Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 11

$$\eqalign{ & {\text{local maximum }}\left( {0,0} \right) \cr & {\text{local minimum }}\left( { - \sqrt 3 , - 9} \right){\text{ and }}\left( {\sqrt 3 , - 9} \right) \cr & {\text{inflection points }}\left( { - 1, - 5} \right){\text{ and }}\left( {1, - 5} \right) \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & x{\text{ - intercepts }}\left( {0,0} \right){\text{, }}\left( { - \sqrt 6 ,0} \right){\text{ and }}\left( {\sqrt 6 ,0} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = {x^4} - 6{x^2} \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^4} - 6{x^2}} \right] \cr & f'\left( x \right) = 4{x^3} - 12x \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find the critical points}} \cr & 4{x^3} - 12x = 0 \cr & 4x\left( {{x^2} - 3} \right) = 0 \cr & {\text{The critical points are }}x = 0{\text{ and }}x = \pm \sqrt 3 \cr & \cr & {\text{Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {4{x^3} - 12x} \right] \cr & f''\left( x \right) = 12{x^2} - 12 \cr & \cr & {\text{Evaluate }}f''\left( x \right){\text{ at the critical points }} \cr & *f''\left( 0 \right) = 12{\left( 0 \right)^2} - 12 = - 12 < 0,{\text{ then}} \cr & {\text{There is a local maximum at }}x = 0 \cr & f\left( 0 \right) = {\left( 0 \right)^4} - 6{\left( 0 \right)^2} \cr & {\text{local maximum }}\left( {0,0} \right) \cr & *f''\left( { - \sqrt 3 } \right) = 12{\left( { - \sqrt 3 } \right)^2} - 12 = 24 > 0,{\text{ then}} \cr & {\text{There is a local minimum at }}x = - \sqrt 3 \cr & f\left( { - \sqrt 3 } \right) = {\left( { - \sqrt 3 } \right)^4} - 6{\left( { - \sqrt 3 } \right)^2} = - 9 \cr & {\text{local minimum }}\left( { - \sqrt 3 , - 9} \right) \cr & *f''\left( {\sqrt 3 } \right) = 12{\left( {\sqrt 3 } \right)^2} - 12 = 24 > 0,{\text{ then}} \cr & {\text{There is a local minimum at }}x = \sqrt 3 \cr & f\left( {\sqrt 3 } \right) = {\left( {\sqrt 3 } \right)^4} - 6{\left( {\sqrt 3 } \right)^2} = - 9 \cr & {\text{local minimum }}\left( {\sqrt 3 , - 9} \right) \cr & \cr & {\text{Set }}f''\left( x \right) = 0{\text{ to locate the inflection points}} \cr & f''\left( x \right) = 12{x^2} - 12 \cr & 12{x^2} - 12 = 0 \cr & 12{x^2} = 12 \cr & x = \pm 1 \cr & f\left( { - 1} \right) = {\left( { - 1} \right)^4} - 6{\left( { - 1} \right)^2} = - 5 \cr & f\left( 1 \right) = {\left( 1 \right)^4} - 6{\left( 1 \right)^2} = - 5 \cr & \cr & {\text{The inflection points are: }}\left( { - 1, - 5} \right){\text{ and }}\left( {1, - 5} \right) \cr & \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & f\left( 0 \right) = {\left( 0 \right)^4} - 6{\left( 0 \right)^2} \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^4} - 6{x^2} \cr & {x^2}\left( {{x^2} - 6} \right) \cr & x = 0,{\text{ }}x = \pm \sqrt 6 \cr & x{\text{ - intercepts }}\left( {0,0} \right){\text{, }}\left( { - \sqrt 6 ,0} \right){\text{ and }}\left( {\sqrt 6 ,0} \right) \cr & \cr & {\text{Summary:}} \cr & {\text{local maximum }}\left( {0,0} \right) \cr & {\text{local minimum }}\left( { - \sqrt 3 , - 9} \right){\text{ and }}\left( {\sqrt 3 , - 9} \right) \cr & {\text{inflection points }}\left( { - 1, - 5} \right){\text{ and }}\left( {1, - 5} \right) \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & x{\text{ - intercepts }}\left( {0,0} \right){\text{, }}\left( { - \sqrt 6 ,0} \right){\text{ and }}\left( {\sqrt 6 ,0} \right) \cr & \cr & {\text{Graph}} \cr} $$