Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 9

$$\eqalign{ & {\text{local maximum }}\left( {1,4} \right) \cr & {\text{local minimum }}\left( {3,0} \right) \cr & {\text{inflection point }}\left( {2,2} \right) \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & x{\text{ - intercepts }}\left( {0,0} \right){\text{ and }}\left( {3,0} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = {x^3} - 6{x^2} + 9x \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^3} - 6{x^2} + 9x} \right] \cr & f'\left( x \right) = 3{x^2} - 12x + 9 \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find the critical points}} \cr & 3{x^2} - 12x + 9 = 0 \cr & {x^2} - 4x + 3 = 0 \cr & \left( {x - 3} \right)\left( {x - 1} \right) = 0 \cr & {\text{The critical points are }}x = 1{\text{ and }}x = 3 \cr & \cr & {\text{Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {3{x^2} - 12x + 9} \right] \cr & f''\left( x \right) = 6x - 12 \cr & \cr & {\text{Evaluate }}f''\left( x \right){\text{ at the critical points }}x = 1{\text{ and }}x = 3 \cr & f''\left( 1 \right) = 6\left( 1 \right) - 12 = 0 - 6 < 0,{\text{ then}} \cr & {\text{There is a local maximum at }}x = 1 \cr & f\left( 1 \right) = {\left( 1 \right)^3} - 6{\left( 1 \right)^2} + 9\left( 1 \right) = 4 \cr & {\text{local maximum }}\left( {1,4} \right) \cr & f''\left( 3 \right) = 6\left( 3 \right) - 12 = 6 > 0,{\text{ then}} \cr & {\text{There is a local minimum at }}x = 3 \cr & f\left( 3 \right) = {\left( 3 \right)^3} - 6{\left( 3 \right)^2} + 9\left( 3 \right) = 0 \cr & {\text{local minimum }}\left( {3,0} \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 \right)^3} - 6{\left( 2 \right)^2} + 9\left( 2 \right) = 2 \cr & {\text{The inflection point is }}\left( {2,2} \right) \cr & \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & f\left( 0 \right) = {\left( 0 \right)^3} - 6{\left( 0 \right)^2} + 9\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 & 0 = {x^3} - 6{x^2} + 9x \cr & x\left( {{x^2} - 6x + 9} \right) = 0 \cr & x{\left( {x - 3} \right)^2} = 0 \cr & x = 0,{\text{ }}x = 3 \cr & x{\text{ - intercepts }}\left( {0,0} \right){\text{ and }}\left( {3,0} \right) \cr & \cr & {\text{Summary:}} \cr & {\text{local maximum }}\left( {1,4} \right) \cr & {\text{local minimum }}\left( {3,0} \right) \cr & {\text{inflection point }}\left( {2,2} \right) \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & x{\text{ - intercepts }}\left( {0,0} \right){\text{ and }}\left( {3,0} \right) \cr & \cr & {\text{Graph}} \cr} $$