Answer
See graph; absolute minimum $0$, no absolute maximum.
Work Step by Step
Step 1. Graph the piecewise function $f(x)=\begin{cases} x+1\hspace1cm -1\leq x\lt0 \\ cos(x)\hspace1cm 0\lt x\leq \pi/2 \end{cases}$
as shown in the figure.
Step 2. We can determine that the function has an absolution minimum of $f(x)=0$ at $x=-1$ and $x=\pi/2$ on its domain. The function does not have an absolute maximum.
Step 3. Theorem 1 states that if a function is continuous on a closed interval, then it attains both an absolute maximum and an absolute minimum. The above results are consistent with Theorem 1. Because the function is not continuous at $x=0$, it does not have an absolute maximum.
