Observe the graph given below: State the value(s) of where the function has removable…

At $x = -2$, the function has a removable discontinuity because $\lim_{x \to -2^-} f(x) = \lim_{x \to -2^+} f(x) = 2$ (both one-sided limits exist and are equal), but $f(-2)$ is undefined.

The graph of the function is not differentiable at $x = -2, 0, 2$ because it is discontinuous at these points (discontinuity at $x = -2$ is removable, and at $x = 0$ and $x = 2$ there are jump discontinuities).