Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 2 - Review Exercises - Page 234: 66


$f^{-1} (x) =\sqrt [3] {x+1}$

Work Step by Step

We have $f^{-1} (x) =\sqrt [3] {x+1}$ and $f(f^{-1} (x))=f(\sqrt [3] {x+1})=(\sqrt [3] {x+1})^3-1=x+1-1=x$ Now, $f(f^{-1} (x))=f^{-1} (x^3+1)= \sqrt [3] {(x^3+1)-1}=\sqrt[3] x^3=x$ Thus, we have $f^{-1} (x) =\sqrt [3] {x+1}$
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