Finding critical numbers of f(x) = sin(x) + cos(x)

Differentiate to get $f'(x) = \cos(x) - \sin(x)$, which is defined for all $x$, so critical numbers occur only where $f'(x) = 0$. Setting $\cos(x) - \sin(x) = 0$ gives $\sin(x) = \cos(x)$, i.e., $\tan(x) = 1$.

On $[0, 2\pi]$, $\tan(x) = 1$ at $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$. These are the two critical numbers.

The critical numbers $x = \pi/4$ and $x = 5\pi/4$ split $[0, 2\pi]$ into three subintervals: $(0, \pi/4)$, $(\pi/4, 5\pi/4)$, and $(5\pi/4, 2\pi)$. Testing a point in each, such as $x=0$, $x=\pi$, and $x=3\pi/2$, gives $f'(0) = 1 > 0$, $f'(\pi) = -1 < 0$, and $f'(3\pi/2) = 1 > 0$.

Since $f'>0$ on $(0,\pi/4)$ and $(5\pi/4, 2\pi)$, $f$ is increasing there. Since $f'<0$ on $(\pi/4, 5\pi/4)$, $f$ is decreasing there.

At $x = \pi/4$, $f'(x)$ changes from positive (on $(0,\pi/4)$) to negative (on $(\pi/4, 5\pi/4)$), so by the First Derivative Test $f$ has a local maximum at $x=\pi/4$.

At $x = 5\pi/4$, $f'(x)$ changes from negative (on $(\pi/4, 5\pi/4)$) to positive (on $(5\pi/4, 2\pi)$), so $f$ has a local minimum at $x = 5\pi/4$. The corresponding values are $f(\pi/4) = \sqrt{2}$ and $f(5\pi/4) = -\sqrt{2}$.