Answer
$f$ has an absolute maximum at $x=2$, 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 g(x).
The point (2,1) is on the graph and its y-coordinate is greater than any other on the graph.
So, g has an absolute maximum at $x=2$.
There is no point on the graph such that it has the lowest possible y-coordinate.
This is because the point (1,-1) is NOT on the graph.
No absolute minimum.
Theorem 1 demands that "If $f$ is continuous...",
so it does not apply here (g is not continuous at $x=1$).
Thus, there are no inconsistencies.
