Chapter 3 - Applications of Differentiation - 3.3 Exercises - Page 183: 46

$$\eqalign{ & \left( {\text{a}} \right)x = \frac{\pi }{6},{\text{ }}x = \frac{{5\pi }}{6} \cr & \left( {\text{b}} \right){\text{Decreasing on: }}\left( {\frac{{5\pi }}{6},\frac{{11\pi }}{6}} \right) \cr & {\text{Increasing on: }}\left( {0,\frac{{5\pi }}{6}} \right){\text{ and }}\left( {\frac{{11\pi }}{6},2\pi } \right),{\text{ }} \cr & \left( {\text{c}} \right){\text{relative maximum}}\left( {\frac{{5\pi }}{6},2} \right) \cr & {\text{ relative minimum}}\left( {\frac{{11\pi }}{6}, - 2} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = \sin x - \sqrt 3 \cos x \cr & \cr & \left( {\text{a}} \right){\text{Calculating the first derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\sin x - \sqrt 3 \cos x} \right] \cr & f'\left( x \right) = \cos x + \sqrt 3 \sin x \cr & {\text{Calculating the critical points, set }}f'\left( x \right) = 0 \cr & \cos x + \sqrt 3 \sin x = 0 \cr & \cos x = - \sqrt 3 \sin x \cr & \frac{{\sin x}}{{\cos x}} = - \frac{1}{{\sqrt 3 }} \cr & \tan x = - \frac{{\sqrt 3 }}{3} \cr & {\text{On the interval }}\left( {0,2\pi } \right){\text{ }}\tan x = - \frac{{\sqrt 3 }}{3},{\text{ for }}x = \frac{{5\pi }}{6},\frac{{11\pi }}{6} \cr & {\text{We obtain the critical points }}x = \frac{{5\pi }}{6},{\text{ }}x = \frac{{11\pi }}{6} \cr & {\text{Set the intervals }}\left( {0,\frac{{5\pi }}{6}} \right),\left( {\frac{{5\pi }}{6},\frac{{11\pi }}{6}} \right),\left( {\frac{{11\pi }}{6},2\pi } \right) \cr & \cr & {\text{Making a table of values }}\left( {{\text{See examples on page 180 }}} \right) \cr} $$ \[\begin{array}{*{20}{c}} {{\text{Interval}}}&{\left( {0,\frac{{5\pi }}{6}} \right)}&{\left( {\frac{{5\pi }}{6},\frac{{11\pi }}{6}} \right)}&{\left( {\frac{{11\pi }}{6},2\pi } \right)} \\ {{\text{Test Value}}}&{\frac{\pi }{3}}&{\frac{{3\pi }}{2}}&{\frac{{23}}{{12}}\pi } \\ {{\text{Sign of }}f'\left( x \right)}&{{\text{ }}2 > 0}&{{\text{ }} - \sqrt 3 < 0}&{{\text{ }}\frac{{\sqrt 6 - \sqrt 2 }}{2} > 0} \\ {{\text{Conclusion}}}&{{\text{Increasing}}}&{{\text{Decreasing}}}&{{\text{Increasing}}} \end{array}\] $$\eqalign{ & *f'\left( x \right){\text{ changes from positive to negative at }}x = \frac{{11\pi }}{6},{\text{ so }}f\left( x \right) \cr & {\text{has a relative minimum at }}\left( {\frac{{11\pi }}{6},f\left( {\frac{\pi }{6}} \right)} \right) \cr & f\left( {\frac{{11\pi }}{6}} \right) = - 2 \cr & *f'\left( x \right){\text{ changes from negative to positive at }}x = \frac{{5\pi }}{6},{\text{ so }}f\left( x \right) \cr & {\text{has a relative maximum at }}\left( {\frac{{5\pi }}{6},f\left( {\frac{{5\pi }}{6}} \right)} \right) \cr & f\left( {\frac{{5\pi }}{6}} \right) = 2 \cr} $$