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

Given $f(x) = \cos^{2} x + \sin x$.
First,we find the derivative $f'(x) = 2 \cos x(-\sin x) + \cos x = -2 \sin x \cos x + \cos x$.
Setting $f'(x) = 0$,we get $\cos x(1 - 2 \sin x) = 0$.
This implies $\cos x = 0$ or $\sin x = \frac{1}{2}$.
For $x \in [0, \pi]$,the critical points are $x = \frac{\pi}{6}$ and $x = \frac{\pi}{2}$.
Now,we evaluate $f(x)$ at the critical points and the endpoints $x = 0$ and $x = \pi$:
$f(0) = \cos^{2} 0 + \sin 0 = 1 + 0 = 1$.
$f(\pi) = \cos^{2} \pi + \sin \pi = (-1)^{2} + 0 = 1$.
$f(\frac{\pi}{6}) = \cos^{2} \frac{\pi}{6} + \sin \frac{\pi}{6} = (\frac{\sqrt{3}}{2})^{2} + \frac{1}{2} = \frac{3}{4} + \frac{1}{2} = \frac{5}{4}$.
$f(\frac{\pi}{2}) = \cos^{2} \frac{\pi}{2} + \sin \frac{\pi}{2} = 0 + 1 = 1$.
Comparing these values,the absolute maximum value is $\frac{5}{4}$ and the absolute minimum value is $1$.