Answer
a. $y(\frac{\pi}{4})=-1$, absolute maximum.
b. See graph; absolute maximum $(\frac{\pi}{4},-1)$.
Work Step by Step
a. Given $y=cot(x)-\sqrt 2 csc(x), 0\lt x\lt\pi$, we have $y'=-csc^2(x)+\sqrt 2 csc(x)cot(x)=csc(x)(\sqrt 2cot(x)-csc(x))$. Letting $y'=0$, we have $\sqrt 2cot(x)=csc(x)$ or $\sqrt 2cos(x)=1$, which gives $x=\frac{\pi}{4}$ in $0\lt x\lt\pi$. We have $y(\frac{\pi}{4})=cot(\frac{\pi}{4})-\sqrt 2 csc(\frac{\pi}{4})=1-2=-1$. Checking signs of $y'$ across $x=\frac{\pi}{4}$, we have $(0)..(+)..(\frac{\pi}{4})..(-)..(\pi)$, indicating $y(\frac{\pi}{4})=-1$ is an absolute maximum.
b. See graph. We can identify an absolute maximum at $(\frac{\pi}{4},-1)$, which agrees with the result from part (a).
