Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 37

$$\eqalign{ & y{\text{ - intercept }}\left( {0,2} \right) \cr & x{\text{ - intercepts: }}\left( { - 2.173,0} \right),\left( {0.3530,0} \right),\left( {7.820,0} \right) \cr & {\text{local maximum at }}\left( { - 1,\frac{{14}}{3}} \right) \cr & {\text{local minimum at }}\left( {5, - \frac{{94}}{3}} \right) \cr & {\text{inflection point : }}\left( {2, - \frac{{40}}{3}} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{1}{3}{x^3} - 2{x^2} - 5x + 2 \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}{3}{\left( 0 \right)^3} - 2{\left( 0 \right)^2} - 5\left( 0 \right) + 2 \cr & f\left( 0 \right) = 2 \cr & y{\text{ - intercept }}\left( {0,2} \right) \cr & {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr & \frac{1}{3}{x^3} - 2{x^2} - 5x + 2 = 0 \cr & {\text{Using the graphing utility we obtain:}} \cr & {x_1} \approx - 2.173,{\text{ }}{x_2} \approx 0.3530,{\text{ }}{x_3} \approx 7.820 \cr & x{\text{ - intercepts: }}\left( { - 2.173,0} \right),\left( {0.3530,0} \right),\left( {7.820,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{3}{x^3} - 2{x^2} - 5x + 2} \right] \cr & f'\left( x \right) = {x^2} - 4x - 5 \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find critical points}} \cr & {x^2} - 4x - 5 = 0 \cr & \left( {x - 5} \right)\left( {x + 1} \right) = 0 \cr & x = - 1,{\text{ }}x = 5 \cr & \cr & *{\text{Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {{x^2} - 4x - 5} \right] \cr & f''\left( x \right) = 2x - 4 \cr & \cr & {\text{Evaluate }}f''\left( x \right){\text{ at the critical points }}x = - 1,{\text{ }}x = 5 \cr & f''\left( { - 1} \right) = 2\left( { - 1} \right) - 4 = - 6 < 0,{\text{ then}} \cr & {\text{There is a local maximum at }}\left( { - 1,f\left( { - 1} \right)} \right) \cr & f\left( { - 1} \right) = \frac{1}{3}{\left( { - 1} \right)^3} - 2{\left( { - 1} \right)^2} - 5\left( { - 1} \right) + 2 = \frac{{14}}{3} \cr & \to {\text{local maximum at }}\left( { - 1,\frac{{14}}{3}} \right) \cr & and \cr & f''\left( 5 \right) = 2\left( 5 \right) - 4 = 6 > 0,{\text{ then}} \cr & {\text{There is a local minimum at }}\left( {5,f\left( 5 \right)} \right) \cr & f\left( 5 \right) = \frac{1}{3}{\left( 5 \right)^3} - 2{\left( 5 \right)^2} - 5\left( 5 \right) + 2 = - \frac{{94}}{3} \cr & \to {\text{local minimum at }}\left( {5, - \frac{{94}}{3}} \right) \cr & \cr & {\text{*Find the Inflection points}}{\text{, set }}f''\left( x \right) = 0 \cr & 2x - 4 = 0 \cr & x = 2 \cr & f\left( 2 \right) = \frac{1}{3}{\left( 2 \right)^3} - 2{\left( 2 \right)^2} - 5\left( 2 \right) + 2 = - \frac{{40}}{3} \cr & \cr & {\text{The inflection point : }}\left( {2, - \frac{{40}}{3}} \right) \cr & \cr & {\text{Graph}} \cr} $$