Question

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

Hint:

To find increasing/decreasing intervals, check the sign of $f'(x)$ in each sub-interval between critical points.

Answer 1

Step 1: Differentiate $f(x)$. \[ f(x) = \sin x + \cos x \] \[ f'(x) = \cos x - \sin x \]

Step 2: Solve $f'(x)=0$. \[ \cos x - \sin x = 0 \implies \tan x = 1 \] \[ x = \frac{\pi}{4}, \; \frac{5\pi}{4} \]

Step 3: Test sign of $f'(x)$. - For $0 < x < \frac{\pi}{4}$: take $x=0$, $f'(0)=1>0$ $\;\implies$ increasing. - For $\frac{\pi}{4} < x < \frac{5\pi}{4}$: take $x=\pi$, $f'(\pi)=-1<0$ $\;\implies$ decreasing. - For $\frac{5\pi}{4} < x < 2\pi$: take $x=2\pi$, $f'(2\pi)=1>0$ $\;\implies$ increasing.

Step 4: Combine intervals. - Increasing on: $\; [0, \tfrac{\pi}{4}] \cup [\tfrac{5\pi}{4}, 2\pi]$ - Decreasing on: $\; [\tfrac{\pi}{4}, \tfrac{5\pi}{4}]$

Final Answer: \[ \boxed{\text{Increasing on } [0, \tfrac{\pi}{4}] \cup [\tfrac{5\pi}{4}, 2\pi]; \text{Decreasing on } [\tfrac{\pi}{4}, \tfrac{5\pi}{4}]} \]