Answer
$$\eqalign{
& {\text{Increasing on: }}\left( {0,\frac{\pi }{4}} \right){\text{ and }}\left( {\frac{{5\pi }}{4},2\pi } \right) \cr
& {\text{Decreasing on: }}\left( {\frac{\pi }{4},\frac{{5\pi }}{4}} \right) \cr} $$
Work Step by Step
\[\begin{align}
& f\left( x \right)=\sin x+\cos x,\text{ }\left[ 0,2\pi \right] \\
& \text{Calculate the first derivative} \\
& f'\left( x \right)=\frac{d}{dx}\left[ \sin x+\cos x \right] \\
& f'\left( x \right)=\cos x-\sin x \\
& \text{Find the critical points, set the first derivative to }0 \\
& \cos x-\sin x=0 \\
& \cos x=\sin x \\
& \text{For the interval }\left[ 0,2\pi \right],\text{ }\cos x=\sin x\text{ when:} \\
& x=\frac{\pi }{4},\text{ }x=\frac{5\pi }{4} \\
& \text{Intervals: }\left( 0,\frac{\pi }{4} \right),\left( \frac{\pi }{4},\frac{5\pi }{4} \right),\left( \frac{5\pi }{4},2\pi \right) \\
& \text{Making a table of values }\left( \text{See examples on page 178 } \right) \\
\end{align}\]
\[\boxed{\begin{array}{*{20}{c}}
{{\text{Interval}}}&{\left( {0,\frac{\pi }{4}} \right)}&{\left( {\frac{\pi }{4},\frac{{5\pi }}{4}} \right)}&{\left( {\frac{{5\pi }}{4},2\pi } \right)} \\
{{\text{Test Value}}}&{x = \frac{\pi }{8}}&{x = \pi }&{x = \frac{{7\pi }}{4}} \\
{{\text{Sign of }}f'\left( x \right)}&{0.54 > 0}&{ - 1 < 0}&{\sqrt 2 > 0} \\
{{\text{Conclusion}}}&{{\text{Increasing}}}&{{\text{Decreasing}}}&{{\text{Increasing}}}
\end{array}}\]
$$\eqalign{
& {\text{By Theorem 3}}{\text{.5 }}f{\text{ is:}} \cr
& {\text{Increasing on: }}\left( {0,\frac{\pi }{4}} \right){\text{ and }}\left( {\frac{{5\pi }}{4},2\pi } \right) \cr
& {\text{Decreasing on: }}\left( {\frac{\pi }{4},\frac{{5\pi }}{4}} \right) \cr} $$
