Answer
$f(x)$ is not continuous on $[-1,3]$. It is discontinuous at $x=2$.
Work Step by Step
The exercises 1-4 mention the appearance of "breaks" as a sign of discontinuities in the graph, which we shall prove later on.
The graph of $f(x)$ in Exercise 1 has a break at $x=2$. $f(2)$, as seen in the graph, does not exist.
We also notice that as $x$ approaches $2$, $f(x)$ approaches $1$. So $\lim_{x\to2}f(x)=1$
Because while $\lim_{x\to2}f(x)$ exists, $f(2)$ does not exist, according to definition, $f(x)$ is discontinuous at $x=2$.
