Question

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

Answer 1

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

Putting this to zero, we get
$f^{{}^{\prime }}\left(x\right)=-2\cos x\sin x+\cos x$
$\Rightarrow -2\cos x\sin x+\cos x=0$
$\Rightarrow \cos x(2\sin x-1)$
$\Rightarrow x=\frac{\pi }{6}$ or $\frac{\pi }{2}$

Now let's evaluate the value of the function at critical points and at extreme points of domain.

$f\left(\frac{\pi }{6}\right)={\cos }^2\left(\frac{\pi }{6}\right)+\sin \left(\frac{\pi }{6}\right)=\frac{5}{4}$
$f\left(\frac{\pi }{2}\right)={\cos }^2\left(\frac{\pi }{2}\right)+\sin \left(\frac{\pi }{2}\right)=1$
$f\left(0\right)={\cos }^2\left(0\right)+\sin \left(0\right)=1$

$f\left(\pi \right)={\cos }^2\left(\pi \right)+\sin \left(\pi \right)=1$
And we can see that Function will have maxima at $x=\frac{\pi }{6}$ and will have minima at $x=\frac{\pi }{2},0,\pi$