Answer
$f^{-1} (x)=5x+4$
Work Step by Step
We have $f^{-1} (x) =5x+4$
and $f(f^{-1} (x))=f(5x+4)= \dfrac{(5x+4)-4}{5}=\dfrac{5x}{5}=x$
Now, $f(f^{-1} (x))=f(\dfrac{x-4}{5})= 5(\dfrac{x-4}{5})+4=x-4+4=x$
Thus, we have $f^{-1} (x)=5x+4$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 2 - Review Exercises - Page 234: 65
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