Answer
${{f}^{-1}}\left( x \right)=\sqrt[3]{x-2}$
Work Step by Step
Inverse of a function $f\left( x \right)$ is defined by a function ${{f}^{-1}}\left( x \right)$ such that the domain $f\left( x \right)$ is equal to the range of ${{f}^{-1}}\left( x \right)$. This means that if the function $f$ is the set of ordered pairs $\left( x,y \right)$ , then ${{f}^{-1}}$ is the set of ordered pairs $\left( y,x \right)$. And if we interchange the values of x and y in the equation function $f$ , it will give the equation for function ${{f}^{-1}}$.
The provided equation is $f\left( x \right)={{x}^{3}}+2$.
Now replace $f\left( x \right)$ with y:
$y={{x}^{3}}+2$
Interchange x and y to provide the inverse function as follows:
$x={{y}^{3}}+2$
Subtract 2 from both sides of the equation.
$x-2={{y}^{3}}$
Take the cubic root of y.
$\begin{align}
& \sqrt[3]{x-2}=\sqrt[3]{{{y}^{3}}} \\
& \sqrt[3]{x-2}=y \\
\end{align}$
Now,
$y={{f}^{-1}}\left( x \right)$
${{f}^{-1}}\left( x \right)=\sqrt[3]{x-2}$
Thus, the inverse of $f\left( x \right)={{x}^{3}}+2$ is ${{f}^{-1}}\left( x \right)=\sqrt[3]{x-2}$
