**Unformatted text preview: **= −2 + x +
=x 1
2 1
2 We now check that f ◦ f −1 (x) = x for all x in the range of f which is also all real numbers.
(Recall that the domain of f −1 ) is the range of f .)
f ◦ f −1 (x) = f (f −1 (x))
1 − 2f −1 (x)
=
5
5
1 − 2 −2x +
=
5
1 + 5x − 1
=
5
5x
=
5
=x 1
2 To check our answer graphically, we graph y = f (x) and y = f −1 (x) on the same set of axes.5
They appear to be reﬂections across the line y = x.
5 Note that if you perform your check on a calculator for more sophisticated functions, you’ll need to take advantage
of the ‘ZoomSquare’ feature to get the correct geometric perspective. 5.2 Inverse Functions 301
y 2 y=x
1 −4 −3 −2 −1 1 2 3 4 x −1 y = f (x)
−2 y = f −1 (x) 2. To ﬁnd g −1 (x), we start with y = g (x). We note that the domain of g is (−∞, 1) ∪ (1, ∞).
y = g (x)
2x
y=
1−x
2y
x=
1−y
x(1 − y ) = 2y switch x and y x − xy = 2y
x = xy + 2y
x = y (x + 2)
y= factor x
x+2 x
We obtain g −1 (x) = x+2 . To check this analytically, we ﬁrst check...

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