use the graph to determine the x - values at which f is discontinuous. for each x - value, determine whether…

Explanation:

Step1: Recall continuity definition

A function $f(x)$ is continuous at $x = a$ if $\lim_{x\rightarrow a^{-}}f(x)=\lim_{x\rightarrow a^{+}}f(x)=f(a)$. Discontinuities occur where this fails.

Step2: Analyze $x=-2$

At $x = - 2$, the left - hand limit $\lim_{x\rightarrow - 2^{-}}f(x)$ exists, the right - hand limit $\lim_{x\rightarrow - 2^{+}}f(x)$ exists, but $f(-2)$ is not defined (open - circle at $x=-2$). The function is continuous from the left since $\lim_{x\rightarrow - 2^{-}}f(x)=f(-2)$ would hold if $f(-2)$ was defined as the left - hand limit value.

Step3: Analyze $x = 2$

At $x = 2$, $\lim_{x\rightarrow 2^{-}}f(x)$ exists, $\lim_{x\rightarrow 2^{+}}f(x)$ exists, and $f(2)$ is defined. But $\lim_{x\rightarrow 2^{-}}f(x)\neq\lim_{x\rightarrow 2^{+}}f(x)$. The function is continuous from the right since $\lim_{x\rightarrow 2^{+}}f(x)=f(2)$.

Step4: Analyze $x = 4$

At $x = 4$, the function has a vertical asymptote. $\lim_{x\rightarrow 4^{-}}f(x)=-\infty$ and $\lim_{x\rightarrow 4^{+}}f(x)$ exists. The function is neither continuous from the left nor from the right.

Answer:

$x=-2$, continuous from the left $x = 2$, continuous from the right $x = 4$, neither