which values of x are point(s) of discontinuity for this function?\ng(x)=\begin{cases}-4, & xleq…

Explanation:

Step1: Check continuity at $x = - 2$

Left - hand limit: $\lim_{x\rightarrow - 2^{-}}g(x)=-4$. Right - hand limit: $\lim_{x\rightarrow - 2^{+}}g(x)=-(-2)^{2}=-4$. Since $\lim_{x\rightarrow - 2^{-}}g(x)=\lim_{x\rightarrow - 2^{+}}g(x)=-4$, the function is continuous at $x = - 2$.

Step2: Check continuity at $x = 0$

Left - hand limit: $\lim_{x\rightarrow0^{-}}g(x)=-(0)^{2}=0$. Right - hand limit: $\lim_{x\rightarrow0^{+}}g(x)=(0)^{2}=0$. Since $\lim_{x\rightarrow0^{-}}g(x)=\lim_{x\rightarrow0^{+}}g(x)=0$, the function is continuous at $x = 0$.

Step3: Check continuity at $x = 2$

Left - hand limit: $\lim_{x\rightarrow2^{-}}g(x)=(2)^{2}=4$. Right - hand limit: $\lim_{x\rightarrow2^{+}}g(x)=2$. Since $\lim_{x\rightarrow2^{-}}g(x)\neq\lim_{x\rightarrow2^{+}}g(x)$, the function is discontinuous at $x = 2$.

Step4: Check continuity at $x=-4$ and $x = 4$

For $x=-4$, $g(x)=-4$ (a constant in its sub - domain), so it is continuous. For $x = 4$, $g(x)=2$ (a constant in its sub - domain), so it is continuous.

Answer:

$x = 2$