Answer
$h(x)$ has an absolute maximum value at $x=4$, where $h(4)=2$, but it does not have any absolute minimum value.
Work Step by Step
$h(x)=1/x$ for $-1\le x\lt0$ and $h(x)=\sqrt x$ for $0\le x\le4$
The graph is sketched below.
- The graph has an absolute maximum value at $x=4$, where $h(4)=2$.
- The graph, however, does not have any absolute minimum value, as we notice that as $x\to0^-$, $h(x)$ approaches $-\infty$.
The reason Theorem 1 cannot be applied here is because $h(x)$ is not continuous on the closed interval $[-1,4]$, having a break at $x=0$. So the answer is still consistent with Theorem 1.
