Let  $$[x]$$  denote the greatest integer function, and let $$m$$  and  $$n$$  respectively be the numbers of the points  where the function $$f(x) = [x] + |x-2|, -2 < x < 3,$$ is not continuous and not differentiable. Then  $$m+n$$  is equal to:

 $$|x - 2|$$ is continuous everywhere and $$[x]$$ is discontinuous at all integers, within the interval $$(-2,3)$$ the integers are $$-1, 0, 1, 2$$.

Because $$f(x)$$ combines these two parts and only $$[x]$$ causes discontinuities, it follows that $$f(x)$$ is discontinuous exactly at $$x = -1, 0, 1, 2$$, giving $$m = 4$$.

 $$x = -1, 0, 1, 2$$, $$f$$ is non-differentiable at those four points. Although $$|x - 2|$$ introduces a corner at $$x = 2$$, this point has already been counted among the discontinuities. No additional points of non-differentiability arise because between consecutive integers both $$[x]$$ (which is constant) and $$|x - 2|$$ (which is linear) are differentiable. Therefore, $$n = 4$$.

$$m + n = 4 + 4 = 8$$.