Answer
See below:
Work Step by Step
(a)
Suppose the above function
$f\left( x \right)=\sqrt[3]{x-1}$.
Take $f\left( x \right)$ with $y$:
$\sqrt[3]{x-1}=y$
Interchange $x$ with $y$:
$\sqrt[3]{y-1}=x$
Solve the obtained equation for the value of $y$:
Take the cube of both sides of the equation:
$\begin{align}
& {{\left( \sqrt[3]{y-1} \right)}^{3}}={{\left( x \right)}^{3}} \\
& {{\left( y-1 \right)}^{3\times \frac{1}{3}}}={{\left( x \right)}^{3}} \\
& y-1={{\left( x \right)}^{3}}
\end{align}$
Add $1$ to both sides of the equation:
$\begin{align}
& y-1+1={{x}^{3}}+1 \\
& y={{x}^{3}}+1
\end{align}$
Take $y$ with ${{f}^{-1}}\left( x \right)$:
${{f}^{-1}}\left( x \right)={{x}^{3}}+1$
So, the inverse function is ${{f}^{-1}}\left( x \right)={{x}^{3}}+1$.
Thus, the equation of ${{f}^{-1}}\left( x \right)$ with the function $f\left( x \right)=\sqrt{x}+2$ is ${{f}^{-1}}\left( x \right)={{x}^{3}}+1$.
(b)
Suppose the above equation
$f\left( x \right)=\sqrt[3]{x-1}$.
Put $f\left( x \right)$ with $y$:
$\sqrt[3]{x-1}=y$
Put $y=1$ to solve for the value of $x$:
$\sqrt[3]{x-1}=1$
Take the cube of both sides:
$\begin{align}
& {{\left( \sqrt[3]{x-1} \right)}^{3}}=1 \\
& x-1=1 \\
& x=2
\end{align}$
So, the point A is $\left( 2,1 \right)$.
Now, put $x=1$ in the equation $\sqrt[3]{x-1}=y$ to solve for the value of $y$:
$\begin{align}
& \sqrt[3]{1-1}=y \\
& y=0
\end{align}$
So, the point $B$ is $\left( 1,0 \right)$.
Put $y=2$ in the equation $\sqrt[3]{x-1}=y$ to solve for the value of $x$:
$\begin{align}
& \sqrt[3]{x-1}=2 \\
& x-1=8 \\
& x=9
\end{align}$
So, the point C is $\left( 9,2 \right)$.
Since inverse of a function exists. Therefore, points to the plot inverse function are
${{A}_{1}}=\left( 1,2 \right)$ , ${{B}_{1}}=\left( 0,1 \right)$ and ${{C}_{1}}=\left( 2,9 \right)$.
Plot all the points on the same graph for both the function and its inverse as shown above:
(c)
The domain and range of the function $f$ are $\left( \infty ,-\infty \right)$ and the domain and range of the function ${{f}^{-1}}$ are $\left( \infty ,-\infty \right)$.
