Question

Find the absolute maximum and absolute minimum values of the function $f$ given by $f\left(x\right)={\sin }^{2}x-\cos x,x\in [0,\pi ]$.

Answer 1

$f\left(x\right)={\sin }^{2}x-\cos x$

$f^{\prime }\left(x\right)=2\sin x\cos x+\sin x$

= $\sin x(2\cos x+1)$

Now, $f^{\prime }\left(x\right)=0$

$\Rightarrow \sin x=0\Rightarrow x=n\pi \Rightarrow x=0,\pi$

and $2\cos x+1=0\Rightarrow \cos x=\frac{-1}{2}=\cos \frac{2\pi }{3}$

$\Rightarrow x=\frac{2\pi }{3}\in [0,\pi ]$

$\therefore f\left(0\right)=-1$

and $f\left(\frac{2\pi }{3}\right)={\left(\sin \frac{2\pi }{3}\right)}^2-\cos \left(\frac{2\pi }{3}\right)=\frac{3}{4}+\frac{1}{2}=\frac{5}{4}$

$f\left(\pi \right)={\sin }^2\pi -\cos \pi =1$

Hence, absolute maximum value is $\frac{5}{4}$ and absolute minimum value is $-1$.