Chapter 3 - Applications of Differentiation - Review Exercises - Page 239: 70

$$\eqalign{ & y{\text{ - intercept }}\left( {0,16} \right) \cr }$$

$$\eqalign{ & f\left( x \right) = {\left( {{x^2} - 4} \right)^2} \cr & {\text{Find the }}y{\text{ intercept, let }}x = 0 \cr & f\left( 0 \right) = {\left( {{0^2} - 4} \right)^2} \cr & f\left( 0 \right) = 16 \cr & y{\text{ - intercept }}\left( {0,16} \right) \cr & {\text{Find the }}x{\text{ intercept, let }}y = 0 \cr & 0 = {\left( {{x^2} - 4} \right)^2} \cr & {\left( {{x^2} - 4} \right)^2} = 0 \cr & {x^2} - 4 = 0 \cr & x = - 2,{\text{ }}x = 2 \cr & x{\text{ - intercepts: }}\left( { - 2,0} \right){\text{ and }}\left( {2,0} \right) \cr & \cr & {\text{*Differentiate}} \cr & y' = \frac{d}{{dx}}\left[ {{{\left( {{x^2} - 4} \right)}^2}} \right] \cr & y' = 2\left( {{x^2} - 4} \right)\left( {2x} \right) \cr & {\text{Let }}y' = 0{\text{ to find critical points}} \cr & 4x\left( {{x^2} - 4} \right) = 0 \cr & x = - 2,{\text{ }}x = 0,{\text{ }}x = 2 \cr & \cr & *{\text{Find the second derivative}} \cr & y'' = \frac{d}{{dx}}\left[ {4x\left( {{x^2} - 4} \right)} \right] \cr & y'' = 4x\left( {2x} \right) + 4\left( {{x^2} - 4} \right) \cr & y'' = 8{x^2} + 4{x^2} - 16 \cr & y'' = 12{x^2} - 16 \cr & {\text{*Evaluate }}y''{\text{ at the critical point }}x = - 2 \cr & {\text{ }}y''\left( { - 2} \right) = 32 > 0,{\text{ there are a relative minimum at }}f\left( { - 2} \right) \cr & {\text{ }}f\left( { - 2} \right) = 0 \cr & {\text{Relative minimum at }}\left( { - 2,0} \right) \cr & {\text{*Evaluate }}y''{\text{ at the critical point }}x = 0 \cr & {\text{ }}y''\left( 0 \right) = - 16 < 0,{\text{ there are a relative maximum at }}f\left( 0 \right) \cr & {\text{ }}f\left( 0 \right) = 16 \cr & {\text{Relative maximum at }}\left( {0,16} \right) \cr & {\text{*Evaluate }}y''{\text{ at the critical point }}x = 2 \cr & {\text{ }}y''\left( 2 \right) = 32 > 0,{\text{ there are a relative minimum at }}f\left( 2 \right) \cr & {\text{ }}f\left( 2 \right) = 0 \cr & {\text{Relative minimum at }}\left( {2,0} \right) \cr & \cr & {\text{Let }}y''\left( x \right) = 0 \cr & 12{x^2} - 16 = 0 \cr & {x^2} = \frac{4}{3} \cr & x = \pm \frac{2}{{\sqrt 3 }} \cr & f\left( {\frac{2}{{\sqrt 3 }}} \right) = \frac{{64}}{9}{\text{ and }}f\left( { - \frac{2}{{\sqrt 3 }}} \right) = \frac{{64}}{9} \cr & {\text{The inflection points are:}} \cr & \left( {\frac{2}{{\sqrt 3 }},\frac{{64}}{9}} \right){\text{ and }}\left( { - \frac{2}{{\sqrt 3 }},\frac{{64}}{9}} \right) \cr & \cr & {\text{*Calculate the asymptotes}} \cr & {\text{No vertical asymptotes as it is a polynomial function}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to \infty } {\left( {{x^2} - 4} \right)^2} = \infty \cr & \mathop {\lim }\limits_{x \to - \infty } {\left( {{x^2} - 4} \right)^2} = \infty \cr & {\text{No horizontal asymptotes}} \cr & \cr } $$