$f(x)$= $(|x-2|)([x^2 -2x-2])$ where, [.]denotes the greatest integer function, then find the number of points of discontinuity in the interval
$(\frac{1}{2},$2).
Since, $|x-2|$ is continuous for all x , $[x^2-2x-2]$ is discontinuous at x=1,2.
So $f(x)$ has a chance to be discontinuous at $x=1,2$ checking it we find $f(x)$ is discontinuous at x=2.
However, I am not able to prove it rigoursly that, this is the only point of discontinuity.
On
Let $$f(x)=x^2-2x-2.$$ its derivative is $$f'(x)=2(x-1)$$
$ f $ is decreasing at $ (\frac 12,1] $ and increasing at $ [1,2)$.
So $$\forall x\in(\frac 12,2)\;$$ $$ f(1)\le f(x)<\max(f(\frac 12),f(2))$$ or $$ -3\le f(x)<-2$$ $$\implies \lfloor x^2-2x-2\rfloor=-3$$ $$\implies |x-2|\lfloor x^2-2x-2\rfloor=3(x-2)$$
which is continuous at $(1/2,2)$.
Since, if $x\in\left[\frac12,2\right]$,$$\lfloor x^2-2x-2\rfloor=\begin{cases}-3&\text{ if }x\in\left[\frac12,2\right)\\-2&\text{ if }x=2,\end{cases}$$this function is discontinuous at $2$ and only at $2$. But it is a bounded function. So, since $\lim_{x\to2}|x-2|=0$,\begin{align}\lim_{x\to2}f(x)&=\lim_{x\to2}|x-2|\lfloor x^2-2x-2\rfloor\\&=0\\&=f(2).\end{align}So, $f$ is continuous everywhere.