Find the absolute maximum and minimum values of the function of given by

Question:

Find the absolute maximum and minimum values of the function of given by

$f(x)=\cos ^{2} x+\sin x, x \in[0, \pi]$

Solution:

Given : $f(x)=\cos ^{2} x+\sin x$

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

For a local maximum or a local minimum, we must have

$f^{\prime}(x)=0$

$\Rightarrow-2 \sin x \cos x+\cos x=0$

$\Rightarrow \cos x(2 \sin x-1)=0$

$\Rightarrow \sin x=\frac{1}{2}$ or $\cos x=0$

$\Rightarrow x=\frac{\pi}{6}$ or $\frac{\pi}{2}$       $[\because x \in(0, \pi)]$

Thus, the critical points of $f$ are $0, \frac{\pi}{6}, \frac{\pi}{2}$ and $\pi$.

Now,

$f(0)=\cos ^{2}(0)+\sin (0)=1$

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

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

$f(\pi)=\cos ^{2}(\pi)+\sin (\pi)=1$

Hence, the absolute maximum value when $x=\frac{\pi}{6}$ is $\frac{5}{4}$ and the absolute minimum value when $x=0, \frac{\pi}{2}, \pi$ is $1 .$

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