Answer
\[\begin{gathered}
\left( {\text{a}} \right)x = \frac{\pi }{6},{\text{ }}x = \frac{{5\pi }}{6} \hfill \\
\left( {\text{b}} \right){\text{Decreasing on: }}\left( {\frac{\pi }{6},\frac{{5\pi }}{6}} \right) \hfill \\
{\text{Increasing on: }}\left( {0,\frac{\pi }{6}} \right){\text{ and }}\left( {\frac{{5\pi }}{6},2\pi } \right),{\text{ }} \hfill \\
\left( {\text{c}} \right){\text{Relative maximum}}\left( {\frac{\pi }{6},\frac{{\pi + 6\sqrt 3 }}{{12}}} \right) \hfill \\
{\text{ Relative minimum}}\left( {\frac{{5\pi }}{6},\frac{{5\pi - 6\sqrt 3 }}{{12}}} \right) \hfill \\
\end{gathered} \]
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{x}{2} + \cos x \cr
& {\text{Calculating the first derivative}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{x}{2} + \cos x} \right] \cr
& f'\left( x \right) = \frac{1}{2} - \sin x \cr
& {\text{Calculating the critical points, set }}f'\left( x \right) = 0 \cr
& f'\left( x \right) = \frac{1}{2} - \sin x \cr
& \sin x = \frac{1}{2} \cr
& {\text{On the interval }}\left( {0,2\pi } \right){\text{ }}\sin x = \frac{1}{2}{\text{ for }}x = \frac{\pi }{6},{\text{ }}x = \frac{{5\pi }}{6} \cr
& {\text{We obtain the critical points }}x = \frac{\pi }{6},{\text{ }}x = \frac{{5\pi }}{6} \cr
& {\text{Set the intervals }}\left( {0,\frac{\pi }{6}} \right),\left( {\frac{\pi }{6},\frac{{5\pi }}{6}} \right),\left( {\frac{{5\pi }}{6},2\pi } \right) \cr
& \cr
& {\text{Making a table of values }}\left( {{\text{See examples on page 180 }}} \right) \cr} $$
\[\begin{array}{*{20}{c}}
{{\text{Interval}}}&{\left( {0,\frac{\pi }{6}} \right)}&{\left( {\frac{\pi }{6},\frac{{5\pi }}{6}} \right)}&{\left( {\frac{{5\pi }}{6},2\pi } \right)} \\
{{\text{Test Value}}}&{\frac{\pi }{{12}}}&{\frac{\pi }{2}}&\pi \\
{{\text{Sign of }}f'\left( x \right)}&{{\text{ }}f'\left( {\frac{\pi }{{12}}} \right) > 0}&{{\text{ }}f'\left( {\frac{\pi }{2}} \right) < 0}&{{\text{ }}f'\left( \pi \right) > 0} \\
{{\text{Conclusion}}}&{{\text{Increasing}}}&{{\text{Decreasing}}}&{{\text{Increasing}}}
\end{array}\]
$$\eqalign{
& f'\left( x \right){\text{ changes from positive to negative at }}x = \frac{\pi }{6},{\text{ so }}f\left( x \right) \cr
& {\text{has a relative maximum at }}\left( {\frac{\pi }{6},f\left( {\frac{\pi }{6}} \right)} \right) \cr
& f\left( {\frac{\pi }{6}} \right) = \frac{{\pi /6}}{2} + \cos \left( {\frac{\pi }{6}} \right) = \frac{\pi }{{12}} + \frac{{\sqrt 3 }}{2} = \frac{{\pi + 6\sqrt 3 }}{{12}} \cr
& f'\left( x \right){\text{ changes from negative to positive at }}x = \frac{{5\pi }}{6},{\text{ so}}f\left( x \right) \cr
& {\text{has a relative minimum at }}\left( {\frac{{5\pi }}{6},f\left( {\frac{{5\pi }}{6}} \right)} \right) \cr
& f\left( {\frac{\pi }{6}} \right) = \frac{{5\pi /6}}{2} + \cos \left( {\frac{{5\pi }}{6}} \right) = \frac{{5\pi }}{{12}} - \frac{{\sqrt 3 }}{2} = \frac{{5\pi - 6\sqrt 3 }}{{12}} \cr} $$
