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

Explanation

Given, `f(x) = cos^(2) x + sin x, x in [0, pi]`
` rArr f'(x) = 2 cos x (-sin x) + cos x`
` =- 2 sin x cos x + cos x `
For maxima/minima f'(x) = 0
` rArr -2 sin x cos x + cos x = 0`
` rArr cos x (-2 sin x + 1) = 0`
`rArr cos x = 0 or sin x = 1/2`
` rArr x = pi/2, pi/6`
For absolute maxima and absolute minima, we find
`f(0), f(pi/6),f(pi/2), f(pi)`.
at ` x = 0, f(0) = cos^(2)0+sin 0=1^(2)+0=1`
at ` x = pi/6, f (pi/6) = cos^(2)(pi/6)+sin.pi/6 = ((sqrt3)/2)^(2)+1/2 = 5/4 = 1.25`
at ` x = pi/2, f (pi/2) = cos^(2)(pi/2) + sin.pi/2= 0^(2) + 1=1`
at ` x = pi, f (pi) = cos^(2) pi + sin pi = (-1)^(2)+0=1`
Therefore, absolute maximum value of f is ` 1.25 at x = pi/6 ` and absolute minimum value is ` 1 at x = 0,pi/2 and pi `.