Answer
$f(x)$ is discontinuous at $x=0$ (removable) and $x=1$ (not removable).
Work Step by Step
Refer back to the graph in Exercise 1, section 2.4.
The graph is defined in the domain $[-1,2]$. We would examine continuity only in this interval.
- At $x=0$, we see that as $\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)=0$, we can conclude that $\lim_{x\to0}f(x)=0$. However, $f(0)=1$. Since $\lim_{x\to0}f(x)\ne f(0)$, $f(x)$ is discontinuous at $x=0$.
Yet, since we can remove the discontinuity if we set $f(0)$ equal to $\lim_{x\to0}f(x)=0$, this discontinuity is removable.
- At $x=1$, we see that as $\lim_{x\to1^-}f(x)=1$, while $\lim_{x\to1^+}f(x)=0$, 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.
