Answer
$$f^{-1}(x)=\sqrt[4] x$$
- Domain: $[0,\infty)$.
- Range: $[0,\infty)$.
Work Step by Step
$$y=f(x)=x^4\hspace{1cm}x\ge0$$
To find its inverse:
1) Solve for $x$ in terms of $y$:
$$y=x^4$$ $$|x|=\sqrt[4]y$$
Since $x\ge0$, we have $|x|=x$, therefore:
$$x=\sqrt[4]y$$
2) Interchange $x$ and $y$:
$$y=\sqrt[4]x$$
Therefore, $$f^{-1}(x)=\sqrt[4] x$$
- Domain: $x$ is defined where $x\ge0$.
So the domain of $f^{-1}$ is $[0,\infty)$.
- Range:
We see here that $\sqrt[4]{x}\ge0$ for all $x\in[0,\infty)$ and it will range to $\infty$.
In other words, the range of $f^{-1}$ is $[0,\infty)$
*Check:
$$f(f^{-1}(x))=(\sqrt[4]x)^4=x$$
$$f^{-1}(f(x))=\sqrt[4]{x^4}=|x|$$
Again, since $x\ge0$, $|x|=x$:
$$f^{-1}(f(x))=x$$
Therefore, $f(f^{-1}(x))=f^{-1}(f(x))=x$
