Let f(x) = begin{cases} x^3+8; x

Question

(A) First, analyze the continuity of $f(x)$ at $x=0$. $f(0^-) = \lim_{x 	o 0^-} (x^3+8) = 8$. $f(0^+) = \lim_{x 	o 0^+} (x^2-4) = -4$. Since $f(0^-)

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$1$

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(A) First, analyze the continuity of $f(x)$ at $x=0$. $f(0^-) = \lim_{x 	o 0^-} (x^3+8) = 8$. $f(0^+) = \lim_{x 	o 0^+} (x^2-4) = -4$. Since $f(0^-) 
eq f(0^+)$, $f(x)$ is discontinuous at $x=0$. Now, consider the composite function $g(f(x))$. At $x=0$, the left-hand limit is $g(f(0^-)) = g(8)$. Since $8 \ge 0$, $g(8) = (8+4)^{1/2} = \sqrt{12} = 2\sqrt{3}$. The right-hand limit is $g(f(0^+)) = g(-4)$. Since $-4 Since $g(f(0^-)) 
eq g(f(0^+))$, $g(f(x))$ is discontinuous at $x=0$. Next, check for points where $f(x)$ takes values that make $g(f(x))$ discontinuous. $g(u)$ is continuous for all $u$ in its domain. We check if $f(x)$ crosses the boundary $u=0$ where $g(u)$ changes definition. For $x For $x \ge 0$, $f(x) = x^2-4 = 0 \implies x = 2$. At $x=2$, $f(2)=0$. $g(f(2)) = g(0) = 2$. The limit exists and is continuous. Thus, the only point of discontinuity is $x=0$. The number of points is $1$.

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 ————

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