Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 27

$$\eqalign{ & {\text{Domain }}\left[ {0,2\pi } \right] \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & x{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Inflection point }}\left( {\pi , - \pi } \right) \cr & {\text{No local extrema}} \cr & {\text{No asymptotes}} \cr} $$

$$\eqalign{ & f\left( x \right) = \sin x - x,{\text{ Domain }}\left[ {0,2\pi } \right] \cr & \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & f\left( 0 \right) = \sin \left( 0 \right) - \left( 0 \right) \cr & f\left( 0 \right) = 0 \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr & \sin x - x = 0 \cr & x = 0 \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\sin x - x} \right] \cr & f'\left( x \right) = \cos x - 1 \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find critical points}} \cr & \cos x - 1 = 0 \cr & \cos x = 1 \cr & {\text{On the interval }}\left[ {0,2\pi } \right]{\text{ }}\cos x = 1{\text{ at }} \cr & x = 0{\text{ and }}x = 2\pi \cr & \cr & *{\text{Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\cos x - 1} \right] \cr & f''\left( x \right) = - \sin x \cr & {\text{*Evaluate }}f''\left( x \right){\text{ at the critical points }}x = 0{\text{ and }}x = 2\pi \cr & f''\left( 0 \right) = - \sin x = 0,{\text{ Inconclusive test}} \cr & {\text{Using the first derivative test}} \cr & f''\left( { - 1} \right) = - \sin \left( { - 1} \right),\, - 1{\text{ is not in the domain}}{\text{, we cannot}} \cr & {\text{apply the first derivative test at }}x = 0 \cr & f''\left( {2\pi } \right) = - \sin \left( {2\pi } \right) = 0,{\text{ Inconclusive test}} \cr & {\text{Using the first derivative test}} \cr & f''\left( {2\pi + 0.01} \right) = - \sin \left( {2\pi + 0.01} \right),\,{\text{ }} \cr & 2\pi + 0.01{\text{ is not in the domain}}{\text{, we cannot apply the first}} \cr & {\text{derivative test}}{\text{.}} \cr & {\text{There are no relative extrema on the interval }}\left[ {0,2\pi } \right]{\text{ for}} \cr & {\text{the function }}f\left( x \right) = \sin x - x \cr & \cr & {\text{*Find the Inflection points}}{\text{, set }}f''\left( x \right) = 0 \cr & - \sin x = 0 \cr & {\text{On the interval }}\left[ {0,2\pi } \right]{\text{ }}\sin x = 0,{\text{ at }} \cr & x = 0,{\text{ }}x = \pi {\text{ and }}x = 2\pi \cr & {\text{No inflection points for }}x = 0{\text{ and }}x = 2\pi {\text{ because we cannot }} \cr & {\text{evaluate before or after these points}} \cr & {\text{Inflection point at }}x = \pi \cr & f\left( \pi \right) = \sin \left( \pi \right) - \left( \pi \right) = - \pi \cr & {\text{Inflection point }}\left( {\pi , - \pi } \right) \cr & \cr & {\text{*Calculate the asymptotes}} \cr & {\text{No vertical asymptotes}}{\text{, the denominator is 1}}{\text{.}} \cr & {\text{No horizontal asymptotes}} \cr & \cr & {\text{Graph}} \cr & \cr} $$