Answer
See explanation.
Work Step by Step
\[
-\frac{1}{2} \sec ^{2} \frac{x}{2}=f^{\prime}(x)
\]
First derivative
Zeros: None (because sec $x$ does not have roots)
(a) Never positive, so the function $f$ is never increasing.
(b) Negative everywhere, thus the function $f$ is decreasing everywhere $(\text { on }(-\pi, \pi))$
$
f^{\prime \prime}(x)=-\sec \frac{x}{2} \cdot \sec \frac{x}{2} \tan \frac{x}{2} \cdot \frac{1}{2}$
Zeros : $0=x$
(c) Positive on $(-\pi, 0)$ so the function $f$ is concave up on this interval.
(d) Negative on $(0, \pi),$ so the function $f$ is concave down on this interval.
Properties of second derivative
(e) $0=x$
The points where the function changes from concave up to concave down or the other way around are inflection points.