Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.1 - Inverse Functions and Their Derivatives - Exercises 7.1 - Page 373: 25


$ f^{-1}=\sqrt[5] x $ Domain $ f^{-1}$ is $(-\infty,\infty)$ Range $ f^{-1}$ is $(-\infty,\infty)$

Work Step by Step

Given $$ f(x) =x^{5}$$ Find the inverse: \begin{aligned} y &=x^{5} \\ x &=y^{5} \\ \sqrt[5]{x} &=\sqrt[5]{y^{5}} \\ \sqrt[5]{x} &=y \\ \sqrt[5]{x} &=f^{-1}(x) \end{aligned} We see that: Domain of $ f^{-1}$ is $(-\infty,\infty)$ and Range of $ f^{-1}$ is $(-\infty,\infty)$ Also, we have: \begin{aligned} f\left(f^{-1}(x)\right) &=f(\sqrt[5]{x}) \\ &=(\sqrt[5]{x})^{5} \\ &=x \\ f^{-1}(f(x)) &=f^{-1}\left(x^{5}\right) \\ &=\sqrt[5]{x^{5}} \\ &=x \end{aligned}
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