The given function is f(x)={\begin{matrix}3(1-\frac{|x|}{2}) & if & |x|\leq 2 \\ 0 & if & |x|>2\end{matrix}
By using the definition of modulus function, we have |x|={\begin{matrix}x, & x\geq 0 \\ -x, & x<0\end{matrix} and also, we know that if ∣x∣≤a,⇒−a≤x≤a and if ∣x∣>a,⇒x<−aorx>a,
\Rightarrow f(x)={\begin{matrix}3(1+\frac{x}{2}), & -2\leq x\leq 0 \\ 3(1-\frac{x}{2}), & 0<x\leq 2 \\ 0, & -2>xorx>2\end{matrix}
Now, f(x-2)={\begin{matrix}3(1+\frac{x-2}{2}), & -2\leq x-2\leq 0 \\ 3(1-\frac{x-2}{2}), & 0<x-2\leq 2 \\ 0, & -2>x-2orx-2>2\end{matrix}
\Rightarrow f(x-2)={\begin{matrix}\frac{3x}{2}, & 0\leq x\leq 2 \\ 6-\frac{3x}{2}, & 2<x\leq 4 \\ 0, & 0>xorx>4\end{matrix}
And, f(x+2)={\begin{matrix}3(1+\frac{x+2}{2}), & -2\leq x+2\leq 0 \\ 3(1-\frac{x+2}{2}), & 0<x+2\leq 2 \\ 0, & -2>x+2orx+2>2\end{matrix}
\Rightarrow f(x+2)={\begin{matrix}6+\frac{3x}{2}, & -4\leq x\leq -2 \\ -\frac{3x}{2}, & -2<x\leq 0 \\ 0, & -4>xorx>0\end{matrix}
Hence, g(x)=f(x+2)-f(x-2)={\begin{matrix}\frac{3x}{2}+6,-4\leq x\leq -2 \\ -\frac{3x}{2},-2<x<2 \\ \frac{3x}{2}-6,2\leq x\leq 4 \\ 0,x\in (-\infty ,-4)\cup (4,\infty )\end{matrix}
The graph of g(x) is given below
From the graph we can observe that the function g(x) is continuous everywhere, hence the number of points where g(x) is not continuos is n=0 and the graph has sharp corners at the points (-4,0),(4,0),(-2,3)&(2,-3) thus the number of points where g(x) is not differentiable are m=4
⇒n+m=4.