Answer
$$\eqalign{
& f\left( x \right){\text{ has a relative minimum at }}x = \frac{{3\pi }}{4}{\text{ and }}x = \frac{{7\pi }}{4} \cr
& f\left( x \right){\text{ has a relative maximum at }}x = \frac{\pi }{4}{\text{ and }}x = \frac{{5\pi }}{4} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \sin x\cos x + 5 \cr
& \left( {\text{a}} \right){\text{Calculating the first derivative}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\sin x\cos x + 5} \right] \cr
& f'\left( x \right) = \sin x\frac{d}{{dx}}\left[ {\cos x} \right] + \cos x\frac{d}{{dx}}\left[ {\sin x} \right] + \frac{d}{{dx}}\left[ 5 \right] \cr
& f'\left( x \right) = - {\sin ^2}x + {\cos ^2}x + 0 \cr
& f'\left( x \right) = {\cos ^2}x - {\sin ^2}x \cr
& f'\left( x \right) = \cos 2x \cr
& {\text{Calculating the critical points, set }}f'\left( x \right) = 0 \cr
& f'\left( x \right) = \cos 2x \cr
& \cos 2x = 0 \cr
& {\text{On the interval }}\left( {0,2\pi } \right)\cos 2x = 0{\text{ for }}x = \frac{\pi }{4},\frac{{3\pi }}{4},\frac{{5\pi }}{4},\frac{{7\pi }}{4} \cr
& \cr
& \left( {\text{b}} \right){\text{Set the intervals: }} \cr
& \left( {0,\frac{\pi }{4}} \right),\left( {\frac{\pi }{4},\frac{{3\pi }}{4}} \right),\left( {\frac{{3\pi }}{4},\frac{{5\pi }}{4}} \right),\left( {\frac{{5\pi }}{4},\frac{{7\pi }}{4}} \right),\left( {\frac{{7\pi }}{4},2\pi } \right) \cr
& {\text{Making a table of values }}\left( {{\text{See examples on page 180 }}} \right) \cr} $$
\[\begin{array}{*{20}{c}}
{{\text{Interval}}}&{{\text{Test Value}}}&{{\text{Sign of }}f'\left( x \right)}&{{\text{Conclusion}}} \\
{\left( {0,\frac{\pi }{4}} \right)}&{x = \frac{\pi }{6}}&{\frac{1}{2} > 0}&{{\text{Increasing}}} \\
{\left( {\frac{\pi }{4},\frac{{3\pi }}{4}} \right)}&{x = \frac{\pi }{2}}&{ - 1 < 0}&{{\text{Decreasing}}} \\
{\left( {\frac{{3\pi }}{4},\frac{{5\pi }}{4}} \right)}&{x = \pi }&{1 > 0}&{{\text{Increasing}}} \\
{\left( {\frac{{5\pi }}{4},\frac{{7\pi }}{4}} \right)}&{x = \frac{{3\pi }}{2}}&{ - 1 < 0}&{{\text{Decreasing}}} \\
{\left( {\frac{{7\pi }}{4},2\pi } \right)}&{x = \frac{{11\pi }}{6}}&{\frac{1}{2} > 0}&{{\text{Increasing}}}
\end{array}\]
$$\eqalign{
& f'\left( x \right){\text{ changes from negative to positive at }}x = \frac{{3\pi }}{4},\frac{{7\pi }}{4},{\text{ so }} \cr
& f\left( x \right){\text{ has a relative minimum at }}x = \frac{{3\pi }}{4}{\text{ and }}x = \frac{{7\pi }}{4} \cr
& \cr
& f'\left( x \right){\text{ changes from positive to negative at }}x = \frac{\pi }{4},\frac{{5\pi }}{4},{\text{ so }} \cr
& f\left( x \right){\text{ has a relative maximum at }}x = \frac{\pi }{4}{\text{ and }}x = \frac{{5\pi }}{4} \cr
& \cr
& \left( {\text{c}} \right){\text{Graph}} \cr} $$
