Answer
$ 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}
