College Algebra, 2013a

5.2 Inverse Functions 385
y = f(x)
y = 1 − 2x
5
x = 1 − 2y
5
5x = 1− 2y
5x − 1 = −2y
5x − 1
−2
= y
switch x and y
y = − 5 2 x + 1 2
We have f −1 (x) =− 5 2 x+ 1 2 . To check this answer analytically, we first check that ( f −1 ◦ f ) (x) =
x for all x in the domain of f, whichisallrealnumbers.
(
f −1 ◦ f ) (x) = f −1 (f(x))
= − 5 2 f(x)+1 2
= − 5 ( ) 1 − 2x
+ 1 2 5 2
= − 1 2 (1 − 2x)+1 2
= − 1 2 + x + 1 2
= x ̌
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
= 1 − 2 ( − 5 2 x + 1 )
2
5
= 1+5x − 1
5
= 5x 5
= x ̌
To check our answer graphically, we graph y = f(x) andy = 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.