Answer
$f$ has an absolute minimum at $x=0$,
$f$ has no absolute maximum.
Th.1 does not apply, so the answer is not inconsistent with it.
Work Step by Step
The graph of f contains points $(x,f(x)).$
Sketch the function $g(x)=|x|$ and reduce the domain to $(-1,2).$
(see image)
Points ($-1,1)$ and $(2,2)$ are excluded from the graph because the domain is an open interval, from either end.
The point ($0,0$) has a lower y-coordinate than any other point on the graph. $f$ has an absolute minimum at $x=0.$
We can not find a point so that its y-coordinate is greater than any other on the graph (the graph approaches, but never reaches the point (2,2).)
No absolute maximum.
Theorem 1 demands that "$f$ is continuous on a closed interval $[a, b]$,...", so it does not apply here (the domain here is an open interval). There, there are no inconsistencies.