Question

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

Answer 1

\[\text { Given }: f\left( x \right) = \cos^2 x + \sin x\]

\[ \Rightarrow f'\left( x \right) = 2 \cos x\left( - \sin x \right) + \cos x = - 2 \sin x \cos x + \cos x\]

\[\text { For a local maximum or a local minimum, we must have }\]

\[ f'\left( x \right) = 0\]

\[ \Rightarrow - 2 \sin x \cos x + \cos x = 0\]

\[ \Rightarrow \cos x \left( 2 \sin x - 1 \right) = 0\]

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

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

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

\[\text { Now }, \]

\[f\left( 0 \right) = \cos^2 \left( 0 \right) + \sin \left( 0 \right) = 1\]

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

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

\[f\left( \pi \right) = \cos^2 \left( \pi \right) + \sin \left( \pi \right) = 1\]

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