Answer
a)
$f$ increasing on $\left(0,\frac{\pi}{4}\right)\cup\left(\frac{5\pi}{4},2\pi\right)$
$f$ decreasing on $\left(\frac{\pi}{4},\frac{5\pi}{4}\right)$
b)
$f(\frac{\pi}{4})$ = $\sqrt 2$ is a local maximum value and
$f(\frac{5\pi}{4})$ = $-\sqrt 2$ is a local minimum value.
c)
$f$ is concave upward on $\left(\frac{3\pi}{4},\frac{7\pi}{4}\right)$
$f$ is concave downward on $\left(0,\frac{3\pi}{4}\right)\cup\left(\frac{7\pi}{4},2\pi\right)$
There are inflection points at $\left(\frac{3\pi}{4},0\right)$ and $\left(\frac{7\pi}{4},0\right)$
Work Step by Step
a)
$f(x) = \sin x+\cos x$, $x\in[0,2\pi]$
$f'(x) = \cos x-\sin x$
$f'(x)=0\Rightarrow \cos x-\sin x=0$
$\cos x = \sin x$
$x = \frac{\pi}{4}$ or $\frac{5\pi}{4}$
When $f'(x)$ $\gt$ $0$
$\cos x-\sin x\gt 0$
$\cos x\gt \sin x$
$0 \lt x \lt \frac{\pi}{4}$ or $\frac{5\pi}{4} \lt x \lt 2\pi$
When $f'(x) \lt 0$
$\cos x \lt \sin x$
$\frac{\pi}{4} \lt x \lt 5\frac{\pi}{4}$
$f$ increasing on $\left(0,\frac{\pi}{4}\right)\cup\left(\frac{5\pi}{4},2\pi\right)$
$f$ decreasing on $\left(\frac{\pi}{4},\frac{5\pi}{4}\right)$
b)
$f$ changes from increasing to decreasing at $x = \frac{\pi}{4}$ and from decreasing to increasing at $x = \frac{5\pi}{4}$. Thus,
$f\left(\frac{\pi}{4}\right) = \sqrt 2$ is a local maximum value and
$f\left(\frac{5\pi}{4}\right) = -\sqrt 2$ is a local minimum value.
c)
$f''(x) = -\sin x-\cos x$
$f''(x)=0\Rightarrow -\sin x-\cos x=0\rightarrow \tan x=-1$
$x = \frac{3\pi}{4}$ or $\frac{7\pi}{4}$
$f$ is concave upward on $\left(\frac{3\pi}{4},\frac{7\pi}{4}\right)$
$f$ is concave downward on $\left(0,\frac{3\pi}{4}\right)\cup\left(\frac{7\pi}{4},2\pi\right)$
There are inflection points at $\left(\frac{3\pi}{4},0\right)$ and $\left(\frac{7\pi}{4},0\right)$
