Chapter 1 | Functions and Graphs
23
Alternatively, we could write
1 x 2 +1
( g ∘ f )( x ) = g ⎛
⎞ ⎠ = 1 f ( x ) =
⎝ f ( x )
.
Since x 2 +1≠0 for all real numbers x , the domain of ( g ∘ f )( x ) is the set of all real numbers. Since 0<1/( x 2 +1) ≤1, the range is, at most, the interval (0, 1]. To show that the range is this entire interval, we let y =1/( x 2 +1) and solve this equation for x to show that for all y in the interval (0, 1], there exists a real number x such that y =1/( x 2 +1). Solving this equation for x , we see that x 2 +1=1/ y , which implies that x = ± 1 y −1. If y is in the interval (0, 1], the expression under the radical is nonnegative, and therefore there exists a real number x such that 1/( x 2 +1) = y . We conclude that the range of g ∘ f is the interval (0, 1]. b. ( g ∘ f )(4) = g ( f (4)) = g (4 2 +1) = g (17) = 1 17 ( g ∘ f ) ⎛ ⎝ − 1 2 ⎞ ⎠ = g ⎛ ⎝ f ⎛ ⎝ − 1 2 ⎞ ⎠ = 4 5 c. We can find a formula for ( f ∘ g )( x ) in two ways. First, we could write ⎞ ⎠ ⎞ ⎠ = g ⎛ ⎝ ⎜ ⎛ ⎝ − 1 2 ⎞ ⎠ 2 +1 ⎞ ⎠ ⎟ = g ⎛ ⎝ 5 4
2
( f ∘ g )( x ) = f ( g ( x )) = f ⎛ ⎝ 1 x ⎞ ⎠ = ⎛ ⎝ 1 x
⎞ ⎠
+1.
Alternatively, we could write
2
( f ∘ g )( x ) = f ( g ( x )) = ( g ( x )) 2 +1= ⎛ ⎝ 1 x
⎞ ⎠
+1.
The domain of f ∘ g is the set of all real numbers x such that x ≠0. To find the range of f , we need to find all values y for which there exists a real number x ≠0 such that ⎛ ⎝ 1 x ⎞ ⎠ 2 +1= y .
Solving this equation for x , we see that we need x to satisfy ⎛ ⎝ 1 x ⎞ ⎠ 2 = y −1,
which simplifies to
1 x = ± y −1.
Finally, we obtain
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