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# Find the absolute maximum and minimum values of the function f given by f (x) = cos ^ 2 x + sin x, x belongs to 0 , pi

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

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Given function is
$f (x) = \cos ^2 x + \sin x$
$f^{'} (x) = 2\cos x(-\sin x) + \cos x \\ f^{'}(x) = 0\\ -2\cos x\sin x + \cos x=0\\ \cos x(1-2\sin x) = 0\\ either\\ \cos x = 0 \ \ \ \ \ \ and \ \ \ \ \ \ \ \sin x = \frac{1}{2} \\ x = \frac{\pi}{2} \ \ \ \ \ \ \ \ \ \ and \ \ \ \ \ \ \ \ \ \ x = \frac{\pi}{6} \ \ \ \ \ as \ x \ \epsilon [0,\pi]$
Now,
$f^{''} (x) = -2(-\sin x)\sin x - 2\cos x\cos x + (-\sin x) \\ f^{''}(x)= 2\sin^2x - 2\cos^2x - \sin x\\ f^{''}(\frac{\pi}{6}) = \frac{-3}{2} < 0$
Hence, the point  $x = \frac{\pi}{6}$  is the point of maxima and the maximum value is
$f (\frac{\pi}{6}) = \cos ^2 \frac{\pi}{6} + \sin \frac{\pi}{6} = \frac{3}{4}+\frac{1}{2} = \frac{5}{4}$
And
$f^{''}(\frac{\pi}{2}) = 1 > 0$
Hence, the point  $x = \frac{\pi}{2}$ is the point of minima and the minimum value is
$f (\frac{\pi}{2}) = \cos ^2 \frac{\pi}{2} + \sin \frac{\pi}{2} =0 + 1 = 1$

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