Question

Find the absolute maximum and absolute minimum values of the function f given by $f\left(x\right)=si{n}^{2}xcosx,xϵ(0,\pi )$.

Answer 1

$f\left(x\right)={\sin }^{2}x\cos x,xϵ(0,\pi )$

For finding critical points $f^{\prime }\left(x\right)=0$

${\sin }^{2}x\times (-\sin x)+\cos x\times 2\sin x\cos x=0$

$2\sin x{\cos }^{2}x={\sin }^{3}x$

$\tan x=\sqrt{2}$

There is only one critical point
Thus,

$f\left({\tan }^{-1}\sqrt{2}\right)=(1-{\cos }^{2}x)\cos x$

$=(1-\frac{1}{{\sec }^{2}x})\times \frac{1}{\sec x}$

$=(1-\frac{1}{1+{\tan }^{2}x})\times \frac{1}{\sqrt{1+{\tan }^{2}x}}$

$\therefore$ Maximum value of $f\left(x\right)=(1-\frac{1}{1+2})\times \frac{1}{\sqrt{1+2}}$

$=\frac{2}{3\sqrt{3}}$ (Answer)