Number of Discontinuity Points of Composite Function

To find the number of points of discontinuity for the composite function $(g \circ f)(x) = g(f(x))$, we first need to explicitly define $g(f(x))$ for different intervals of $x$.

The given functions are: $f(x) = \begin{cases} x^3 + 8, & x < 0 \\ x^2 - 4, & x \ge 0 \end{cases}$

$g(x) = \begin{cases} (x - 8)^{1/3}, & x < 0 \\ (x + 4)^{1/2}, & x \ge 0 \end{cases}$

---

Step 1: Evaluate $g(f(x))$ for $x < 0$

When $x < 0$, $f(x) = x^3 + 8$. We split this based on the sign of $f(x)$:

1. When $f(x) < 0$: $x^3 + 8 < 0 \implies x^3 < -8 \implies x < -2$ Since $f(x) < 0$, we use the definition of $g(y)$ for $y < 0$: $g(f(x)) = (f(x) - 8)^{1/3} = (x^3 + 8 - 8)^{1/3} = (x^3)^{1/3} = x$

2. When $f(x) \ge 0$: $x^3 + 8 \ge 0 \implies x \ge -2$ Combined with $x < 0$, this gives the interval $x \in [-2, 0)$. Since $f(x) \ge 0$, we use the definition of $g(y)$ for $y \ge 0$: $g(f(x)) = (f(x) + 4)^{1/2} = (x^3 + 8 + 4)^{1/2} = \sqrt{x^3 + 12}$

---

Step 2: Evaluate $g(f(x))$ for $x \ge 0$

When $x \ge 0$, $f(x) = x^2 - 4$. We split this based on the sign of $f(x)$:

1. When $f(x) < 0$: $x^2 - 4 < 0 \implies -2 < x < 2$ Combined with $x \ge 0$, this gives the interval $x \in [0, 2)$. Since $f(x) < 0$, we use the definition of $g(y)$ for $y < 0$: $g(f(x)) = (f(x) - 8)^{1/3} = (x^2 - 4 - 8)^{1/3} = (x^2 - 12)^{1/3}$

2. When $f(x) \ge 0$: $x^2 - 4 \ge 0 \implies x \ge 2 \text{ or } x \le -2$ Combined with $x \ge 0$, this gives the interval $x \in [2, \infty)$. Since $f(x) \ge 0$, we use the definition of $g(y)$ for $y \ge 0$: $g(f(x)) = (f(x) + 4)^{1/2} = (x^2 - 4 + 4)^{1/2} = \sqrt{x^2} = x \quad (\text{since } x \ge 2)$

---

Step 3: Piecewise Definition of $g(f(x))$

Combining all the intervals, we get:

$$g(f(x)) = \begin{cases} x, & x < -2 \\ \sqrt{x^3 + 12}, & -2 \le x < 0 \\ (x^2 - 12)^{1/3}, & 0 \le x < 2 \\ x, & x \ge 2 \end{cases}$$

---

Step 4: Check Continuity at Critical Points

1. At $x = -2$: $\lim_{x \to -2^-} g(f(x)) = -2$ $\lim_{x \to -2^+} g(f(x)) = \sqrt{(-2)^3 + 12} = \sqrt{4} = 2$ Since $\lim_{x \to -2^-} g(f(x)) \neq \lim_{x \to -2^+} g(f(x))$, $g(f(x))$ is discontinuous at $x = -2$.

2. At $x = 0$: $\lim_{x \to 0^-} g(f(x)) = \sqrt{0^3 + 12} = \sqrt{12} = 2\sqrt{3}$ $\lim_{x \to 0^+} g(f(x)) = (0^2 - 12)^{1/3} = (-12)^{1/3} = -\sqrt[3]{12}$ Since $\lim_{x \to 0^-} g(f(x)) \neq \lim_{x \to 0^+} g(f(x))$, $g(f(x))$ is discontinuous at $x = 0$.

3. At $x = 2$: $\lim_{x \to 2^-} g(f(x)) = (2^2 - 12)^{1/3} = (-8)^{1/3} = -2$ $\lim_{x \to 2^+} g(f(x)) = 2$ Since $\lim_{x \to 2^-} g(f(x)) \neq \lim_{x \to 2^+} g(f(x))$, $g(f(x))$ is discontinuous at $x = 2$.

Within each open interval $(-\infty, -2)$, $(-2, 0)$, $(0, 2)$, and $(2, \infty)$, the functions composing $g(f(x))$ are continuous.

---

Conclusion

The points of discontinuity for $g(f(x))$ are $x = -2$, $x = 0$, and $x = 2$.

Therefore, the number of points where the function $g(f(x))$ is discontinuous is 3.