Answer
$ f^{-1}(x)=\sqrt[4]{x} $
Domain of $ f^{-1}$ is $[0,\infty)$
Range of $ f^{-1}$ is $[0,\infty)$
Work Step by Step
Given $$ f(x) =x^{4}$$
Find the inverse:
\begin{aligned} y
&=x^{4} \\ x^4
&=y \\ x
&=\sqrt[4]{y} \\
\text {Switch }& x \ and\ y, \text{}\\
y=&\sqrt[4]{x} =f^{-1}(x) \end{aligned}
We see that:
Domain of $ f^{-1}$ is $[0,\infty)$
and
Range of $ f^{-1}$ is $[0,\infty)$
Confirm the inverse:
\begin{aligned} f\left(f^{-1}(x)\right)
&=f(\sqrt[4]{x}) \\ &=(\sqrt[4]{x})^{4} \\ &=x \\ f^{-1}(f(x)) &=f^{-1}\left(x^{4}\right) \\ &=\sqrt[4]{x^{4}} \\ &=x \end{aligned}
