Answer
See below:
Work Step by Step
(a)
Assume the above function
$f\left( x \right)=\sqrt{\left( x-1 \right)}$.
Put $f\left( x \right)$ with $y$:
$\sqrt{\left( x-1 \right)}=y$
Interchange $x$ with $y$:
$\sqrt{\left( y-1 \right)}=x$
Now solve the obtained equation for the value of $y$:
Square both sides of the equation:
$\begin{align}
& {{\left( \sqrt{\left( y-1 \right)} \right)}^{2}}={{\left( x \right)}^{2}} \\
& {{\left( y-1 \right)}^{2\times \frac{1}{2}}}={{x}^{2}} \\
& y-1={{x}^{2}}
\end{align}$
Add $1$ to both sides of the equation:
$\begin{align}
& y-1+1={{x}^{2}}+1 \\
& y={{x}^{2}}+1
\end{align}$
Put $y$ with ${{f}^{-1}}\left( x \right)$:
${{f}^{-1}}\left( x \right)={{x}^{2}}+1$
So, the inverse function is ${{f}^{-1}}\left( x \right)={{x}^{2}}+1$ , $x\ge 0$.
Thus, the equation of ${{f}^{-1}}\left( x \right)$ with the function $f\left( x \right)=\sqrt{\left( x-1 \right)}$ ${{f}^{-1}}\left( x \right)={{x}^{2}}+1$ , $x\ge 0$.
(b)
Assume the above function
$f\left( x \right)=\sqrt{\left( x-1 \right)}$.
Put $f\left( x \right)$ with $y$:
$\sqrt{\left( x-1 \right)}=y$
Put $y=1$ to solve for the value of $x$:
$\sqrt{\left( x-1 \right)}=1$
Square both sides:
$\begin{align}
& {{\left( \sqrt{\left( x-1 \right)} \right)}^{2}}={{\left( 1 \right)}^{2}} \\
& x-1=1
\end{align}$
Add $1$ to both sides of the equation:
$\begin{align}
& x=2
\end{align}$
So, the point A is $\left( 2,1 \right)$.
Now, put $x=5$ in the equation $\sqrt{\left( x-1 \right)}=y$ to solve for the value of $y$:
$\begin{align}
& \sqrt{\left( 5-1 \right)}=y \\
& y=\sqrt{4} \\
& =2
\end{align}$
Therefore, the point $B$ is $\left( 5,2 \right)$.
Put $y=0$ in the equation $\sqrt{\left( x-1 \right)}=y$ to solve the value of $x$:
$\begin{align}
& \sqrt{\left( x-1 \right)}=y \\
& x-1=0 \\
& x=1
\end{align}$
Put, the point C is $\left( 1,0 \right)$.
But, inverse of a function exists. Therefore, points to plot the inverse function are
${{A}_{1}}=\left( 1,2 \right)$ , ${{B}_{1}}=\left( 2,5 \right)$ , and ${{C}_{1}}=\left( 0,1 \right)$.
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 $\left[ 1,\infty \right)$ and $\left[ 0,\infty \right)$ , respectively, and domain and range of the function ${{f}^{-1}}$ are $\left[ 0,\infty \right)$ and $\left[ 1,\infty \right)$ , respectively.
