Answer
$$\eqalign{
& {\text{The inflection point occurs at }}x = 2\pi \cr
& f\left( {2\pi } \right) = \sin \frac{{2\pi }}{2} = 0 \cr
& {\text{Inflection point }}\left( {2\pi ,0} \right) \cr
& {\text{Concave downward}}:{\text{ }}\left( {0,2\pi } \right) \cr
& {\text{Concave upward}}:{\text{ }}\left( {2\pi ,4\pi } \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \sin \frac{x}{2},{\text{ }}\left[ {0,4\pi } \right] \cr
& {\text{Calculate the first and second derivatives}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\sin \frac{x}{2}} \right] \cr
& f'\left( x \right) = \cos \left( {\frac{x}{2}} \right)\frac{d}{{dx}}\left[ {\frac{x}{2}} \right] \cr
& f'\left( x \right) = \cos \left( {\frac{x}{2}} \right)\left( {\frac{1}{2}} \right) \cr
& f'\left( x \right) = \frac{1}{2}\cos \left( {\frac{x}{2}} \right) \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{2}\cos \left( {\frac{x}{2}} \right)} \right] \cr
& f''\left( x \right) = - \frac{1}{2}\sin \left( {\frac{x}{2}} \right)\frac{d}{{dx}}\left[ {\frac{x}{2}} \right] \cr
& f''\left( x \right) = - \frac{1}{4}\sin \left( {\frac{x}{2}} \right) \cr
& {\text{Set the second derivative to }}0 \cr
& - \frac{1}{4}\sin \left( {\frac{x}{2}} \right) = 0 \cr
& \sin \left( {\frac{x}{2}} \right) = 0 \cr
& {\text{On the interval }}\left[ {0,4\pi } \right]{\text{ }}\sin \left( {\frac{x}{2}} \right) = 0{\text{ at }}x = 0,{\text{ 2}}\pi ,{\text{ 4}}\pi \cr
& {\text{Set the intervals }}\left( {0,2\pi } \right),\left( {2\pi ,4\pi } \right) \cr
& {\text{Making a table of values }}\left( {{\text{See examples on page 188 }}} \right) \cr} $$
\[\boxed{\begin{array}{*{20}{c}}
{{\text{Interval}}}&{\left( {0,2\pi } \right)}&{\left( {2\pi ,4\pi } \right)} \\
{{\text{Test Value}}}&{x = \frac{\pi }{2}}&{x = 3\pi } \\
{{\text{Sign of }}f''\left( x \right)}&{f''\left( {\frac{\pi }{2}} \right) = - \frac{{\sqrt 2 }}{8} < 0}&{f''\left( {\frac{{3\pi }}{2}} \right) = \frac{1}{4} > 0} \\
{{\text{Conclusion}}}&{{\text{Concave downward}}}&{{\text{Concave upward}}}
\end{array}}\]
$$\eqalign{
& {\text{The inflection point occurs at }}x = 2\pi \cr
& f\left( {2\pi } \right) = \sin \frac{{2\pi }}{2} = 0 \cr
& {\text{Inflection point }}\left( {2\pi ,0} \right) \cr
& {\text{Concave downward}}:{\text{ }}\left( {0,2\pi } \right) \cr
& {\text{Concave upward}}:{\text{ }}\left( {2\pi ,4\pi } \right) \cr} $$
