Answer
See below:
Work Step by Step
(a)
Assume the above function
$f\left( x \right)=\sqrt{x}+2$.
Put $f\left( x \right)$ with $y$:
$\sqrt{x}+2=y$
Interchange x with $y$:
$\sqrt{y}+2=x$
Solve the obtained equation for the value of $y$:
And subtract $2$ to both sides of the equation:
$\begin{align}
& \sqrt{y}-2+2=x-2 \\
& \sqrt{y}=x-2
\end{align}$
Square both sides of the equation:
$\begin{align}
& {{\left( \sqrt{y} \right)}^{2}}={{\left( x-2 \right)}^{2}} \\
& {{\left( y \right)}^{2\times \frac{1}{2}}}={{\left( x-2 \right)}^{2}} \\
& y={{\left( x-2 \right)}^{2}}
\end{align}$
Put $y$ with ${{f}^{-1}}\left( x \right)$:
${{f}^{-1}}\left( x \right)={{\left( x-2 \right)}^{2}}$
So, the inverse function is ${{f}^{-1}}\left( x \right)={{\left( x-2 \right)}^{2}}$ , $x\ge 0$.
Thus, the equation of ${{f}^{-1}}\left( x \right)$ with the function $f\left( x \right)=\sqrt{x}+2$ is ${{f}^{-1}}\left( x \right)={{\left( x-2 \right)}^{2}}$ , $x\ge 0$.
(b)
Assume the above equation
$f\left( x \right)=\sqrt{x}+2$
Put $f\left( x \right)$ with $y$:
$\sqrt{x}+2=y$
Put $y=2$ to solve for the value of $x$:
$\begin{align}
& \sqrt{x}+2=2 \\
& \sqrt{x}=0
\end{align}$
Square both sides:
$\begin{align}
& {{\left( \sqrt{x} \right)}^{2}}={{\left( 0 \right)}^{2}} \\
& x=0
\end{align}$
So, the point A is $\left( 0,2 \right)$.
Now, put $x=1$ in the equation $\sqrt{x}+2=y$ to solve for the value of $y$:
$\begin{align}
& \sqrt{x}+2=y \\
& y=\sqrt{1}+2 \\
& =3
\end{align}$
So, the point $B$ is $\left( 1,3 \right)$.
Put $y=4$ in the equation $\sqrt{x}+2=y$ to solve for the value of $x$:
$\begin{align}
& \sqrt{x}+2=4 \\
& \sqrt{x}=2 \\
& x=4
\end{align}$
So, the point C is $\left( 4,4 \right)$.
Since inverse of a function exists. Therefore, points to plot the inverse function are
${{A}_{1}}=\left( 2,0 \right)$ , ${{B}_{1}}=\left( 3,1 \right)$ and $C=\left( 4,4 \right)$.
Thus 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 $\left[ 0,\infty \right)$ and $\left[ 2,\infty \right)$ , respectively, and domain and range of the function ${{f}^{-1}}$ are $\left[ 2,\infty \right)$ and $\left[ 0,\infty \right)$ , respectively.
