Use the triangle inequality and the reverse triangle inequality to find an upper bound for the set of all numbers of the form $\left\lvert\frac{x^2-3}{x-2}\right\rvert$ as $x$ ranges over the interval defined by $\lvert x-1\rvert < \frac 23$.
I know you need to use the reverse inequality as well but I'm not too sure what to do
We will use the triangle inequality on the numerator, and the reverse triangle inequality on the denominator.
Note that $|x^2 - 3| \le x^2 + 3$, we know that $|x-1| < \frac{2}{3} $, so that $\frac{1}{3} < x < \frac{5}{3}$ . Therefore it is now an easy task to find an upper bound for $ x^2 + 3 $
Lets use the reverse triangle inequality now.
Note that the reverse triangle inequality says that $||a| - |b|| \le |a-b|$ for all real a and b.
We will apply this to the denominator as I said, so that $||x|- |2|| \le |x-2| $,
this is actually a very vacuous application, because $x$ is always positive, for $|x-1| < \frac{2}{3} $ so that $|x| = x$ and $|2| = 2$ and the inequality we have produced is $|x- 2| \le |x-2| $
Regardless, we will now find a lower bound for $|x-2|$ .
We note that $\frac{1}{3} < x < \frac{5}{3}$ , so that $\frac{-7}{3} < x-2 < \frac{-1}{3}$ .
This is sufficient to show that $|x-2| > \frac{1}{3} $
Now since $|x-2| > \frac{1}{3} , |x-2| > 0 , \frac{1}{3} > 0$ we know that
$\frac{1}{|x-2|} < 3$
So we have bounded both $\frac{1}{|x-2|}$ and $|x^2 + 3|$ above using the triangle inequality and sorta using the reverse triangle inequality.
Could you perhaps elaborate on why you insist on using the reverse triangle inequality in this question?