Answer
$$\eqalign{
& {\text{Concave downward}}:{\text{ }}\left( {0,\frac{\pi }{2}} \right){\text{ and }}\left( {\frac{{3\pi }}{2},2\pi } \right) \cr
& {\text{Concave upward}}:{\text{ }}\left( {\frac{\pi }{2},\frac{{3\pi }}{2}} \right) \cr
& {\text{Inflection points: }}\left( {\frac{\pi }{2},\frac{\pi }{2}} \right),\left( {\frac{{3\pi }}{2},\frac{{3\pi }}{2}} \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = x + 2\cos x,{\text{ }}\left[ {0,2\pi } \right] \cr
& {\text{*Calculate the first and second derivatives}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {x + 2\cos x} \right] \cr
& f'\left( x \right) = 1 - 2\sin x \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {1 - 2\sin x} \right] \cr
& f''\left( x \right) = - 2\cos x \cr
& {\text{*Set }}f''\left( x \right) = 0 \cr
& - 2\cos x = 0 \cr
& \cos x = 0 \cr
& {\text{On the interval }}\left[ {0,2\pi } \right]{\text{ }}\cos x = 0{\text{ at }}x = \frac{\pi }{2},{\text{ }}x = \frac{{3\pi }}{2} \cr
& {\text{The function is continuous for all the domain of }}f\left( x \right). \cr
& {\text{We obtain the intervals}} \cr
& \left( {0,\frac{\pi }{2}} \right),\left( {\frac{\pi }{2},\frac{{3\pi }}{2}} \right),\left( {\frac{{3\pi }}{2},2\pi } \right) \cr
& {\text{Making a table of values }}\left( {{\text{See examples on page 188 }}} \right) \cr} $$
\[\boxed{\begin{array}{*{20}{c}}
{{\text{Interval}}}&{{\text{Test Value}}}&{{\text{Sign of }}f''\left( x \right)}&{{\text{Conclusion}}} \\
{\left( {0,\frac{\pi }{2}} \right)}&{x = \frac{\pi }{4}}& - &{{\text{C}}{\text{. downward}}} \\
{\left( {\frac{\pi }{2},\frac{{3\pi }}{2}} \right)}&{x = \pi }& + &{{\text{C}}{\text{. upward}}} \\
{\left( {\frac{{3\pi }}{2},2\pi } \right)}&{x = \frac{{7\pi }}{4}}& - &{{\text{C}}{\text{. downward}}}
\end{array}}\]
$$\eqalign{
& {\text{Summary:}} \cr
& {\text{The inflection points are at }}x = \frac{\pi }{2},{\text{ }}x = \frac{{3\pi }}{2} \cr
& f\left( {\frac{\pi }{2}} \right) = \frac{\pi }{2} \to {\text{Point}}\left( {\frac{\pi }{2},\frac{\pi }{2}} \right) \cr
& f\left( {\frac{{3\pi }}{2}} \right) = \frac{{3\pi }}{2} \to {\text{Point}}\left( {\frac{{3\pi }}{2},\frac{{3\pi }}{2}} \right) \cr
& {\text{Concave downward}}:{\text{ }}\left( {0,\frac{\pi }{2}} \right){\text{ and }}\left( {\frac{{3\pi }}{2},2\pi } \right) \cr
& {\text{Concave upward}}:{\text{ }}\left( {\frac{\pi }{2},\frac{{3\pi }}{2}} \right) \cr} $$
