Answer
$h(x)$ is continuous on $[-1,3]$.
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 $h(x)$ in Exercise 3 does not have breaks.
- At $x=-1$, $\lim_{x\to(-1)^+}h(x)=h(-1)=2$. So $h(x)$ is right-continuous at $x=-1$
- At $x=3$, $\lim_{x\to3^-}h(x)=h(3)=2$. So $h(x)$ is left-continuous at $x=3$
- For all interior points of the interval $[-1,3]$, $\lim_{x\to c}h(x)=h(c)$. So $h(x)$ is continuous at all interior points of the interval $[-1,3]$.
Therefore, $h(x)$ is continuous on $[-1,3]$.
