Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 38

$$\eqalign{ & y{\text{ - intercept }}\left( {0,2} \right) \cr & x{\text{ - intercepts: }}\left( { - 4.299,0} \right),\left( {1.085,0} \right),\left( {3.214,0} \right) \cr & {\text{local maximum at }}\left( { - \sqrt 5 ,\frac{{3 + 2\sqrt 5 }}{3}} \right) \cr & {\text{local minimum at }}\left( {\sqrt 5 ,\frac{{3 - 2\sqrt 5 }}{3}} \right) \cr & {\text{inflection point : }}\left( {0,1} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{1}{{15}}{x^3} - x + 1 \cr & {\text{Domain: }}\left( { - \infty ,\infty } \right) \cr & \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & f\left( 0 \right) = \frac{1}{{15}}{\left( 0 \right)^3} - \left( 0 \right) + 1 \cr & f\left( 0 \right) = 1 \cr & y{\text{ - intercept }}\left( {0,1} \right) \cr & {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr & \frac{1}{{15}}{x^3} - x + 1 = 0 \cr & {\text{Using the graphing utility we obtain:}} \cr & {x_1} \approx - 4.299,{\text{ }}{x_2} \approx 1.085,{\text{ }}{x_3} \approx 3.214 \cr & x{\text{ - intercepts: }}\left( { - 4.299,0} \right),\left( {1.085,0} \right),\left( {3.214,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{{15}}{x^3} - x + 1} \right] \cr & f'\left( x \right) = \frac{1}{5}{x^2} - 1 \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find critical points}} \cr & \frac{1}{5}{x^2} - 1 = 0 \cr & {x^2} = 5 \cr & x = \pm \sqrt 5 \cr & \cr & *{\text{Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{5}{x^2} - 1} \right] \cr & f''\left( x \right) = \frac{2}{5}x \cr & \cr & {\text{Evaluate }}f''\left( x \right){\text{ at the critical points }}x = \pm \sqrt 5 \cr & f''\left( { - \sqrt 5 } \right) = \frac{2}{5}\left( { - \sqrt 5 } \right) < 0,{\text{ then}} \cr & {\text{There is a local maximum at }}\left( { - \sqrt 5 ,f\left( { - \sqrt 5 } \right)} \right) \cr & f\left( { - \sqrt 5 } \right) = \frac{1}{{15}}{\left( { - \sqrt 5 } \right)^3} - \left( { - \sqrt 5 } \right) + 1 = \frac{{3 + 2\sqrt 5 }}{3} \cr & \to {\text{local maximum at }}\left( { - \sqrt 5 ,\frac{{3 + 2\sqrt 5 }}{3}} \right) \cr & and \cr & f''\left( {\sqrt 5 } \right) = \frac{2}{5}\left( {\sqrt 5 } \right) > 0,{\text{ then}} \cr & {\text{There is a local minimum at }}\left( {\sqrt 5 ,f\left( {\sqrt 5 } \right)} \right) \cr & f\left( {\sqrt 5 } \right) = \frac{1}{{15}}{\left( {\sqrt 5 } \right)^3} - \left( {\sqrt 5 } \right) + 1 = \frac{{3 - 2\sqrt 5 }}{3} \cr & \to {\text{local minimum at }}\left( {\sqrt 5 ,\frac{{3 - 2\sqrt 5 }}{3}} \right) \cr & \cr & {\text{*Find the Inflection points}}{\text{, set }}f''\left( x \right) = 0 \cr & \frac{2}{5}x = 0 \cr & x = 0 \cr & f\left( 0 \right) = \frac{1}{{15}}{\left( 0 \right)^3} - \left( 0 \right) + 1 = 1 \cr & \cr & {\text{The inflection point : }}\left( {0,1} \right) \cr & \cr & {\text{Graph}} \cr} $$