Section 2.8 Function Operations and Composition
87. (a) ( f
g )( x) f ( g ( x)) f
1
1 x
x 1 x2
90. 12x
The domain and range of g are (, 0) (0, ) . The domain of f is (, 2) (2, ), and the range of f is (, 0) 0 x
(0, ) . So, 1 2
1
Thus, x 0 and x
domain of f ,0
2 1
0 xx 0 or (using test intervals).
( f g )(2) f g (2) f (0) 0 and
. Therefore, the 2
g is
0, 12
1 2
88. (a) ( f
(2, ).
g )( x) f ( g ( x)) f
1
1 x
x 14x
1 x4
The domain and range of g are (, 0) (0, ). The domain of f is (, 4) (4, ), and the range of f is (, 0)
x 14x
(0, ). So,
,0 (b)
0,
1 4
1 4
g)(1) f g (1) f (2) 1 and
(f
g)(2) f g (2) f (0) 0. –2
–1
0
1
2
f x
–1
2
0
–2
1
g x
0
2
1
2
0
0
1
–2
1
0
f
92.
0 x 0
gx
91. Answers will vary. In general, composition of functions is not commutative. Sample answer: f g x f 2x 3 3 2 x 3 2 6 x 9 2 6 x 11 g f x g 3x 2 2 3x 2 3 6x 4 3 6x 7 Thus, f gx is not equivalent to g f x. f
gx f g x f
g
1 x4
1 ( x 4)
x 4 The domain and range of g are (, 0) (0, ) . The domain of f is
93.
f
g x
g f ( x)
1
3
2
7
2
1
5
2
3
2
7
5
x3 x
1
4x 2 2 4
4x 2 2
1
4
f
4x x
gx f g x 3 1 x 3
f x
3
g x f g x 4 1 4x 2 2 1 4 x 42 2 x 2 2 x 2 2 x
1
94.
x
x3 7 7
7
g fx g f x
(, 4) (4, ) , and the range of f is (, 0) (0, ) . Therefore, the domain of g f is (, 4) (4, ). 89. g f (2) g (1) 2 and g f (3) g (2) 5 Since g f (1) 7 and f (1) 3, g (3) 7.
3
f x g f x g x3 3
1
x 7
x 7 7 x
, .
( g f )( x) g ( f ( x)) g
3
x3 7 7
or 0 x 1 4 or 1 4 x 0 x 1 4 (using test intervals). Thus, x ≠ 0 and x ≠ 41 . Therefore, the domain of f g is
(f
x
, .
(b) ( g f )( x) g ( f ( x)) g 1 x 2 1 1 ( x 2) x 2 The domain and range of g are (, 0) (0, ). The domain of f is (, 2) (2, ), and the range of f is (, 0) (0, ). Therefore, the domain of g f is (, 2)
f ( x) is odd, so f (1) f (1) (2) 2. Because g ( x) is even, g (1) g (1) 2 and g (2) g (2) 0. ( f g)(1) 1, so f (2) 1. f ( x) is odd, so f g (1) 1 and f (2) f (2) 1. Thus,
12x
or x
247
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1
x x
4 3
3 1
g f x g f x 33x 3 x1 x 3