Answer
$ a.\quad$
Absolute maximum at $(0,0)$
Absolute minimum at $(\displaystyle \frac{\pi}{3},-\ln 2)$
$ b.\quad$
Absolute maximum at $(1,2)$
Absolute minimum at $(\displaystyle \frac{1}{2},0.769)$ and $(2,0.769)$
Work Step by Step
$ a.\quad$
Apply the chain rule:
$\displaystyle \frac{d}{dx}[f(x)]=\frac{1}{\cos x}\cdot(-\sin x)=-\tan x$
$f'(0)=0$, when$\quad -\tan x=0 \qquad $that is, when $x=0.$
$f(0)=\ln[\cos 0]=\ln 1=0$
$f(-\displaystyle \frac{\pi}{4})=\ln[\cos(-\frac{\pi}{4})]=\ln\frac{1}{\sqrt{2}}=\ln 2^{-1/2}=-\frac{1}{2}\ln 2$
$f(\displaystyle \frac{\pi}{3})=\ln[\cos(\frac{\pi}{3})]=\ln\frac{1}{2}=\ln 2^{-1}=-\ln 2$
For the first derivative test,
tan is negative on $(-\displaystyle \frac{\pi}{4},0)$ and positive on $(0,\displaystyle \frac{\pi}{3})$.
$f'=-\tan x$, so f increases on $(-\displaystyle \frac{\pi}{4},0)$ and decreases on $(0,\displaystyle \frac{\pi}{3})$.
Thus, we have:
Absolute maximum at $(0,0)$
Absolute minimum at $(\displaystyle \frac{\pi}{3},-\ln 2)$
$ b.\quad$
Apply the chain rule:
$\displaystyle \frac{d}{dx}[f(x)]=-\sin(\ln x)\cdot\frac{1}{x}=-\frac{\sin(\ln x)}{x}$
$f'(0)=0$ when $\ln x=0$, or, when $x=1$
$f(\displaystyle \frac{1}{2})\approx 0.769$
$f(1)=1$
$f(2)\approx 0.769$
For the first derivative test, we evaluate
$f'(0.7)\approx+0.4988,$
$f'(1.5)\approx-0.26296$
So f increases on $(1/2,1)$ and decreases on $(1,2)$
Thus, we have:
Absolute maximum at $(1,2)$
Absolute minimum at $(\displaystyle \frac{1}{2},0.769)$ and $(2,0.769)$
