Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 13

$$\eqalign{ & {\text{local maximum }}\left( { - 6,0} \right) \cr & {\text{local minimum }}\left( {2, - 256} \right) \cr & {\text{inflection point }}\left( { - 2, - 128} \right) \cr & y{\text{ - intercept }}\left( {0, - 216} \right) \cr & x{\text{ - intercepts }}\left( { - 6,0} \right){\text{ and }}\left( {6,0} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = \left( {x - 6} \right){\left( {x + 6} \right)^2} \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\left( {x - 6} \right){{\left( {x + 6} \right)}^2}} \right] \cr & f'\left( x \right) = 2\left( {x - 6} \right)\left( {x + 6} \right) + {\left( {x + 6} \right)^2} \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find the critical points}} \cr & 2\left( {x - 6} \right)\left( {x + 6} \right) + {\left( {x + 6} \right)^2} = 0 \cr & \left( {x + 6} \right)\left[ {2\left( {x - 6} \right) + \left( {x + 6} \right)} \right] = 0 \cr & \left( {x + 6} \right)\left( {2x - 12 + x + 6} \right) = 0 \cr & \left( {x + 6} \right)\left( {3x - 6} \right) = 0 \cr & {\text{The critical points are }}x = - 6{\text{ and }}x = 2 \cr & \cr & {\text{Find the second derivative}} \cr & f'\left( x \right) = 2\left( {x - 6} \right)\left( {x + 6} \right) + {\left( {x + 6} \right)^2} \cr & f'\left( x \right) = 3{x^2} + 12x - 36 \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {3{x^2} + 12x - 36} \right] \cr & f''\left( x \right) = 6x + 12 \cr & \cr & {\text{Evaluate }}f''\left( x \right){\text{ at the critical points }}x = - 6{\text{ and }}x = 2 \cr & *f''\left( { - 6} \right) = 6\left( { - 6} \right) + 12 = - 24 < 0,{\text{ then}} \cr & {\text{There is a local maximum at }}x = - 6 \cr & f\left( { - 6} \right) = \left( { - 6 - 6} \right){\left( { - 6 + 6} \right)^2} = 0 \cr & {\text{local maximum }}\left( { - 6,0} \right) \cr & *f''\left( 2 \right) = 6\left( 2 \right) + 12 = 24 > 0,{\text{ then}} \cr & {\text{There is a local minimum at }}x = 2 \cr & f\left( 2 \right) = \left( {2 - 6} \right){\left( {2 + 6} \right)^2} = - 256 \cr & {\text{local minimum }}\left( {2, - 256} \right) \cr & \cr & {\text{Set }}f''\left( x \right) = 0{\text{ to locate the inflection points}} \cr & f''\left( x \right) = 6x + 12 \cr & 6x + 12 = 0 \cr & x = - 2 \cr & f\left( { - 2} \right) = \left( { - 2 - 6} \right){\left( { - 2 + 6} \right)^2} = - 128 \cr & {\text{The inflection point is }}\left( { - 2, - 128} \right) \cr & \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & f\left( 0 \right) = \left( {0 - 6} \right){\left( {0 + 6} \right)^2} \cr & f\left( 0 \right) = - 216 \cr & y{\text{ - intercept }}\left( {0, - 216} \right) \cr & {\text{Find the }}x{\text{ intercept}}{\text{, let }}f\left( x \right) = 0 \cr & \left( {x - 6} \right){\left( {x + 6} \right)^2} = 0 \cr & x = - 6,{\text{ }}x = 6 \cr & x{\text{ - intercepts }}\left( { - 6,0} \right){\text{ and }}\left( {6,0} \right) \cr & \cr & {\text{Summary:}} \cr & {\text{local maximum }}\left( { - 6,0} \right) \cr & {\text{local minimum }}\left( {2, - 256} \right) \cr & {\text{inflection point }}\left( { - 2, - 128} \right) \cr & y{\text{ - intercept }}\left( {0, - 216} \right) \cr & x{\text{ - intercepts }}\left( { - 6,0} \right){\text{ and }}\left( {6,0} \right) \cr & \cr & {\text{Graph}} \cr} $$