It is given that f (x) = cos² x + sin x, x ∈ [0, π]

f’(x) = 2cosx( - sinx) + cosx

= - 2sinxcosx + cosx

Now, if f’(x) = 0

⇒ 2sinxcosx = cosx

⇒ cosx(2sinx - 1)=0

⇒ sin x = or cosx = 0

⇒ x =

Now, evaluating the value of f at critical points x = and x = and at the end points of the interval [0,π], (ie, at x = 0 and x =π), we get,

f

f(0)=

f(π)=cos²π + sinπ = ( - 1)² + 0 =1

f

Therefore, the absolute maximum value of f is occurring at x = and the absolute minimum value of f is 1 occuring at x =1, and π.