Answer
$f$ has an absolute maximum at $x=0$ and no absolute minimum.
Th. 1 does not apply, so the answer is not inconsistent with it.
Work Step by Step
The graph of f contains the points $(x,f(x)).$
Sketch the function $g(x)=\displaystyle \frac{6}{x^{2}-2}$ and reduce the domain to $(-1,1).$
(See image.)
Points ($-1,-6)$ and $(1,-6)$ are excluded from the graph because the domain is an open interval, from either end.
The point ($0,-3$) has a higher y-coordinate than any other point on the graph. $f$ has an absolute maximum at $x=0.$
We can not find a point so that its y-coordinate is lower (or equal) than any other on the graph, because the endpoints are excluded. Thus, there is no absolute minimum.
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). Thus, there are no inconsistencies.
