Finding and classifying critical numbers of f(x) = cos(2x)

Differentiate $f(x) = \cos(2x)$ using the chain rule to get $f'(x) = -2\sin(2x)$. This is defined everywhere, so critical numbers occur only where $f'(x) = 0$, i.e., where $\sin(2x) = 0$. On $(0, \pi)$, this happens when $2x = \pi$, giving $x = \pi/2$ (note $x=0$ and $x=\pi$ are endpoints, not interior critical numbers).

To classify $x = \pi/2$, examine the sign of $f'(x) = -2\sin(2x)$ on either side of it. For $x$ in the interval $(\pi/4, \pi/2)$, we have $2x$ in $(\pi/2, \pi)$, so $\sin(2x) > 0$, making $f'(x) < 0$. For $x$ in the interval $(\pi/2, 3\pi/4)$, we have $2x$ in $(\pi, 3\pi/2)$, so $\sin(2x) < 0$, making $f'(x) > 0$. Since $f'$ changes from negative to positive, $x = \pi/2$ is a relative minimum.