Answer
See below:
Work Step by Step
(a)
Consider the function:
$f\left( x \right)={{\left( x-1 \right)}^{2}}$
Follow the steps to determine the inverse ${{f}^{-1}}\left( x \right)$ of the function $f\left( x \right)={{\left( x-1 \right)}^{2}}$.
Step 1: Replace the function $f\left( x \right)$ with y in the equation of $f\left( x \right)$.
$y={{\left( x-1 \right)}^{2}}$
Step 2: Interchange the variables x and y.
$x={{\left( y-1 \right)}^{2}}$
Step 3: Solve the equation for y.
The variable y has to be isolated. So, take square root on both sides of the equation. So, the equation becomes,
$\begin{align}
& \pm \sqrt{x}=\sqrt{{{\left( y-1 \right)}^{2}}} \\
& \pm \sqrt{x}=y-1 \\
\end{align}$
Add $1$ to both sides of the equation. So,
$\pm \sqrt{x}+1=y-1+1$
Combine like terms.
$\pm \sqrt{x}+1=y$
As $x\le 1$ , the equation is $y=-\sqrt{x}+1$. 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,
${{f}^{-1}}\left( x \right)=-\sqrt{x}+1$
Therefore, the inverse function is ${{f}^{-1}}\left( x \right)=-\sqrt{x}+1$.
(b) Given above.
(c)
The domain and range of the functions in interval notation are:
\[\begin{align}
& \text{domain of }f=\text{range of }{{f}^{-1}}=\left( -\infty ,1 \right] \\
& \text{range of }f=\text{domain of }{{f}^{-1}}=\left[ 0,\infty \right)
\end{align}\]