Answer
$h$ has an absolute maximum at $x=4$, and no absolute minimum.
Th. 1 does not apply, so the answer is not inconsistent with it.
Work Step by Step
See image for the graph of $h(x)$.
The point ($4,2$) is on the graph and its y-coordinate is greater than any other on the graph.
So, $h$ has an absolute maximum at $x=2$.
There is no point on the graph such that it has the lowest possible y-coordinate.
No absolute minimum.
Theorem 1 demands that "$f$ is continuous...",
so it does not apply here (h is not continuous at $x=0$).
Thus, there are no inconsistencies.