Answer
a. $f(\pi/4)=0$ local minimum.
$f(\pi/4)=0$ absolute minimum, no absolute maximum in the interval.
b. See graph and explanations.
Work Step by Step
a. Step 1. Given the function $f(x)=sec^{2}x-2tan(x), -\pi/2\leq x\lt\pi/2$, we have $f'(x)=2\tan\left(x\right)\sec^{2}x-2\sec^{2}x=2sec^2x(tan(x)-1))$.
Step 2. The extrema happen when $f'(x)=0$, undefined, or at endpoints; thus we have $tan(x)=1$ and the critical point within the domain is $x=\pi/4$.
Step 3. We have $f(\pi/4)=sec^{2}(\pi/4)-2tan(\pi/4)=0$. Test signs of $f'(x)$ across critical points $..(-)..(\pi/4)..(+)..$; thus we have $f(\pi/4)=0$ as a local minimum.
Step 4. We can identify $f(\pi/4)=0$ as an absolute minimum and there is no absolute maximum in the interval.
b. See graph. We can see that when $f'\lt0$, $f$ decreases; while when $f'\gt0$, $f$ increases (both with the tangent slopes corresponding to the values of $f'$). When $f'=0$, $f$ will be at an extrema with tangent slopes of zero. We can also observe the change of signs in $f'$ on different sides of critical points where $f'=0$.
