Let $$f(x) = \begin{cases} x^3 + 8, & x < 0 \\ x^2 - 4, & x \ge 0 \end{cases}$$ and $$g(x) = \begin{cases} (x - 8)^{1/3}, & x < 0 \\ (x + 4)^{1/2}, & x \ge 0 \end{cases}$$. Then the number of points, where the function $$g \circ f$$ is discontinuous, is __________.

$$\lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)$$, $$f(x)$$ is discontinuous at $$x = 0$$

$$\lim_{x \to 0^-} g(x) \neq \lim_{x \to 0^+} g(x)$$, $$g(x)$$ is discontinuous at its own boundary $$x = 0$$

The definition of $$g(x)$$ changes behavior at its input value $$0$$. Therefore, $$g(f(x))$$ will potentially change its functional form whenever $$f(x) = 0$$

$$x^3 + 8 = 0 \implies x^3 = -8 \implies x = -2$$

$$x^2 - 4 = 0 \implies x^2 = 4 \implies x = 2 \quad (\text{as } x \ge 0)$$

Thus, the potential critical transition points for $$g(f(x))$$ are $$x = -2, 0, 2$$

$$\lim_{x \to -2^-} g(f(x)) = \lim_{f(x) \to 0^-} (f(x) - 8)^{1/3} = (-8)^{1/3} = -2$$

$$\lim_{x \to -2^+} g(f(x)) = \lim_{f(x) \to 0^+} (f(x) + 4)^{1/2} = (4)^{1/2} = 2$$

Hence, discontinuous at $$x = -2$$

$$\lim_{x \to 0^-} g(f(x)) = (8 + 4)^{1/2} = \sqrt{12}$$

$$\lim_{x \to 0^+} g(f(x)) = (-4 - 8)^{1/3} = (-12)^{1/3}$$

Hence, discontinuous at $$x = 0$$

$$\lim_{x \to 2^-} g(f(x)) = (0 - 8)^{1/3} = -2$$

$$\lim_{x \to 2^+} g(f(x)) = (0 + 4)^{1/2} = 2$$

Hence, discontinuous at $$x = 2$$