Answer
$f^{-1}(x)=\displaystyle \frac{1}{2}x+\frac{3}{2}$
(the inverse is blue on the graph)
Work Step by Step
Step 1: Replace $f(x)$ with $y$.
$y=2x-3$
Step 2: Interchange $x$ and $y$.
$x=2y-3$
Step 3: Solve the equation for $y$.
$ x=2y-3,\qquad$ ... add $3$,
$ x+3=2y,\qquad$ ... divide with 2
$\displaystyle \frac{1}{2}x+\frac{3}{2}=y$
Step 4: Replace y with the notation $f^{-1}(x)$.
$f^{-1}(x)=\displaystyle \frac{1}{2}x+\frac{3}{2}$
Graphing $f(x)=2x-3, \left[\begin{array}{lll}
x & f(x) & (x,y)\\
0 & -3 & (0,-3)\\
3 & 3 & (3,3)
\end{array}\right], $
the graph of $f(x)$ is a line passing through $(0,-3)$ and $(3,3)$.
The graph of $f^{-1}(x)$ is a line passing through points (y,x) of the above table,
$(-3,0)$ and $(3,3)$.
