Answer
$f^{-1}(x)=\sqrt[5]x$
- Domain: $(-\infty,\infty)$
- Range: $(-\infty,\infty)$
Work Step by Step
$$y=f(x)=x^{5}\hspace{1cm}$$
To find its inverse:
1) Solve for $x$ in terms of $y$:
$$y=x^{5}$$
$$x=\sqrt[5]{y}$$
2) Interchange $x$ and $y$:
$$y=\sqrt[5]x$$
Therefore, $$f^{-1}(x)=\sqrt[5]x$$
- Domain: $x$ is defined in $R$. So the domain of $f^{-1}$ is $(-\infty,\infty)$.
- Range: $x$ ranges from $-\infty$ to $\infty$. All the same, $\sqrt[5]x$ also ranges from $-\infty$ to $\infty$. So the range of $f^{-1}$ is $(-\infty,\infty)$
*Check:
$f(f^{-1}(x))=(\sqrt[5]x)^5=(x^{1/5})^5=x^{\frac{1}{5}\times5}=x^1=x$
$f^{-1}(f(x))=\sqrt[5]{x^5}=(x^5)^{1/5}=x^{5\times\frac{1}{5}}=x^1=x$
Therefore, $f(f^{-1}(x))=f^{-1}(f(x))=x$
