1. Understand the function $f(x)$

Given

For $|x|\le 2$,

So $f$ is a triangular function supported on $[-2,2]$.

More explicitly,

• for $-2\le x\le 0$, $|x|=-x$, so $f(x)=3+\frac{3x}{2};$
• for $0\le x\le 2$, $|x|=x$, so $f(x)=3-\frac{3x}{2};$
• outside $[-2,2]$, $f(x)=0$.

Thus $f$ is continuous everywhere, but not differentiable at $x=-2,\;0,\;2.$

2. Form $g(x)=f(x+2)-f(x-2)$

We determine where each term is nonzero.

• $f(x+2)\neq 0$ when $|x+2|\le 2 \implies -4\le x\le 0.$
• $f(x-2)\neq 0$ when $|x-2|\le 2 \implies 0\le x\le 4.$

So $g(x)$ is supported on $[-4,4]$, with a joining point at $x=0$.

3. Find piecewise expression for $g$

Case 1: $-4\le x\le -2$

Then $x+2\in[-2,0]$, so $f(x+2)=3+\frac{3}{2}(x+2)=6+\frac{3x}{2}.$ Also $x-2<-2$, so $f(x-2)=0$. Hence $g(x)=6+\frac{3x}{2}.$

Case 2: $-2\le x\le 0$

Then $x+2\in[0,2]$, so $f(x+2)=3-\frac{3}{2}(x+2)=-\frac{3x}{2}.$ And still $x-2\in[-4,-2]$, so $f(x-2)=0$. Thus $g(x)=-\frac{3x}{2}.$

Case 3: $0\le x\le 2$

Then $x+2\in[2,4]$, so $f(x+2)=0$ except at endpoint $x=0$, but formula-wise on this interval it is $0$. Also $x-2\in[-2,0]$, so $f(x-2)=3+\frac{3}{2}(x-2)=\frac{3x}{2}.$ Hence $g(x)= -\frac{3x}{2}.$

Case 4: $2\le x\le 4$

Then $x+2>2$, so $f(x+2)=0$. Also $x-2\in[0,2]$, so $f(x-2)=3-\frac{3}{2}(x-2)=6-\frac{3x}{2}.$ Thus $g(x)= -6+\frac{3x}{2}.$

Outside these intervals, $g(x)=0$.

So,

4. Count points where $g$ is not continuous

Possible trouble points are the junctions: $x=-4,-2,2,4.$ Check continuity:

• At $x=-4$: left value $=0$, right value $=6+\frac{3(-4)}{2}=0$. Continuous.

• At $x=-2$: from left $=6+\frac{3(-2)}{2}=3$, from right $=-\frac{3(-2)}{2}=3$. Continuous.

• At $x=2$: from left $=-\frac{3(2)}{2}=-3$, from right $=-6+\frac{3(2)}{2}=-3$. Continuous.

• At $x=4$: from left $=-6+\frac{3(4)}{2}=0$, right value $=0$. Continuous.

Also at $x=0$, from both sides $g(x)=-\frac{3x}{2}\to 0$, so continuous.

Therefore, $n=0.$

5. Count points where $g$ is not differentiable

On each open interval, $g$ is linear, so differentiable there. Possible non-differentiability points are the junctions: $x=-4,-2,2,4.$ (At $x=0$, same formula $-\frac{3x}{2}$ on both sides, so differentiable.)

Now compare left and right derivatives:

• At $x=-4$: left derivative $=0$, right derivative $=\frac{3}{2}$. Not differentiable.

• At $x=-2$: left derivative $=\frac{3}{2}$, right derivative $=-\frac{3}{2}$. Not differentiable.

• At $x=2$: left derivative $=-\frac{3}{2}$, right derivative $=\frac{3}{2}$. Not differentiable.

• At $x=4$: left derivative $=\frac{3}{2}$, right derivative $=0$. Not differentiable.

Thus, $m=4.$

6. Final value

$n+m=0+4=4.$

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