Answer
See screenshot of a geogebra project.
Instructions and observations on each step given below.
.
Work Step by Step
$CAS$ used: geogebra online (www.geogebra.org/classic).
In the algebra window (to the left) enter the following:
$ a.\quad$
Define $f(x)$ by entering
$f(x)=\displaystyle \frac{3x+2}{2x-11},\quad {-2}\leq x\leq 2$
Next cell:
$f'=Derivative(f)$
Results in$\displaystyle \quad \frac{3(2x-11)-2(3x+2)}{2}$
Enter $\displaystyle \quad Simplify (\frac{3(2x-11)-2(3x+2)}{2})$
which results in $\displaystyle \frac{-37}{(2x-11)^{2}}$
We know that $f$ is one-to-one on the given interval, because it passes the horizontal line test.
$ b.\quad$
$y=\displaystyle \frac{3x+2}{2x-11}$
$2xy-11y=3x+2$
$2xy-3x=11y+2$
$x(2y-3)=11y+2$
$x=\displaystyle \frac{11y+2}{2y-3}\qquad\Rightarrow f^{-1}(x)=\frac{11x+2}{2x-3}$
enter:$\quad Solve(-2\displaystyle \leq\frac{11x+2}{2x-3}\leq 2)$
to find the domain of $g=f^{-1}:\qquad [-1.14,0.27]$
Next cell, enter:$\displaystyle \quad g(x)=\frac{11x+2}{2x-3},-1.14\leq x\leq 0.27$
$ c.\quad$
Enter:$ \displaystyle \quad P=(\frac{1}{2},f(\frac{1}{2}))\quad $(results in $(0.5,-0.35)$
the point on the graph of $f$ for $x_{0}=\displaystyle \frac{1}{2}$
Enter $\quad Q=(-0.35,0.5)$
(the point on the graph of $g=f^{-1}$, symmetric to the point P)
Next, enter
$h:=Tangent(P,f)\quad $resulting in $y=-0.37x-0.17$
$ d.\quad$
Next, enter
$ i:=Tangent((Q,g)\quad $resulting in $y=-2.7x-0.45$
(the tangent to the inverse of $f$ at the point $(f(x_{0}),x_{0})$
Note that $-2.7\times(-0.37)=1,$
which is in line with Theorem 1, by which $f^{-1}(-0.35)=\displaystyle \frac{1}{f'(0.5)}$
$ e.\quad$
Enter the equation of the identity function, $\quad y=x.$
Enter $j=Segment(P,Q)$
With right-clicks on objects in the graphing window (to the right),
set colors, dashed lines and captions, so that objects are distinct upon viewing.
Note that $f$ and $g=f^{-1}$ are symmetric about the line $y=x$
as are the tangents h and i.
