Answer
$f(x)$ has an absolute minimum value at both $x=0$ and $x=\pi/2$, where $f(x)=0$, but it does not have any absolute maximum values.
Work Step by Step
$f(x)=x+1$ for $-1\le x\lt0$ and $f(x)=\cos x$ for $0\lt x\le\pi/2$
The graph is sketched below.
- The graph has an absolute minimum value at both $x=0$ and $x=\pi/2$, where $f(x)=0$.
- The graph, however, does not have any absolute maximum values.
We might think there exists an absolute maximum value at $x=0$, which has the value $f(x)=1$. Yet from the given equation, $f(x)$ is not defined for $x=0$. Considering the true domain $[-1,0)\cup(0,\pi/2]$, we do not find any points $x=c$ where $f(x)\le f(c)$.
The reason Theorem 1 does not work here is because the domain $[-1,0)\cup(0,\pi/2]$ is still an open domain, where the point $x=0$ is left out. So the answer is still consistent with Theorem 1.
