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
$f(x)=\displaystyle \sqrt{3x-2}, \frac{3}{2}\leq x\leq 4$
Next cell:
$f'=Derivative(f)$
Results in$\displaystyle \quad \frac{3}{2}(3x-2)^{-1/2}$
We know that f is one-to-one on the given interval, because it passes the horizontal line test.
$ b.\quad$
Next cell, we find the inverse of f. Enter:
$g(x)=Invert(f)$
Results in$\quad \displaystyle \frac{x^{2}+2}{3}, \quad \frac{3}{2}\leq g(x)\leq 4$
$ c.\quad$
Enter:$ \quad P=(3,f(3))\quad $(results in $(3,2.65)$
the point on the graph of $f$ for $x_{0}=3$
Next, enter
$h:=Tangent(P,f)$
resulting in $y=0.57x+0.94$
$ d.\quad$
Next, enter
$i:=Tangent((f(3),3),g)$
resulting in $y=1.76x-1.67$,
(the tangent to the inverse of f at the point $(f(x_{0}),x_{0})$
Note that $1.76\times 0.57=1,$
which is in line with Theorem 1, by which $f^{-1}(2.65)=\displaystyle \frac{1}{f'(3)}$
$ e.\quad$
Enter the equation of the identity function, $\quad y=x.$
Enter $\quad Q=(f(3),3)$
Enter $j=Segment(P,Q)$
You may want to find the domain of $g=f^{-1}$, which is $[1.58,3.16]$
found by entering$\quad Solve(\displaystyle \frac{3}{2}\leq\frac{x^{2}+2}{3}\leq 4)$
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.
