Answer
See below:
Work Step by Step
(a)
Consider the function:
$f\left( x \right)={{x}^{3}}-1$
Follow the steps to determine the inverse ${{f}^{-1}}\left( x \right)$ of the function $f\left( x \right)={{x}^{3}}-1$.
Step 1: Replace the function $f\left( x \right)$ with y in the equation of $f\left( x \right)$.
$y={{x}^{3}}-1$
Step 2: Interchange the variables x and y. So, the equation is:
$x={{y}^{3}}-1$
Step 3: Solve the equation for y.
The variable y has to be isolated. So, add $1$ to both sides of the equation. So, the equation becomes,
$x+1={{y}^{3}}-1+1$
Combine like terms.
$x+1={{y}^{3}}$
Take cubic root on both sides of the equation. So,
$\begin{align}
& {{\left( x+1 \right)}^{{1}/{3}\;}}={{\left( {{y}^{3}} \right)}^{{1}/{3}\;}} \\
& \sqrt[3]{x+1}=y
\end{align}$
The equation is $y=\sqrt[3]{x+1}$. Because the obtained equation defines y as a function of x, then the inverse function exists for the function f.
Step 4: Replace y by ${{f}^{-1}}\left( x \right)$ in the equation obtained in step 3.
Thus, the inverse function${{f}^{-1}}\left( x \right)$ is obtained as,
$\sqrt[3]{x+1}={{f}^{-1}}\left( x \right)$
Therefore, the inverse function is ${{f}^{-1}}\left( x \right)=\sqrt[3]{x+1}$.
(b) Given above.
(c)
The domain and range of the functions in interval notation are:
\[\begin{align}
& \text{domain of }f=\text{range of }{{f}^{-1}}=\left( -\infty ,\infty \right) \\
& \text{range of }f=\text{domain of }{{f}^{-1}}=\left( -\infty ,\infty \right)
\end{align}\]
