Order Precalculus Function Skills

Order Precalculus Function Skills

Recall that a function is a set of ordered pairs in which each first

component is paired with exactly one second component. That is, in a function

no two ordered pairs have the same first component but different second

components.

(A) True, because {(–4,5),(–1,5),(0,5),(5,5)} is a set of ordered pairs in

which each first component is paired with exactly one second

component, so it is a function.

(B) False, because, for instance, if x is 8, then y

2 = 16 and y = ±4, so the set

{(x, y) | y

2 = 2x} contains the ordered pairs (8,4) and (8,–4); thus, it is

not a function.

(C) True. If (4,a) and (4,b) are elements of a function, then a = b because

the first component of a function cannot be paired with two different

second components.

(D) False. A function is a set of ordered pairs. The domain of a function is

the set composed of the first elements of the function; the domain is

not a set of ordered pairs.

Order Precalculus Function Skills

(E) True. In the function f = {(x, y) | y = 5x 3}, y = f (x) is the image of x

under f.

67. y = f = 28 – 10 = 21 – 10 = 11

68. y = f (–5) = (–5)

2 1 = 25 1 = 26

69. y = f (–1) = 4(–1)

5 2(–1)

4 – 3(–1)

3 – 5(–1)

2 (–1) 5 = 0

70.

71.

72.

73.

74. The domain of a function f = {(x, y) | y = f (x)} is the set of all possible x

values of f and the range is the set of all possible y values of f. Because the

square root of a negative number is not a real number, to find the domain of f

solve 3x –15 ≥ 0 for x. Thus, Df = {x | x ≥ 5}. Because the square root always

returns a nonnegative number, Rf = {y | y ≥ 0}.

75. Df = {4,5,5.2,10}; Rf =

76. g(x) = is undefined when x = –4 because division by zero would

occur. Thus, Dg ={x | x ≠ –4}. Solving y = g(x) = for x yields x = ,

which is undefined when y = 0. Thus, Rg = {y | y ≠ 0}.

77. The absolute value is defined for all real numbers and always returns a

nonnegative number, so Dh = R and Rh = {y | y ≥ 0}.

78. Because the square root of a negative number is not a real number, to find

the domain of f determine when x

2 – 9, which equals the product (x 3)(x – 3), is

nonnegative. The product is zero at x = –3 or x = 3. It is positive when both

factors have the same sign, which occurs to the left of –3 and to right of 3.

Therefore, Df = {x | x ≤ – 3 or x ≥ 3}. Because the square root always returns a

nonnegative number, Rf = {y | y ≥ 0}.

79. (f g)(x) = f(x) g(x) = x

4

, x ≠ 0

80. (f – g)(x) = f(x) – g(x) = – x

4

, x ≠ 0

81. (fg)(x) = f(x)g(x) = (x

4) = x

3

, x ≠ 0

82.

Order Precalculus Function Skills

83. (f g)(x) = (x

2 3) ( – 3) = x

2 , x ≥ 0.

Thus, (f g)(4) = 4

2 = 16 2 = 18.

84. (f – g)(x) = (x

2 3) – ( – 3) = x

2 – 6, x ≥ 0.

Thus, (f – g)(6) = 6

2 6 = 36 – 6 = 42 – .

85. (fg)(–1) is undefined because –1 is not in the domain of fg.

86. . Notice that in addition to being nonnegative, the

domain of excludes the value 9 because it must exclude values of x for which

– 3 = 0. Thus, (9) is undefined because 9 is not in the domain of .

87. The composition of two functions f and g is the function f g defined by (f

g)(x) = f(g(x)), provided that g(x) ∈ Df

. Thus, the function g takes x into g(x)

and the function f takes g(x) into f(g(x)). Obviously, any value of x for which

g(x) is not in the domain of f cannot be in the domain of (f g)(x).

Looking at the ordered pairs of f and g, you can see that the function g takes –

4 and 3 into 2 and –4, respectively; and the function f takes 2 and –4 into 5 and –

3, respectively. Thus, (a) f g = {(–4,5),(3,–3)}; (b) (f g)(–4) = 5.

88. From the ordered pairs of f and g, you can see that the function f takes –5,

–4, –1, and 6 into 1, –3, 3, and –4, respectively; and the function g takes 1, –3, 3,

and –4 into 9, –3, –4, and 2, respectively. Thus, (a) g f = {(–5,9),(–4,–3),(–1,–

4),(6,2)}; (b) (g f)(6) = 2.

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