. b. Find the average speed of 145 kilograms per mole gas molecules at 30°C. c. How will the average velocity change as the molar mass of gas increases? d. Decompose the function into two separate functions.

SOLUTION: a. {m| m > 0, m R}; The molar mass of the gas cannot be negative or zero. b.

c. As the denominator increases, the value under the radical sign will also decrease. Likewise, as a value decreases, the square root of that value will also decrease. Thus, the velocity will decrease.

d. Sample answer: v(m) = f [g(x)]; f (m) = ; g(m) =

.

Find functions f , g , and h such that a(x) = [f o g o h](x).

68.

SOLUTION: With this type of problem, look for the first thing that must be done to a value when it is plugged into the composition function. This will usually represent the innermost function in the composition, in this case, h(x).

The innermost expression in the composition is , so let h(x) = x

− 7. Now we have [f o g o h](x) = [f o g](h(x)) = [ + 4]2.

The next inner expression is the value that is being squared. Let g(x) =

+ 4. Finally, let f (x) = x2.

Therefore, f (x) = x2; g(x) = + 4; h(x) = x – 7

Check the solution.

69.

SOLUTION: With this type of problem, look for the first thing that must be done to a value when it is plugged into the composition function. This will usually represent the innermost function in the composition, in this case, h(x). The innermost expression in the composition is the term under the

radical sign, so let h(x) = x − 5. Now we have [f o g o h](x) = [f o g](h

(x)) = .

The next inner expression is the expression left under he radical sign.

Let g(x) = x2 + 8. Finally, let f (x) = .

Therefore, f (x) = ; g(x) = x2 + 8; h(x) = x – 5

Check the solution.

70. a(x) =

SOLUTION: With this type of problem, look for the first thing that must be done to a value when it is plugged into the composition function. This will usually represent the innermost function in the composition, in this case, h(x). The innermost expression is x − 3, so let h(x) = x − 3. Now we have [f

o g o h](x) = [f o g](h(x)) = . The next inner expression is x2 +

4, so let g(x) = x2 + 4. Finally, let f (x) = .


Check the solution.

71.


The innermost expression is + 3, so let this be h(x). Now we have

[f o g o h](x) = [f o g](h(x)) = . The next inner expression is x2

+ 1, so let this be g(x). Finally, let f (x) = .

Therefore, f (x) = ; g(x) = x2 + 1; h(x) = + 3.

Check the solution.

For each pair of functions, find f o g and g o f .

72. f (x) = x2 – 6x + 5

g(x) = + 3

SOLUTION:

To find f o g, you must first be able to evaluate g(x) = + 3. Since we cannot take the square root of a negative number, x + 4 ≥ 0,

which implies x ≥ –4. Then you must be able to evaluate f (x) = x2 – 6x + 5 for each of these g(x)-values, which can be done for all real g(x)-

values. Therefore, the domain of f o g is {x| x ≥ −4, x R}. Now find [f o g](x).

Therefore, [f o g](x) = x for x ≥ −4.

To find g o f , you must first be able to evaluate f (x) = x2 – 6x + 5,

which can be done for all real numbers. Then you must be able to

evaluate g(x) = + 3, for each of these f (x)-values, which can only be done when g(x) ≥ –4. This means that we must exclude from

the domain those values for which x2 – 6x + 5< –4.

Since (x – 3)2 will never be less than 0, we do not need to restrict the

domain of g ◦ f . The domain of g ◦ f is all real numbers. Now find [g o f](x).

Therefore, [g o f ](x) = .

73. f (x) = x2 + 8x – 3

g(x) = – 4

SOLUTION:

To find f o g, you must first be able to evaluate g(x) = – 4. Since we cannot take the square root of a negative number, x + 19 ≥ 0,

which implies x ≥ –19. Then you must be able to evaluate f (x) = x2 +

8x – 3 for each of these g(x)-values, which can be done for all real g(x)-values. Therefore, the domain of f o g is {x| x ≥ −19, x R}. Now find [f o g](x).


To find g o f , you must first be able to evaluate f (x) = x2 + 8x – 3, which can be done for all real numbers. Then you must be able to

evaluate g(x) = – 4, for each of these f (x)-values, which can only be done when f (x) ≥ –19. This means that we must exclude from

the domain those values for which x2 + 8x – 3 < –19.

Since (x + 4)2 will never be less than –16, we do not need to restrict the



74. f (x) =

g(x) =

SOLUTION:

To find f o g, you must first be able to evaluate g(x) = ,

which can be done for all real numbers since 16 + x2 will never be

negative. Then you must be able to evaluate f (x) = for each of these g(x)-values, which is true for all real g(x)-values, since g(x) will

always be greater than –6. Therefore, the domain of f o g is all real numbers. Now find [f o g](x).

Therefore, .

To find g o f , you must first be able to evaluate f (x) = , which can only be done for x ≥ –6. Then you must be able to evaluate g(x) =

, for each of these f (x)-values, which can be done for all f

(x)-values. Therefore initially, it appears that the domain of g ◦ f is {x| x≥ −6, x R}. Now find [g o f ](x).

Notice that is only defined for x ≥ –22. Combining this restriction with the restriction the x ≥ –6, the true domain of g ◦ f is {x|

x ≥ −22, x R}. Therefore, [g o f ](x) = for x ≥ −22.



1. f (x) = x2 + 4

g(x) =

SOLUTION:


.

2. f (x) = 8 – x3 g(x) = x – 3

SOLUTION:


.

3. f (x) = x2 + 5x + 6 g(x) = x + 2

SOLUTION:


.

4. f (x) = x – 9 g(x) = x + 5

SOLUTION:


.

5. f (x) = x2 + x g(x) = 9x

SOLUTION:




6. f (x) = x – 7 g(x) = x + 7

SOLUTION:


.

7. f (x) =

g(x) = x3 + x

SOLUTION:



8. f (x) =

g(x) =

SOLUTION:




9. f (x) =

g(x) = 4

SOLUTION:


10. f (x) =

g(x) = x4

SOLUTION:



11. f (x) =

g(x) = − 3

SOLUTION:


12. f (x) =

g(x) =

SOLUTION:


(4, ).


SOLUTION: a.











SOLUTION: a.


Ff ).



15. f (x) = 2x – 3 g(x) = 4x – 8

SOLUTION:

16. f (x) = –2x2 – 5x + 1 g(x) = –5x + 6

SOLUTION:

17. f (x) = 8 – x2

g(x) = x2 + x + 1

SOLUTION:

18. f (x) = x2 – 16

g(x) = x2 + 7x + 11

SOLUTION:

19. f (x) = 3 – x2

g(x) = x3 + 1

SOLUTION:

20. f (x) = 2 + x4

g(x) = −x2

SOLUTION:

Find f o g .

21. f (x) =

g(x) = x2 – 4

SOLUTION:







22. f (x) =

g(x) = x2 + 6

SOLUTION:









Therefore, .

23. f (x) =

g(x) = x2 – 4

SOLUTION:







Therefore, .

24. f (x) = x2 − 9

g(x) =

SOLUTION:






25. f (x) =

g(x) =

SOLUTION:




which = 0.



26. f (x) =

g(x) =

SOLUTION:







27. f (x) =

g(x) = x2 + 4x – 1

SOLUTION:







Therefore, .

28. f (x) =

g(x) = x2 + 8

SOLUTION:







Therefore, .










30. h(x) = + 7

SOLUTION:


31. h(x) = – 8

SOLUTION:


32. h(x) = | 4x + 8 | − 9

SOLUTION:


33. h(x) = [[−3(x – 9)]]


34. h(x) =

SOLUTION:


35.

SOLUTION:


36. h(x) =

SOLUTION:


37. h(x) =

SOLUTION:


38. h(x) =

SOLUTION:


39. h(x) =

SOLUTION:




.


SOLUTION: a.













43.

SOLUTION:


convert g(x) to .






44.

SOLUTION:


convert g(x) to .






45.

SOLUTION:








46. h(x) = + 9x

SOLUTION:




Replace c with .



47.











(x) is .



48. h(x) = +











of f (x) is .









49. f (x) – g(x) = x2 + x – 6, g(x) = x + 4

SOLUTION:

50. f (x) + g(x) = + – , g(x) = 2x

SOLUTION:

51. g(x) – f (x) + = 9x2 + 4x, g(x) =

SOLUTION:

52. g(x) = f (x) – 18x2 + , g(x) =

SOLUTION:

Find [f o g o h](x).53. f (x) = x + 8

g(x) = x2 – 6

h(x) = + 3

SOLUTION:








54. f (x) = x2 – 2 g(x) = 5x + 12

h(x) =

SOLUTION:






55. f (x) =

g(x) = x2 – 3

h(x) =

SOLUTION:










56. f (x) =

g(x) = x2 – 4x + 1

h(x) = x + 2

SOLUTION:












b. (x) =

SOLUTION: a.

b.


SOLUTION: a.

b.





b.

SOLUTION: a.

b.


SOLUTION: a.




61. (f + g)(2)

SOLUTION:

62. (f – g)(−6)

SOLUTION:

63. (f · g)(4)

SOLUTION:

64. (−2)

SOLUTION:

65. [f o g](−4)

SOLUTION:

66. [g o f ](6)

SOLUTION:







.


68.







Check the solution.

69.



(x)) = .




Check the solution.

70. a(x) =





Check the solution.

71.






Check the solution.


72. f (x) = x2 – 6x + 5

g(x) = + 3

SOLUTION:












73. f (x) = x2 + 8x – 3

g(x) = – 4

SOLUTION:











74. f (x) =

g(x) =

SOLUTION:

To find f o g, you must first be able to evaluate g(x) = ,

which can be done for all real numbers since 16 + x2 will never be

negative. Then you must be able to evaluate f (x) = for each of these g(x)-values, which is true for all real g(x)-values, since g(x) will

always be greater than –6. Therefore, the domain of f o g is all real numbers. Now find [f o g](x).

Therefore, .

To find g o f , you must first be able to evaluate f (x) = , which can only be done for x ≥ –6. Then you must be able to evaluate g(x) =

, for each of these f (x)-values, which can be done for all f

(x)-values. Therefore initially, it appears that the domain of g ◦ f is {x| x≥ −6, x R}. Now find [g o f ](x).

Notice that is only defined for x ≥ –22. Combining this restriction with the restriction the x ≥ –6, the true domain of g ◦ f is {x|

x ≥ −22, x R}. Therefore, [g o f ](x) = for x ≥ −22.

eSolutions Manual - Powered by Cognero Page 1

1-6 Function Operations and Composition of Functions



1. f (x) = x2 + 4

g(x) =

SOLUTION:


.

2. f (x) = 8 – x3 g(x) = x – 3

SOLUTION:


.

3. f (x) = x2 + 5x + 6 g(x) = x + 2

SOLUTION:


.

4. f (x) = x – 9 g(x) = x + 5

SOLUTION:


.

5. f (x) = x2 + x g(x) = 9x

SOLUTION:




6. f (x) = x – 7 g(x) = x + 7

SOLUTION:


.

7. f (x) =

g(x) = x3 + x

SOLUTION:



8. f (x) =

g(x) =

SOLUTION:




9. f (x) =

g(x) = 4

SOLUTION:


10. f (x) =

g(x) = x4

SOLUTION:



11. f (x) =

g(x) = − 3

SOLUTION:


12. f (x) =

g(x) =

SOLUTION:


(4, ).


SOLUTION: a.











SOLUTION: a.


Ff ).



15. f (x) = 2x – 3 g(x) = 4x – 8

SOLUTION:

16. f (x) = –2x2 – 5x + 1 g(x) = –5x + 6

SOLUTION:

17. f (x) = 8 – x2

g(x) = x2 + x + 1

SOLUTION:

18. f (x) = x2 – 16

g(x) = x2 + 7x + 11

SOLUTION:

19. f (x) = 3 – x2

g(x) = x3 + 1

SOLUTION:

20. f (x) = 2 + x4

g(x) = −x2

SOLUTION:

Find f o g .

21. f (x) =

g(x) = x2 – 4

SOLUTION:







22. f (x) =

g(x) = x2 + 6

SOLUTION:









Therefore, .

23. f (x) =

g(x) = x2 – 4

SOLUTION:







Therefore, .

24. f (x) = x2 − 9

g(x) =

SOLUTION:






25. f (x) =

g(x) =

SOLUTION:




which = 0.



26. f (x) =

g(x) =

SOLUTION:







27. f (x) =

g(x) = x2 + 4x – 1

SOLUTION:







Therefore, .

28. f (x) =

g(x) = x2 + 8

SOLUTION:







Therefore, .










30. h(x) = + 7

SOLUTION:


31. h(x) = – 8

SOLUTION:


32. h(x) = | 4x + 8 | − 9

SOLUTION:


33. h(x) = [[−3(x – 9)]]


34. h(x) =

SOLUTION:


35.

SOLUTION:


36. h(x) =

SOLUTION:


37. h(x) =

SOLUTION:


38. h(x) =

SOLUTION:


39. h(x) =

SOLUTION:




.


SOLUTION: a.













43.

SOLUTION:


convert g(x) to .






44.

SOLUTION:


convert g(x) to .






45.

SOLUTION:








46. h(x) = + 9x

SOLUTION:




Replace c with .



47.











(x) is .



48. h(x) = +











of f (x) is .









49. f (x) – g(x) = x2 + x – 6, g(x) = x + 4

SOLUTION:

50. f (x) + g(x) = + – , g(x) = 2x

SOLUTION:

51. g(x) – f (x) + = 9x2 + 4x, g(x) =

SOLUTION:

52. g(x) = f (x) – 18x2 + , g(x) =

SOLUTION:

Find [f o g o h](x).53. f (x) = x + 8

g(x) = x2 – 6

h(x) = + 3

SOLUTION:








54. f (x) = x2 – 2 g(x) = 5x + 12

h(x) =

SOLUTION:






55. f (x) =

g(x) = x2 – 3

h(x) =

SOLUTION:










56. f (x) =

g(x) = x2 – 4x + 1

h(x) = x + 2

SOLUTION:












b. (x) =

SOLUTION: a.

b.


SOLUTION: a.

b.





b.

SOLUTION: a.

b.


SOLUTION: a.




61. (f + g)(2)

SOLUTION:

62. (f – g)(−6)

SOLUTION:

63. (f · g)(4)

SOLUTION:

64. (−2)

SOLUTION:

65. [f o g](−4)

SOLUTION:

66. [g o f ](6)

SOLUTION:







.


68.







Check the solution.

69.



(x)) = .




Check the solution.

70. a(x) =





Check the solution.

71.






Check the solution.


72. f (x) = x2 – 6x + 5

g(x) = + 3

SOLUTION:












73. f (x) = x2 + 8x – 3

g(x) = – 4

SOLUTION:











74. f (x) =

g(x) =

SOLUTION:

To find f o g, you m

1-6 Function Operations and Composition of...

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