#### Answer

See below:

#### Work Step by Step

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