Question

Find the absolute maximum and minimum values of the function $f$ given by $f(x)=cos^2x+sinx,x∈[0,π]$

Answer 1

$f(x)=cos^2x+sinx$
$f'(x)=2cosx(-sinx)+cosx$
$=-2sinx\,cosx+cosx$
Now,$f(x)=0$
⇒2sinx\,cosx=cosx$$⇒cosx(2sinx-1)
$⇒sinx=\frac{1}{2}\, or\, cosx=0$
$⇒x=\frac{π}{6},or \frac{π}{2} as x∈[0,π]$
Now, evaluating the value of $f$ at critical points $x=\frac{π}{2}$ and $x=\frac{π}{6}$ and at the end points of the interval $[0,π]$ (i.e., at $x = 0$ and $x = π$), we have:
$f(\frac{π}{6})=cos^2\frac{π}{6}+sin\frac{π}{6}$
$=(\sqrt{\frac{3}{2}})^2+\frac{1}{2}=\frac{5}{4}$
$f(0)=cos^20+sin0=1+0=1$
$f(π)=cos^2π+sinπ=(-1)^2+0=1$
$f(\frac{π}{2})=cos^2\frac{π}{2}+sin\frac{π}{2}=0+1=1$
Hence, the absolute maximum value of f is $\frac{5}{4}$ is occurring at $x=\frac{π}{6}$ and the absolute minimum value of $f$ is 1 occurring at $x=0,π/2$,and $π$.