Question

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

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

Answer 1

$$\begin{gathered} \mathrm{Given}:f\left(x\right)={\cos }^{2}x+\sin x \\ \Rightarrow f\text{'}\left(x\right)=2\cos x\left(-\sin x\right)+\cos x=-2\sin x\cos x+\cos x \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maximum}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minimum},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow -2\sin x\cos x+\cos x=0 \\ \Rightarrow \cos x\left(2\sin x-1\right)=0 \\ \Rightarrow \sin x=\frac{1}{2}\mathrm{or}\cos x=0 \\ \Rightarrow x=\frac{\mathrm{\pi }}{6}\mathrm{or}\frac{\mathrm{\pi }}{2}\left[\because x\in \left(0,\mathrm{\pi }\right)\right] \\ \mathrm{Thus},\mathrm{the}\mathrm{critical}\mathrm{points}\mathrm{of}f\mathrm{are}0,\frac{\mathrm{\pi }}{6},\frac{\mathrm{\pi }}{2}\mathrm{and}\mathrm{\pi }. \\ \mathrm{Now}, \\ f\left(0\right)={\cos }^2\left(0\right)+\sin \left(0\right)=1 \\ f\left(\frac{\mathrm{\pi }}{6}\right)={\cos }^2\left(\frac{\mathrm{\pi }}{6}\right)+\sin \left(\frac{\mathrm{\pi }}{6}\right)=\frac{5}{4} \\ f\left(\frac{\mathrm{\pi }}{2}\right)={\cos }^2\left(\frac{\mathrm{\pi }}{2}\right)+\sin \left(\frac{\mathrm{\pi }}{2}\right)=1 \\ f\left(\mathrm{\pi }\right)={\cos }^2\left(\mathrm{\pi }\right)+\sin \left(\mathrm{\pi }\right)=1 \\ \mathrm{Hence},\mathrm{the}\mathrm{absolute}\mathrm{maximum}\mathrm{value}\mathrm{when}x=\frac{\mathrm{\pi }}{6}\mathrm{is}\frac{5}{4}\mathrm{and}\mathrm{the}\mathrm{absolute}\mathrm{minimum}\mathrm{value}\mathrm{when}x=0,\frac{\mathrm{\pi }}{2},\mathrm{\pi }\mathrm{is}1. \end{gathered}$$