Answer
$f(x)$ is discontinuous at $x=1$ (not removable) and $x=2$ (removable).
Work Step by Step
Refer back to the graph in Exercise 1, section 2.4.
The graph is defined in the domain $[-1,3]$. We would examine continuity only in this interval.
- At $x=1$, we see that as $\lim_{x\to1^-}f(x)=2$, while $\lim_{x\to1^+}f(x)=1$, so we can conclude that $\lim_{x\to1}f(x)$ does not exist here. The graph is discontinuous at $x=1$ as a result.
Since $\lim_{x\to1}f(x)$ does not even exist, this discontinuity is not removable.
- At $x=2$, we see that as $\lim_{x\to2^-}f(x)=\lim_{x\to2^+}f(x)=1$, so we can conclude that $\lim_{x\to2}f(x)=1$. However, $f(2)=2$. Since $\lim_{x\to2}f(x)\ne f(2)$, $f(x)$ is discontinuous at $x=2$.
Yet, since we can remove the discontinuity if we set $f(2)$ equal to $\lim_{x\to2}f(x)=1$, this discontinuity is removable.
