Find the intervals in which the function $f$ given by $f(x) = \sin x + \cos x$,$0 \leq x \leq 2 \pi$ is increasing or decreasing.

(N/A) We have $f(x) = \sin x + \cos x$.
Taking the derivative,we get $f^{\prime}(x) = \cos x - \sin x$.
Setting $f^{\prime}(x) = 0$,we have $\cos x = \sin x$,which implies $\tan x = 1$. Since $0 \leq x \leq 2 \pi$,the solutions are $x = \frac{\pi}{4}$ and $x = \frac{5 \pi}{4}$.
The points $x = \frac{\pi}{4}$ and $x = \frac{5 \pi}{4}$ divide the interval $[0, 2 \pi]$ into three disjoint intervals: $[0, \frac{\pi}{4})$,$(\frac{\pi}{4}, \frac{5 \pi}{4})$,and $(\frac{5 \pi}{4}, 2 \pi]$.
We analyze the sign of $f^{\prime}(x)$ in these intervals:

Interval	Sign of $f^{\prime}(x)$	Nature of function
$[0, \frac{\pi}{4})$	$f^{\prime}(x) > 0$	$f$ is increasing
$(\frac{\pi}{4}, \frac{5 \pi}{4})$	$f^{\prime}(x) < 0$	$f$ is decreasing
$(\frac{5 \pi}{4}, 2 \pi]$	$f^{\prime}(x) > 0$	$f$ is increasing

Thus,$f$ is increasing in $[0, \frac{\pi}{4}) \cup (\frac{5 \pi}{4}, 2 \pi]$ and decreasing in $(\frac{\pi}{4}, \frac{5 \pi}{4})$.