College Algebra, 2013a

5.2 Inverse Functions 387
=
=
2x
2x +2(1− x)
2x
2x +2− 2x
= 2x 2
= x ̌
Next, we check g ( g −1 (x) ) = x for all x in the range of g. From the graph of g in Example
5.2.1, we have that the range of g is (−∞, −2) ∪ (−2, ∞). This matches the domain we get
from the formula g −1 (x) = , as it should.
x
x+2
(
g ◦ g
−1 ) (x) = g ( g −1 (x) )
( ) x
= g
x +2
( ) x
2
x +2
= ( ) x
1 −
=
=
( x
2
1 −
x +2
)
x +2
( x
x +2
2x
(x +2)− x
) ·
(x +2)
(x +2)
clear denominators
= 2x 2
= x ̌
Graphing y = g(x) andy = g −1 (x) on the same set of axes is busy, but we can see the symmetric
relationship if we thicken the curve for y = g −1 (x). Note that the vertical asymptote
x = 1 of the graph of g corresponds to the horizontal asymptote y = 1 of the graph of g −1 ,
as it should since x and y are switched. Similarly, the horizontal asymptote y = −2 ofthe
graph of g corresponds to the vertical asymptote x = −2 of the graph of g −1 .