Let f: R → R be a function defined as f(x) = \ cases 3(1 - (| x...

The given function is f\left(x\right) = \left\{\begin{cases}3\left(1 - \dfrac{| x |}{2}\right) & if & | x | \le 2 \\ 0 & if & | x | > 2\end{cases}\right)</p><p>By using the definition of modulus function, we have\left|x\right| = \left\{$\begin{cases}x, & x \ge 0 \\ - x, & x < 0\end{cases}$\right) and also, we know that if $\left|x\right| \le a, ⇒ - a \le x \le a and if$\left|x\right| > a, ⇒ x a,$</p><p>$⇒ f\left(x\right) = \left\{$\begin{cases}3\left(1 + \dfrac{x}{2}\right), & - 2 \le x \le 0 \\ 3\left(1 - \dfrac{x}{2}\right), & 0 xorx > 2\end{cases}$\right)

Now, f\left(x - 2\right) = \left\{\begin{cases}3\left(1 + \dfrac{x - 2}{2}\right), & - 2 \le x - 2 \le 0 \\ 3\left(1 - \dfrac{x - 2}{2}\right), & 0 x - 2orx - 2 > 2\end{cases}\right)</p><p>⇒ f\left(x - 2\right) = \left\{$\begin{cases}\dfrac{3x}{2}, & 0 \le x \le 2 \\ 6 - \dfrac{3x}{2}, & 2 xorx > 4\end{cases}$\right)

And, f\left(x + 2\right) = \left\{\begin{cases}3\left(1 + \dfrac{x + 2}{2}\right), & - 2 \le x + 2 \le 0 \\ 3\left(1 - \dfrac{x + 2}{2}\right), & 0 x + 2orx + 2 > 2\end{cases}\right)</p><p>⇒ f\left(x + 2\right) = \left\{$\begin{cases}6 + \dfrac{3x}{2}, & - 4 \le x \le - 2 \\ - \dfrac{3x}{2}, & - 2 xorx > 0\end{cases}$\right)

Hence, g\left(x\right) = f\left(x + 2\right) - f\left(x - 2\right) = \left\{$\begin{array}{ll}\dfrac{3x}{2} + 6, - 4 \le x \le - 2 \\ - \dfrac{3x}{2}, - 2 < x < 2 \\ \dfrac{3x}{2} - 6, 2 \le x \le 4 \\ 0, x \in ( - \infty , - 4 ) \cup ( 4, \infty )\end{array}$\right)</p><p>The graph ofg$\left(x\right)$ 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) \text{ and } \left(2, - 3\right)$thus the number of points where$g$\left(x\right)$ is not differentiable are $m = 4$

$⇒ n + m = 4 .$