Answer
$g(x)$ has an absolute maximum value at $x=2$, where $g(2)=1$, but it does have any absolute minimum value.
Work Step by Step
$g(x)=-x$ for $0\le x\lt1$ and $g(x)=x-1$ for $1\le x\le2$
The graph is sketched below.
- The graph has an absolute maximum value at $x=2$, where $g(2)=1$.
- The graph, however, does not have any absolute minimum value.
Looking at the graph, it does seem to reach the lowest point at $x=1$, where the value of $g(x)=-1$. However, a closer look at the given function shows that $g(1)=1-1=0$ and this is nowhere the absolute minimum value of $g(x)$.
The problem here is that as $x\to1^-$, $g(x)$ does approach $-1$. But at $x=1$, $g(x)$ "jumps" to $0$ and continues from there. Therefore, there is no lowest point for $g(x)$, and it has no absolute minimum value as a result.
The reason Theorem 1 cannot be applied here is because $g(x)$ is not continuous on the closed interval $[0,2]$, having a break at $x=1$. So the answer is still consistent with Theorem 1.
