Chapter 4 - Applications of the Derivative - 4.3 Graphing Functions - 4.3 Exercises - Page 268: 12

$$\eqalign{ & {\text{local maximum }}\left( {0,0} \right) \cr & {\text{local minimum }}\left( { - 1, - 1} \right){\text{ and }}\left( { - 1, - 1} \right) \cr & {\text{inflection points }}\left( {0,0} \right){\text{ and }}\left( { \pm \sqrt {\frac{3}{5}} , = - \frac{{81}}{{125}}} \right) \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & x{\text{ - intercepts }}\left( {0,0} \right){\text{, }}\left( { - \sqrt {\frac{3}{2}} ,0} \right){\text{ and }}\left( {\sqrt {\frac{3}{2}} ,0} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = 2{x^6} - 3{x^4} \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {2{x^6} - 3{x^4}} \right] \cr & f'\left( x \right) = 12{x^5} - 12{x^3} \cr & {\text{Let }}f'\left( x \right) = 0{\text{ to find the critical points}} \cr & 12{x^5} - 12{x^3} = 0 \cr & 12{x^3}\left( {{x^2} - 1} \right) = 0 \cr & {\text{The critical points are }}x = 0{\text{ and }}x = \pm 1 \cr & \cr & {\text{Find the second derivative}} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {12{x^5} - 12{x^3}} \right] \cr & f''\left( x \right) = 60{x^4} - 36{x^2} \cr & \cr & {\text{Evaluate }}f''\left( x \right){\text{ at the critical points }} \cr & *f''\left( 0 \right) = 60{\left( 0 \right)^4} - 36{\left( 0 \right)^2} = 0,{\text{ }} \cr & {\text{The second derivative test is inconclusive}} \cr & {\text{Use the first derivative test}}{\text{,}} \cr & f'\left( { - 0.5} \right) = 12{\left( { - 0.5} \right)^5} - 12{\left( { - 0.5} \right)^3} = \frac{9}{8} > 0 \cr & f'\left( {0.5} \right) = 12{\left( { - 0.5} \right)^5} - 12{\left( { - 0.5} \right)^3} = - \frac{9}{8} < 0 \cr & {\text{By the first derivative test}}{\text{, we conclude that}} \cr & {\text{There is a local maximum at }}x = 0 \cr & f\left( 0 \right) = 2{\left( 0 \right)^6} - 3{\left( 0 \right)^4} = 0 \cr & {\text{local maximum }}\left( {0,0} \right) \cr & \cr & *f''\left( { - 1} \right) = 60{\left( { - 1} \right)^4} - 36{\left( { - 1} \right)^2} = 24 < 0,{\text{ then}} \cr & {\text{There is a local minimum at }}x = - 1 \cr & f\left( { - 1} \right) = 2{\left( { - 1} \right)^6} - 3{\left( { - 1} \right)^4} = - 1 \cr & {\text{local minimum }}\left( { - 1, - 1} \right) \cr & *f''\left( 1 \right) = 60{\left( 1 \right)^4} - 36{\left( 1 \right)^2} = 24 < 0,{\text{ then}} \cr & {\text{There is a local minimum at }}x = - 1 \cr & f\left( 1 \right) = 2{\left( 1 \right)^6} - 3{\left( 1 \right)^4} = - 1 \cr & {\text{local minimum }}\left( {1, - 1} \right) \cr & \cr & {\text{Set }}f''\left( x \right) = 0{\text{ to locate the inflection points}} \cr & f''\left( x \right) = 60{x^4} - 36{x^2} \cr & 60{x^4} - 36{x^2} = 0 \cr & 12{x^2}\left( {5{x^2} - 3} \right) = 0 \cr & x = 0,{\text{ }}x = \pm \sqrt {\frac{3}{5}} \cr & f\left( 0 \right) = 2{\left( 0 \right)^6} - 3{\left( 0 \right)^4} = 0 \cr & f\left( { - \sqrt {\frac{3}{5}} } \right) = 2{\left( { - \sqrt {\frac{3}{5}} } \right)^6} - 3{\left( { - \sqrt {\frac{3}{5}} } \right)^4} = - \frac{{81}}{{125}} \cr & f\left( {\sqrt {\frac{3}{5}} } \right) = 2{\left( {\sqrt {\frac{3}{5}} } \right)^6} - 3{\left( {\sqrt {\frac{3}{5}} } \right)^4} = - \frac{{81}}{{125}} \cr & \cr & {\text{The inflection points are: }}\left( {0,0} \right){\text{ and }}\left( { \pm \sqrt {\frac{3}{5}} , = - \frac{{81}}{{125}}} \right) \cr & \cr & {\text{Find the }}y{\text{ intercept}}{\text{, let }}x = 0 \cr & f\left( 0 \right) = 2{\left( 0 \right)^6} - 3{\left( 0 \right)^4} \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 & 2{x^6} - 3{x^4} = 0 \cr & {x^4}\left( {2{x^2} - 3} \right) = 0 \cr & x = 0,{\text{ }}x = \pm \sqrt {\frac{3}{2}} \cr & x{\text{ - intercepts }}\left( {0,0} \right){\text{, }}\left( { - \sqrt {\frac{3}{2}} ,0} \right){\text{ and }}\left( {\sqrt {\frac{3}{2}} ,0} \right) \cr & \cr & {\text{Summary:}} \cr & {\text{local maximum }}\left( {0,0} \right) \cr & {\text{local minimum }}\left( { - 1, - 1} \right){\text{ and }}\left( { - 1, - 1} \right) \cr & {\text{inflection points }}\left( {0,0} \right){\text{ and }}\left( { \pm \sqrt {\frac{3}{5}} , = - \frac{{81}}{{125}}} \right) \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & x{\text{ - intercepts }}\left( {0,0} \right){\text{, }}\left( { - \sqrt {\frac{3}{2}} ,0} \right){\text{ and }}\left( {\sqrt {\frac{3}{2}} ,0} \right) \cr & \cr & {\text{Graph}} \cr} $$