Let $$f : R \to R$$ be a function defined as $$f(x) = \begin{cases} 3\left(1 - \frac{|x|}{2}\right) & \text{if } |x| \le 2 \\ 0 & \text{if } |x| > 2 \end{cases}$$
Let $$g : R \to R$$ be given by $$g(x) = f(x+2) - f(x-2)$$. 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 function $$f(x) = \begin{cases} 3\left(1 - \frac{|x|}{2}\right) & \text{if } |x| \le 2 \\ 0 & \text{if } |x| > 2 \end{cases}$$ is a triangular function with peak value 3 at $$x = 0$$, decreasing linearly to 0 at $$x = \pm 2$$.

We can write $$f$$ piecewise as: $$f(x) = 0$$ for $$x < -2$$; $$f(x) = 3 + \frac{3x}{2}$$ for $$-2 \le x \le 0$$; $$f(x) = 3 - \frac{3x}{2}$$ for $$0 \le x \le 2$$; $$f(x) = 0$$ for $$x > 2$$.

Now $$g(x) = f(x+2) - f(x-2)$$. The function $$f(x+2)$$ is nonzero when $$|x+2| \le 2$$, i.e., $$-4 \le x \le 0$$. The function $$f(x-2)$$ is nonzero when $$|x-2| \le 2$$, i.e., $$0 \le x \le 4$$.

Computing $$f(x+2)$$ piecewise: For $$-4 \le x \le -2$$, we have $$x + 2 \in [-2, 0]$$, so $$f(x+2) = 3 + \frac{3(x+2)}{2} = 6 + \frac{3x}{2}$$. For $$-2 \le x \le 0$$, we have $$x + 2 \in [0, 2]$$, so $$f(x+2) = 3 - \frac{3(x+2)}{2} = -\frac{3x}{2}$$.

Computing $$f(x-2)$$ piecewise: For $$0 \le x \le 2$$, we have $$x - 2 \in [-2, 0]$$, so $$f(x-2) = 3 + \frac{3(x-2)}{2} = \frac{3x}{2}$$. For $$2 \le x \le 4$$, we have $$x - 2 \in [0, 2]$$, so $$f(x-2) = 3 - \frac{3(x-2)}{2} = 6 - \frac{3x}{2}$$.

Therefore $$g(x)$$ is: $$g(x) = 0$$ for $$x < -4$$; $$g(x) = 6 + \frac{3x}{2}$$ for $$-4 \le x \le -2$$; $$g(x) = -\frac{3x}{2}$$ for $$-2 \le x \le 0$$; $$g(x) = -\frac{3x}{2}$$ for $$0 \le x \le 2$$; $$g(x) = -6 + \frac{3x}{2}$$ for $$2 \le x \le 4$$; $$g(x) = 0$$ for $$x > 4$$.

Notice that on $$[-2, 2]$$, $$g(x) = -\frac{3x}{2}$$ is a single linear piece, so $$g$$ is smooth there (including at $$x = 0$$).

Checking continuity at each breakpoint: At $$x = -4$$: left limit is 0 and $$g(-4) = 6 - 6 = 0$$. Continuous. At $$x = -2$$: left value $$= 6 - 3 = 3$$ and right value $$= 3$$. Continuous. At $$x = 0$$: both sides give 0. Continuous. At $$x = 2$$: left value $$= -3$$ and right value $$= -6 + 3 = -3$$. Continuous. At $$x = 4$$: left value $$= -6 + 6 = 0$$ and right value $$= 0$$. Continuous.

So $$g$$ is continuous everywhere, meaning $$n = 0$$.

Checking differentiability: At $$x = -4$$: left derivative is 0, right derivative is $$\frac{3}{2}$$. Not differentiable. At $$x = -2$$: left derivative is $$\frac{3}{2}$$, right derivative is $$-\frac{3}{2}$$. Not differentiable. At $$x = 0$$: both sides have derivative $$-\frac{3}{2}$$. Differentiable. At $$x = 2$$: left derivative is $$-\frac{3}{2}$$, right derivative is $$\frac{3}{2}$$. Not differentiable. At $$x = 4$$: left derivative is $$\frac{3}{2}$$, right derivative is 0. Not differentiable.

So $$g$$ is not differentiable at 4 points: $$x = -4, -2, 2, 4$$, giving $$m = 4$$.

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