$f ( x )=\left\{\begin{array}{cc}3\left(1-\frac{| x |}{2}\right) & \text { if }| x | \leq 2 \text { } \\ 0 & \text { if }| x |>2 \text { }\end{array}\right.$ Let $g: R \rightarrow R$ be given by $g(x)=f(x+2)-f(x-2)$. If $n$ and $m$ denote the number of points in $R$ where $\mathrm{g}$ is not continuous and not differentiable, respectively, then $\mathrm{n}+\mathrm{m}$ is equal to $....$
$\quad \quad \quad \quad \quad \quad -\frac{3 x}{2} \quad -2\,$
$\quad \quad \quad \quad \quad \quad 0 \quad \quad \quad x \in(-\infty,-4) \cup(0,+\infty)$
$f(x+2)\rightarrow\frac{3 x}{2}+6 \quad \quad 0 \leq x \leq 2$
$\quad \quad \quad \quad \quad -\frac{3 x}{2}+6 \quad 2\,<\,x \leq 4$
$\quad \quad \quad \quad \quad 0 \quad \quad \quad \quad x \in(-\infty, 0) \cup(4,+\infty)$
$g(x)=f(x+2)-f(x-2) \rightarrow\frac{3 x}{2}+6 \quad -4 \leq x \leq-2$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -\frac{3 x}{2} \quad \quad -2\, < \,x\, < \,2$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \frac{3 x}{2}-6\quad \quad 2 \leq x \leq 4$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0 \quad \quad \quad \quad x \in(-\infty,-4) \cup(4,+\infty)$
$n=0$
$m=4 \Rightarrow(n+m=4)$
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