Let the functions $$f : R \to R$$ and $$g : R \to R$$ be defined as:
$$f(x) = \begin{cases} x+2, & x < 0 \\ x^2, & x \geq 0 \end{cases}$$ and $$g(x) = \begin{cases} x^3, & x < 1 \\ 3x-2, & x \geq 1 \end{cases}$$
Then, the number of points in $$R$$ where $$(f \circ g)(x)$$ is NOT differentiable is equal to:

We have $$f(x) = \begin{cases} x+2, & x < 0 \\ x^2, & x \geq 0 \end{cases}$$ and $$g(x) = \begin{cases} x^3, & x < 1 \\ 3x-2, & x \geq 1 \end{cases}$$. We need to find where $$(f \circ g)(x) = f(g(x))$$ is not differentiable.

For $$x < 0$$: $$g(x) = x^3 < 0$$, so $$f(g(x)) = x^3 + 2$$. For $$0 \leq x < 1$$: $$g(x) = x^3 \geq 0$$, so $$f(g(x)) = (x^3)^2 = x^6$$. For $$x \geq 1$$: $$g(x) = 3x - 2 \geq 1 > 0$$, so $$f(g(x)) = (3x-2)^2$$.

At $$x = 0$$: From the left, $$\lim_{x \to 0^-}(x^3 + 2) = 2$$. From the right, $$\lim_{x \to 0^+} x^6 = 0$$. Also $$f(g(0)) = f(0) = 0$$. Since the left-hand limit does not equal the function value, $$f \circ g$$ is not continuous (hence not differentiable) at $$x = 0$$.

At $$x = 1$$: From the left, $$\lim_{x \to 1^-} x^6 = 1$$. From the right, $$(3(1)-2)^2 = 1$$. So $$f \circ g$$ is continuous at $$x = 1$$. The left derivative is $$6x^5|_{x=1} = 6$$ and the right derivative is $$2(3x-2)(3)|_{x=1} = 6$$. Since both derivatives are equal, $$f \circ g$$ is differentiable at $$x = 1$$.

Therefore, there is exactly 1 point in $$\mathbb{R}$$ where $$(f \circ g)(x)$$ is not differentiable.