Answer
The absolute maximum $g(-\pi/3)=2$ and the absolute minimum $g(0)=1$ on the given interval.
See graph.
Work Step by Step
Step 1. Given the function $g(x)=sec(x), -\pi/3\leq x\leq \pi/6$, we have $g(-\pi/3)=2, g(\pi/6)=2\sqrt 3/3$,
Step 2. $g'(x)=sec(x)tan(x)$, let $g'(x)=0$, we get a critical point at $x=0$ and $g(0)=1$
Step 3. We can identify the absolute maximum $g(-\pi/3)=2$ and the absolute minimum $g(0)=1$ on the given interval.
Step 4. See graph.
