## Thomas' Calculus 13th Edition

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