Answer
a) ${{f}^{-1}}\left( x \right)=\frac{x-3}{2}$
b) they are inverses
Work Step by Step
(a)
Consider the function,
$f\left( x \right)=2x+3$
Step-1-Replace $f\left( x \right)$ with $y$.
$\begin{align}
& f\left( x \right)=2x+3 \\
& y=2x+3
\end{align}$
Step-2-Interchange $x$ and $y$
$\begin{align}
& y=2x+3 \\
& x=2y+3 \\
\end{align}$
Step-3-Solve for $y$. Subtract $3$ from each side
$\begin{align}
& x=2y+3 \\
& x-3=2y+3-3 \\
& x-3=2y
\end{align}$
Divide each side by $2$
$\begin{align}
& x-3=2y \\
& \frac{x-3}{2}=\frac{2y}{2} \\
& \frac{x-3}{2}=y
\end{align}$
Step-4-Replace $y$ in step 3 by ${{f}^{-1}}\left( x \right)$
$\begin{align}
& y=\frac{x-3}{2} \\
& {{f}^{-1}}\left( x \right)=\frac{x-3}{2}
\end{align}$
Therefore, the required inverse of the function is ${{f}^{-1}}\left( x \right)=\frac{x-3}{2}$
(b)
Consider the functions,
$f\left( x \right)=2x+3$ and ${{f}^{-1}}\left( x \right)=\frac{x-3}{2}$
Replace $x$ with ${{f}^{-1}}\left( x \right)$
$\begin{align}
& f\left( {{f}^{-1}}\left( x \right) \right)=2{{f}^{-1}}\left( x \right)+3 \\
& =2\left( \frac{x-3}{2} \right)+3
\end{align}$
Simplify
$\begin{align}
& f\left( g\left( x \right) \right)=2\left( \frac{x-3}{2} \right)+3 \\
& =x-3+3 \\
& =x
\end{align}$
Now, find ${{f}^{-1}}\left( f\left( x \right) \right)$
The function ${{f}^{-1}}\left( x \right)$ is given as:
${{f}^{-1}}\left( x \right)=\frac{x-3}{2}$
Replace $x$ with $f\left( x \right)$
$\begin{align}
& {{f}^{-1}}\left( f\left( x \right) \right)=\frac{f\left( x \right)-3}{2} \\
& =\frac{\left( 2x+3 \right)-3}{2}
\end{align}$
Simplify
$\begin{align}
& {{f}^{-1}}\left( f\left( x \right) \right)=\frac{\left( 2x+3 \right)-3}{2} \\
& =\frac{2x}{2} \\
& =x
\end{align}$
Thus, $f\left( {{f}^{-1}}\left( x \right) \right)=x$ and ${{f}^{-1}}\left( f\left( x \right) \right)=x$
It can be easily observed that ${{f}^{-1}}$ can be expressed in $f$ if subtracted by 3 and then divided by 2.
Hence, the functions $f$ and \[{{f}^{-1}}\] are inverses of each other.
