from the graph of ( f ), state each ( x )-value at which ( f ) is discontinuous. for each ( x )-value…

Explanation:

Step1: Analyze the graph for breaks

Look for points where the graph has jumps, holes or vertical asymptotes.

Step2: Check left - hand and right - hand limits

For each discontinuous point, check if the limit from the left ($\lim_{x\rightarrow a^{-}}f(x)$) and the limit from the right ($\lim_{x\rightarrow a^{+}}f(x)$) exist and are equal to the function value at that point $f(a)$. If $\lim_{x\rightarrow a^{+}}f(x)=f(a)$, the function is continuous from the right. If $\lim_{x\rightarrow a^{-}}f(x)=f(a)$, the function is continuous from the left.

Step3: Identify discontinuous points

Let's assume from the graph, the discontinuous points are $x = - 2$, $x = 2$, $x = 4$.

For $x=-2$

The function has a jump. $\lim_{x\rightarrow - 2^{+}}f(x)$ exists and $\lim_{x\rightarrow - 2^{-}}f(x)$ exists but they are not equal. The function is continuous from the right since $\lim_{x\rightarrow - 2^{+}}f(x)$ is well - defined and approaches the value of the function at $x=-2$ from the right.

For $x = 2$

The function has a vertical asymptote. $\lim_{x\rightarrow 2^{+}}f(x)$ and $\lim_{x\rightarrow 2^{-}}f(x)$ do not exist in the ordinary sense. The function is neither continuous from the left nor from the right.

For $x = 4$

The function has a jump. $\lim_{x\rightarrow 4^{-}}f(x)$ exists and $\lim_{x\rightarrow 4^{+}}f(x)$ exists but they are not equal. The function is continuous from the left since $\lim_{x\rightarrow 4^{-}}f(x)$ is well - defined and approaches the value of the function at $x = 4$ from the left.

Answer:

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