Chapter 3 - Applications of Differentiation - 3.6 Exercises - Page 212: 22

Graph

$$\eqalign{ & y = - 2{x^4} + 3{x^2} \cr & {\text{Domain: }}\left( { - \infty ,\infty } \right) \cr & {\text{Find the }}y{\text{ intercept, let }}x = 0 \cr & y = - 2{\left( 0 \right)^4} + 3{\left( 0 \right)^2} \cr & y = 0 \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Find the }}x{\text{ intercept, let }}y = 0 \cr & - 2{x^4} + 3{x^2} = 0 \cr & {x^2}\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),\left( { \pm \sqrt {\frac{3}{2}} ,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & y' = \frac{d}{{dx}}\left[ { - 2{x^4} + 3{x^2}} \right] \cr & y' = - 8{x^3} + 6x \cr & {\text{Let }}y' = 0{\text{ to find critical points}} \cr & - 8{x^3} + 6x = 0 \cr & - 2x\left( {4{x^2} - 3} \right) = 0 \cr & x = 0,{\text{ }}x = - \frac{{\sqrt 3 }}{2},{\text{ }}x = \frac{{\sqrt 3 }}{2} \cr & *{\text{Find the second derivative}} \cr & y'' = \frac{d}{{dx}}\left[ { - 8{x^3} + 6x} \right] \cr & y'' = - 24{x^2} + 6 \cr & {\text{Evaluate }}y''\left( x \right){\text{ at }}x = 0,{\text{ }}x = - \sqrt 3 ,{\text{ }}x = \sqrt 3 \cr & y'' = - 24{\left( 0 \right)^2} + 6 > 0,{\text{ Relative minimum at }}\left( {0,f\left( 0 \right)} \right) \cr & y'' = - 24{\left( { - \frac{{\sqrt 3 }}{2}} \right)^2} + 6 < 0,{\text{ R}}{\text{. maximum at: }}\left( { - \frac{{\sqrt 3 }}{2},f\left( { - \frac{{\sqrt 3 }}{2}} \right)} \right) \cr & y'' = - 24{\left( {\frac{{\sqrt 3 }}{2}} \right)^2} + 6 < 0,{\text{Relative maximum at }}\left( {\frac{{\sqrt 3 }}{2},f\left( {\frac{{\sqrt 3 }}{2}} \right)} \right) \cr & \left( {0,f\left( 0 \right)} \right) = \left( {0,0} \right),{\text{ Relative minimum}} \cr & \left( { - \frac{{\sqrt 3 }}{2},f\left( { - \frac{{\sqrt 3 }}{2}} \right)} \right) = \left( { - \frac{{\sqrt 3 }}{2},\frac{9}{8}} \right),{\text{ Relative maximum}} \cr & \left( {\frac{{\sqrt 3 }}{2},f\left( {\frac{{\sqrt 3 }}{2}} \right)} \right) = \left( {\frac{{\sqrt 3 }}{2},\frac{9}{8}} \right),{\text{ Relative maximum}} \cr & \cr & {\text{Let }}y''\left( x \right) = 0 \cr & - 24{x^2} + 6 = 0 \cr & {x^2} = \frac{1}{4} \cr & x = - \frac{1}{2},{\text{ }}x = \frac{1}{2} \cr & {\text{Inflection points at }}x = - \frac{1}{2}{\text{ and }}x = \frac{1}{2} \cr & \cr & {\text{*Calculate the asymptotes}} \cr & {\text{No vertical asymptotes, the denominator is 1}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to \infty } \left( { - 2{x^4} + 3{x^2}} \right) = - \infty \cr & \mathop {\lim }\limits_{x \to - \infty } \left( { - 2{x^4} + 3{x^2}} \right) = - \infty \cr & {\text{No horizontal asymptotes}} \cr & \cr & {\text{Graph}} \cr} $$