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

We have, $f(x) = {\cos ^2}x + \sin x,x \in [0,\;\pi ]$
Therefore, $f'(x) = 2\cos x( - \sin x) + \cos x = \cos x( - 2\sin x + 1)$

To find maximum or minimum values, $f'(x) = 0$
$\Rightarrow \cos x( - 2\sin x + 1) = 0 \Rightarrow \cos x = 0$, or $- 2\sin x + 1 = 0$
$\Rightarrow x = \cfrac{\pi }{2}$ or

$\sin x = \cfrac{1}{2} \Rightarrow x = \cfrac{\pi }{6},\cfrac{{5\pi }}{6}$

For absolute maximum and minimum values of $f(x)$ , we evaluate $f(0),f(\pi ),f(\pi /2),f(\pi /6),f(5\pi /6)$
Now $f(0) = {\cos ^2}0 + {\sin ^2}0 = {(1)^2} + 0 = 1,$

$f\left( {\cfrac{\pi }{6}} \right) = {\cos ^2}\cfrac{\pi }{6} + \sin \cfrac{\pi }{6} = {\left( {\cfrac{{\sqrt 3 }}{2}} \right)^2} + \cfrac{1}{2} = \cfrac{3}{4} + \cfrac{1}{2} = \cfrac{5}{4},$

$f\left( {\cfrac{{5\pi }}{6}} \right) = {\cos ^2}\left( {\cfrac{{5\pi }}{6}} \right) + \sin \left( {\cfrac{{5\pi }}{6}} \right) = {\left( {\cfrac{{ - \sqrt 3 }}{2}} \right)^2} + \cfrac{1}{2} = \cfrac{3}{4} + \cfrac{1}{2} = \cfrac{5}{4}$

$f\left( {\cfrac{\pi }{2}} \right) = {\cos ^2}\cfrac{\pi }{2} + \sin \cfrac{\pi }{2} = 0 + 1 = 1$ and
$f(\pi ) = {\cos ^2}\pi + \sin \pi = {( - 1)^2} + 0 = 1.$

Therefore we can say that absolute maximum and minimum values are $\cfrac{5}{4}$ and $1$ respectively