Answer
$$f^{-1}(x)=\sqrt[3]{x-1}$$
- Domain: $(-\infty,\infty)$
- Range: $(-\infty,\infty)$
Work Step by Step
$$y=f(x)=x^{3}+1\hspace{1cm}$$
To find its inverse:
1) Solve for $x$ in terms of $y$:
$$y=x^3+1$$ $$x^3=y-1$$ $$x=\sqrt[3]{y-1}$$
2) Interchange $x$ and $y$:
$$y=\sqrt[3]{x-1}$$
Therefore, $$f^{-1}(x)=\sqrt[3]{x-1}$$
- 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, $\sqrt[3]{x-1}$ also ranges from $-\infty$ to $\infty$. So the range of $f^{-1}$ is $(-\infty,\infty)$
*Check:
$f(f^{-1}(x))=(\sqrt[3]{x-1})^3+1=(x-1)+1=x$
$f^{-1}(f(x))=\sqrt[3]{(x^3+1)-1}=\sqrt[3]{x^3}=x$
Therefore, $f(f^{-1}(x))=f^{-1}(f(x))=x$
