Answer
The domain is restricted to $[0,\infty)$, and the inverse is: $f^{-1}(x)=\sqrt [4] x$.
(Other answers are possible.)
Work Step by Step
If a function $f(x)$ is one-to-one, then for all $y=f(x)$ there is only one $x$. When finding the inverse of the function, the $x$ and $y$ values get switched. The domain of the function $f(x)$ becomes the range of the inverse function and the range of the inverse function $f^{-1}(x)$ becomes the domain of the function $f(x)$.
We have:
$y=x^4 \implies x=\sqrt [4] y$
Switch $x$ to $f^{-1} (x)$ and $y$ to $x$ in the function to obtain the inverse.
Then the inverse is $f^{-1}(x)=\sqrt [4] x$
We see that the inverse is undefined for negative $x$ values. Thus, we have the domain that is restricted to $[0,\infty)$,
