1. Given function

We have

where $[x]$ is the greatest integer function.

We need points in $(-2,2)$ where $y=|f(x)|$ is

• not continuous: number $m$
• not differentiable: number $n$

Then find $m+n$.

2. First write $f(x)$ interval-wise using $[x]$

In $(-2,2)$, the integer jumps of $[x]$ occur at $x=-1,0,1$.

For $-2<x<0$

Here $f(x)=x[x]$.

• If $-2<x<-1$, then $[x]=-2$, so $f(x)=x(-2)=-2x$
• If $-1\le x<0$, then $[x]=-1$, so $f(x)=x(-1)=-x$

For $0\le x\le 2$

Here $f(x)=(x-1)[x]$.

• If $0\le x<1$, then $[x]=0$, so $f(x)=0$
• If $1\le x<2$, then $[x]=1$, so $f(x)=x-1$

So

Hence

Since in each interval the expression is already nonnegative, we get

3. Check continuity at possible problem points $x=-1,0,1$

These are the only possible points where continuity/differentiability may fail.

At $x=-1$

Left limit:

Right limit:

Since LHL $\ne$ RHL, $|f(x)|$ is not continuous at $x=-1$.

At $x=0$

Left limit:

Right limit:

Also, $|f(0)|=0$ So $|f(x)|$ is continuous at $x=0$.

At $x=1$

Left limit:

Right limit:

And $|f(1)|=|0|=0$ So $|f(x)|$ is continuous at $x=1$.

Thus the number of discontinuity points is $m=1.$

4. Check differentiability

A function not continuous at a point cannot be differentiable there. So at $x=-1$, $|f(x)|$ is not differentiable.

Now check the continuous junctions $x=0$ and $x=1$.

At $x=0$

From the left, $|f(x)|=-x$, so left derivative is

From the right, $|f(x)|=0$, so right derivative is

Since $-1\ne 0$, $|f(x)|$ is not differentiable at $x=0$.

At $x=1$

From the left, $|f(x)|=0$, so left derivative is $0$ From the right, $|f(x)|=x-1$, so right derivative is $1$ Since $0\ne 1$, $|f(x)|$ is not differentiable at $x=1$.

Inside each open interval $(-2,-1),(-1,0),(0,1),(1,2)$, the function is linear, hence differentiable.

Therefore, the number of non-differentiability points is $n=3.$

5. Compute required value

$m+n=1+3=4$

So the required integer is $\boxed{4}$

6. Comparison with stored answer

Stored correct answer = $4$.

Our derived answer also is $4$, so they agree.