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

Question

(C) First, check the continuity of $g(x)$ at $x=0$. $g(0^-) = \lim_{x 	o 0^-} (x-8)^{1/3} = (-8)^{1/3} = -2$. $g(0^+) = \lim_{x 	o 0^+} (x+4)^{1/2}

Vedclass · Accepted Answer

$3$

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(C) First, check the continuity of $g(x)$ at $x=0$. $g(0^-) = \lim_{x 	o 0^-} (x-8)^{1/3} = (-8)^{1/3} = -2$. $g(0^+) = \lim_{x 	o 0^+} (x+4)^{1/2} = (4)^{1/2} = 2$. Since $g(0^-) 
eq g(0^+)$, $g(x)$ is discontinuous at $x=0$.Next, check the continuity of $g(f(x))$.$1$. At $x=0$: $f(0^-) = 8$ and $f(0^+) = -4$. Then $g(f(0^-)) = g(8) = (8+4)^{1/2} = \sqrt{12} = 2\sqrt{3}$ and $g(f(0^+)) = g(-4) = (-4-8)^{1/3} = (-12)^{1/3} = -\sqrt[3]{12}$. Since $g(f(0^-)) 
eq g(f(0^+))$, $g(f(x))$ is discontinuous at $x=0$.$2$. Where $f(x) = 0$: $g(f(x))$ is discontinuous if $f(x) = 0$ because $g(u)$ is discontinuous at $u=0$.For $x For $x \ge 0$, $f(x) = x^2-4 = 0 \implies x = 2$. At $x=2$, $f(x)$ is continuous and $f(2)=0$. Since $g(u)$ is discontinuous at $u=0$, $g(f(x))$ is discontinuous at $x=2$.Thus, the points of discontinuity are $x = -2, 0, 2$. The number of points is $3$.

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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