Answer
(a) $f$ is decreasing on the interval $(\frac{\pi}{4}, \frac{5\pi}{4})$
$f$ is increasing on the intervals $(0, \frac{\pi}{4}) \cup (\frac{5\pi}{4}, 2\pi)$
(b) The local maximum is $f(\frac{\pi}{4}) = \sqrt{2}$
The local minimum is $f(\frac{5\pi}{4}) = -\sqrt{2}$
(c) The graph is concave down on these intervals:
$(0, \frac{3\pi}{4})\cup (\frac{7\pi}{4}, 2\pi)$
The graph is concave up on this interval:
$(\frac{3\pi}{4}, \frac{7\pi}{4})$
The points of inflection are $(\frac{3\pi}{4},0)$ and $(\frac{7\pi}{4},0)$
Work Step by Step
(a) $f(x) = sin~x+cos~x,~~~~0 \leq x \leq 2\pi$
We can find the points where $f'(x) = 0$:
$f'(x) = cos~x-sin~x = 0$
$cos~x = sin~x$
$tan~x = 1$
$x = \frac{\pi}{4}, \frac{5\pi}{4}$
When $\frac{\pi}{4} \lt x \lt \frac{5\pi}{4}~~$ then $f'(x) \lt 0$
$f$ is decreasing on the interval $(\frac{\pi}{4}, \frac{5\pi}{4})$
When $0 \lt x \lt \frac{\pi}{4}$ and $\frac{5\pi}{4} \lt x \lt 2\pi~~$ then $f'(x) \gt 0$
$f$ is increasing on the intervals $(0, \frac{\pi}{4}) \cup (\frac{5\pi}{4}, 2\pi)$
(b) $f(0) = sin~0+cos~0 = 1$
$f(\frac{\pi}{4}) = sin~\frac{\pi}{4}+cos~\frac{\pi}{4} = \sqrt{2}$
$f(\frac{5\pi}{4}) = sin~\frac{5\pi}{4}+cos~\frac{5\pi}{4} = -\sqrt{2}$
$f(2\pi) = sin~2\pi+cos~2\pi = 1$
The local maximum is $f(\frac{\pi}{4}) = \sqrt{2}$
The local minimum is $f(\frac{5\pi}{4}) = -\sqrt{2}$
(c) We can find the points where $f''(x) = 0$:
$f''(x) = -sin~x-cos~x = 0$
$sin~x+cos~x = 0$
$sin~x = -cos~x$
$tan~x = -1$
$x = \frac{3\pi}{4}, \frac{7\pi}{4}$
The graph is concave down when $f''(x) \lt 0$
The graph is concave down on these intervals:
$(0, \frac{3\pi}{4})\cup (\frac{7\pi}{4}, 2\pi)$
The graph is concave up when $f''(x) \gt 0$
The graph is concave up on this interval:
$(\frac{3\pi}{4}, \frac{7\pi}{4})$
$f(\frac{3\pi}{4}) = sin~\frac{3\pi}{4}+cos~\frac{3\pi}{4} = 0$
$f(\frac{7\pi}{4}) = sin~\frac{7\pi}{4}+cos~\frac{7\pi}{4} = 0$
The points of inflection are $(\frac{3\pi}{4},0)$ and $(\frac{7\pi}{4},0)$
