Answer
See explanation.
Work Step by Step
\[
f^{\prime}(x)=2 \cos 2 x \sin 2 x \cdot 2=2 \sin 4 x
\]
First derivative
Zeros: $x=\pi, x=\frac{3 \pi}{4},x= \pm \frac{\pi}{2}, x=\frac{\pi}{4}$ and $x=0$
(a) Positive on $\left(\frac{\pi}{2}, \frac{3 \pi}{4}\right)$ , $\left(0, \frac{\pi}{4}\right),$ so the function $f$ is increasing on these intervals
(b) Negative on $\left(\frac{3 \pi}{4}, \pi\right)$ , $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$, so the function $f$ is decreasing on these intervals.
Properties of first derivative
\[
8 \cos 4 x=f^{\prime \prime}(x)
\]
Second derivative
Zeros : $x=\frac{3 \pi}{8},x=\frac{\pi}{8}, x=\frac{7 \pi}{8} $ and $x=\frac{5 \pi}{8}$
(c) Positive on $\left(\frac{3 \pi}{8}, \frac{5 \pi}{8}\right)$ , $\left[0, \frac{\pi}{8}\right)$ , and $\left(\frac{7 \pi}{8}, \pi\right]$
so the function $f$ is concave up on these intervals.
(d) Negative on $\left(\frac{5 \pi}{8}, \frac{7 \pi}{8}\right),$ and $\left(\frac{\pi}{8}, \frac{3 \pi}{8}\right)$ so the function $f$ is concave down on these intervals.
Properties of second derivative
(e) $x=\frac{7 \pi}{8}, x=\frac{5 \pi}{8}, x=\frac{3 \pi}{8}$ and $,x=\frac{\pi}{8}$
The points where the function changes from concave up to concave down or the other way around are inflection points.