Answer
$g^{-1}(7) = 2$
Work Step by Step
Let $g(x) = y$
$y = 4x - 1$
$x = 4y - 1$
$x + 1 = 4y$
$y = \frac{x+1}{4}$
$g^{-1}(x) = \frac{x+1}{4}$
$g^{-1}(7) = \frac{7+1}{4}$
$= \frac{8}{4}$
$= 2$
College Algebra (6th Edition)
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-32178-228-3
ISBN 13:
978-0-32178-228-1
Chapter 2 - Functions and Graphs - Exercise Set 2.7 - Page 310: 62
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