which values of x are point(s) of discontinuity for this function?\n g(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: $\lim_{x\rightarrow - 2^{+}}g(x)=-(-2)^{2}=-4$. $g(-2)=-4$. So, it 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: $\lim_{x\rightarrow0^{+}}g(x)=(0)^{2}=0$. But the function is not defined at $x = 0$. So, it is discontinuous 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 the left - hand and right - hand limits are not equal, it is discontinuous at $x = 2$.

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

The function is well - behaved and continuous in the intervals where $x=-4$ and $x = 4$ are located. For $x=-4$, $g(x)=-4$ (constant in $x\leq - 2$ interval). For $x = 4$, $g(x)=2$ (constant in $x\geq2$ interval).

Answer:

$x = 0$, $x = 2$