Answer
$k(x)$ is not continuous on $[-1,3]$. It is discontinuous at $x=1$.
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 $k(x)$ in Exercise 4 has a break at $x=1$. $k(1)$, as seen from the graph, equals $0$.
We also notice that as $x$ approaches $1$ from the left, $k(x)$ approaches $1.5$; but as $x$ approaches $1$ from the right, $k(x)$ approaches $0$. In other words, $\lim_{x\to1^-}k(x)\ne\lim_{x\to1^+}k(x)$, so $\lim_{x\to1}k(x)$ does not exist.
Therefore, since $\lim_{x\to1}k(x)$ does not exist while $k(1)$ does, $k(x)$ is discontinuous at $x=1$.
