Answer
$$\eqalign{
& {\text{Concave downward}}:{\text{ }}\left( {0,1.82347} \right){\text{ and }}\left( {\pi ,4.4597} \right) \cr
& {\text{Concave upward}}:{\text{ }}\left( {1.82347,\pi } \right){\text{ and}}\left( {4.4597,2\pi } \right) \cr
& {\text{Inflection points: }}\left( {1.82347,1.4523} \right) \cr
& \left( {\pi ,0} \right),{\text{ }}\left( {4.4597, - 1.4523} \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 2\sin x + \sin 2x,{\text{ }}\left[ {0,2\pi } \right] \cr
& {\text{Calculate the first and second derivatives}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {2\sin x + \sin 2x} \right] \cr
& f'\left( x \right) = 2\cos x + 2\cos 2x \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {2\cos x + 2\cos 2x} \right] \cr
& f''\left( x \right) = - 2\sin x - 4\sin 2x \cr
& {\text{Set }}f''\left( x \right) = 0 \cr
& - 2\sin x - 4\sin 2x = 0 \cr
& {\text{Solving the equation we obtain:}} \cr
& x = 0,{\text{ }}x = 1.82347,x = \pi ,{\text{ }}x = 4.4597 \cr
& {\text{The function is continuous for all the domain of }}f\left( x \right). \cr
& {\text{We obtain the intervals}} \cr
& \left( {0,1.82347} \right),\left( {1.82347,\pi } \right),\left( {\pi ,4.4597} \right),\left( {4.4597,2\pi } \right) \cr
& {\text{Making a table of values }}\left( {{\text{See examples on page 188 }}} \right) \cr} $$
\[\boxed{\begin{array}{*{20}{c}}
{{\text{Interval}}}&{{\text{Test Value}}}&{{\text{Sign of }}f''\left( x \right)}&{{\text{Conclusion}}} \\
{\left( {0,1.82347} \right)}&{x = 0.9}& - &{{\text{C}}{\text{. downward}}} \\
{\left( {1.82347,\pi } \right)}&{x = 2}& + &{{\text{C}}{\text{. upward}}} \\
{\left( {\pi ,4.4597} \right)}&{x = 4}& - &{{\text{C}}{\text{. downward}}} \\
{\left( {4.4597,2\pi } \right)}&{x = 5}& + &{{\text{C}}{\text{. upward}}}
\end{array}}\]
$$\eqalign{
& {\text{Summary:}} \cr
& {\text{The inflection points are at:}} \cr
& x = 1.82347 \to f\left( {1.82347} \right) = 1.4523,{\text{ Point }}\left( {1.82347,1.4523} \right) \cr
& x = \pi \to f\left( \pi \right) = 0,{\text{ Point }}\left( {\pi ,0} \right) \cr
& x = 4.4597 \to f\left( {4.4597} \right) = - 1.4523,{\text{Point }}\left( {4.4597, - 1.4523} \right) \cr
& {\text{Concave downward}}:{\text{ }}\left( {0,1.82347} \right){\text{ and }}\left( {\pi ,4.4597} \right) \cr
& {\text{Concave upward}}:{\text{ }}\left( {1.82347,\pi } \right){\text{ and}}\left( {4.4597,2\pi } \right) \cr} $$
