Answer
$$ f^{-1}(x) =\sqrt[3 ]{x-1}$$
Domain of $ f^{-1}$ is $(-\infty,\infty)$
Range of $ f^{-1}$ is $(-\infty,\infty)$
Work Step by Step
Given $$ f(x) =x^{3}+1$$
Find the inverse:
\begin{aligned} y &=x^{3}+1 \\ x^3 &=y-1 \\ x&=\sqrt[3]{y-1} \\
\text {Switch }& x \ and\ y: \text{}\\
y=&\sqrt[3]{x-1} =f^{-1}(x) \end{aligned}
We see that:
Domain of $ f^{-1}$ is $(-\infty,\infty)$
and
Range of $ f^{-1}$ is $(-\infty,\infty)$
Confirm the inverse:
\begin{aligned} f\left(f^{-1}(x)\right)
&=f(\sqrt[3]{x-1}) \\ &=(\sqrt[3]{x-1})^{3}+1 \\ &=x +1-1 \\&=x \\ f^{-1}(f(x)) &=f^{-1}\left(x^{3}+1\right) \\ &=\sqrt[3]{x^{3}+1-1} \\ &=\sqrt[3]{x^{3} } \\ &=x \end{aligned}
