Let $f:R\to R$ be a function defined as $f\left(x\right)=\left\{\begin{array}{ccc}3\left(1-\frac{\left|x\right|}{2}\right)& \text{if}& \left|x\right|\le 2\\ 0& \text{if}& \left|x\right|>2\end{array}\right..$

Let $g:R\to R$ be given by $g\left(x\right)=f(x+2)-f(x-2).$ If $n$ and $m$ denote the number of points in $R$ where $g$ is not continuous and not differentiable, respectively, then $n+m$ is equal to ________.

The given function is $f\left(x\right)=\left\{\begin{array}{ccc}3\left(1-\frac{\left|x\right|}{2}\right)& if& \left|x\right|\le 2\\ 0& if& \left|x\right|>2\end{array}\right.$

By using the definition of modulus function, we have $\left|x\right|=\left\{\begin{array}{ll} x,& x\ge 0\\ -x, & x<0\end{array}\right.$ and also, we know that if $\left|x\right|\le a, \Rightarrow -a\le x\le a$ and if $\left|x\right|>a, \Rightarrow x<-a or x>a,$

$\Rightarrow f\left(x\right)=\left\{\begin{array}{cc}3\left(1+\frac{x}{2}\right),& -2\le x\le 0\\ 3\left(1-\frac{x}{2}\right),& 0<x\le 2\\ 0,& -2>x or x>2\end{array}\right.$

Now, $f\left(x-2\right)=\left\{\begin{array}{cc}3\left(1+\frac{x-2}{2}\right),& -2\le x-2\le 0\\ 3\left(1-\frac{x-2}{2}\right),& 0<x-2\le 2\\ 0,& -2>x-2 or x-2>2\end{array}\right.$

$\Rightarrow f\left(x-2\right)=\left\{\begin{array}{cc}\frac{3x}{2},& 0\le x\le 2\\ 6-\frac{3x}{2},& 2<x\le 4\\ 0,& 0>x or x>4\end{array}\right.$

And, $f\left(x+2\right)=\left\{\begin{array}{cc}3\left(1+\frac{x+2}{2}\right),& -2\le x+2\le 0\\ 3\left(1-\frac{x+2}{2}\right),& 0<x+2\le 2\\ 0,& -2>x+2 or x+2>2\end{array}\right.$

$\Rightarrow f\left(x+2\right)=\left\{\begin{array}{cc}6+\frac{3x}{2},& -4\le x\le -2\\ -\frac{3x}{2},& -2<x\le 0\\ 0,& -4>x or x>0\end{array}\right.$

Hence, $g\left(x\right)=f\left(x+2\right)-f\left(x-2\right)=\left\{\begin{array}{l}\frac{3x}{2}+6, -4\le x\le -2\\ -\frac{3x}{2}, -2<x<2\\ \frac{3x}{2}-6, 2\le x\le 4\\ 0, x\in (-\infty ,-4)\cup (4,\infty )\end{array}\right.$

The graph of $g\left(x\right)$ is given below

From the graph we can observe that the function $g\left(x\right)$ is continuous everywhere, hence the number of points where $g\left(x\right)$ is not continuos is $n=0$ and the graph has sharp corners at the points $\left(-4, 0\right), \left(4, 0\right), \left(-2, 3\right) \& \left(2, -3\right)$ thus the number of points where $g\left(x\right)$ is not differentiable are $m=4$

$\Rightarrow n+m=4.$