Let f(x) = cases x^3 + 8 ; & x Then the number of points, where the...

The possible points of discontinuity for the composite function $g(f(x))$ are the points where $f(x)$ is discontinuous and the points where $f(x)$ is equal to a point of discontinuity of $g(x)$.

First, we find the points of discontinuity of $f(x)$ and $g(x)$.

For $f(x)$, the only possible point of discontinuity is at $x = 0$.
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x^3 + 8) = 8$
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2 - 4) = -4$
Since $\lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)$, $f(x)$ is discontinuous at $x = 0$.

For $g(x)$, the only possible point of discontinuity is at $x = 0$.
$\lim_{x \to 0^-} g(x) = \lim_{x \to 0^-} (x - 8)^{1/3} = -2$
$\lim_{x \to 0^+} g(x) = \lim_{x \to 0^+} (x + 4)^{1/2} = 2$
Since $\lim_{x \to 0^-} g(x) \neq \lim_{x \to 0^+} g(x)$, $g(x)$ is discontinuous at $x = 0$.

Next, we find the points where $f(x)$ equals the point of discontinuity of $g(x)$, which is $f(x) = 0$.
For $x < 0$, $x^3 + 8 = 0 \Rightarrow x = -2$.
For $x \geq 0$, $x^2 - 4 = 0 \Rightarrow x = 2$.

Thus, the possible points of discontinuity for $g(f(x))$ are $x = -2$, $x = 0$, and $x = 2$. We check the continuity at each of these points.

At $x = -2$:
As $x \to -2^-$, $f(x) \to 0^-$, so $\lim_{x \to -2^-} g(f(x)) = \lim_{y \to 0^-} g(y) = -2$
As $x \to -2^+$, $f(x) \to 0^+$, so $\lim_{x \to -2^+} g(f(x)) = \lim_{y \to 0^+} g(y) = 2$
Since the left-hand limit and right-hand limit are not equal, $g(f(x))$ is discontinuous at $x = -2$.

At $x = 0$:
$\lim_{x \to 0^-} g(f(x)) = g(8) = (8 + 4)^{1/2} = 2\sqrt{3}$
$\lim_{x \to 0^+} g(f(x)) = g(-4) = (-4 - 8)^{1/3} = (-12)^{1/3}$
Since the limits are not equal, $g(f(x))$ is discontinuous at $x = 0$.

At $x = 2$:
As $x \to 2^-$, $f(x) \to 0^-$, so $\lim_{x \to 2^-} g(f(x)) = \lim_{y \to 0^-} g(y) = -2$
As $x \to 2^+$, $f(x) \to 0^+$, so $\lim_{x \to 2^+} g(f(x)) = \lim_{y \to 0^+} g(y) = 2$
Since the limits are not equal, $g(f(x))$ is discontinuous at $x = 2$.

Therefore, there are $3$ points of discontinuity.

Answer: $3$