University Calculus: Early Transcendentals (3rd Edition)

The function has the absolute maximum value at $x=0$, where $f(0)=3$, but it does not have any absolute minimum value.
$$y=\frac{6}{x^2+2}\hspace{1cm}-1\lt x\lt1$$ The graph is sketched below. - The graph obviously has an absolute maximum value, which is $f(0)=3$. - The graph, however, does not have an absolute minimum value. The graph does seem to reach the lowest point at $x=-1$ and $x=1$, but since we consider here the interval $(-1,1)$, we do not include the point where $x=-1$ or $x=1$. And since there are no other minimum values, the graph does not have any absolute minimum in the defined interval. This answer is still consistent with Theorem 1, because Theorem 1 requires that the function $f$ in question be examined in a CLOSED interval, not an open one like in this case.