Chapter 3 - Applications of Differentiation - Review Exercises - Page 239: 68

$$\eqalign{ & y{\text{ - intercept }}\left( {0,0} \right) \cr & x{\text{ - intercepts: }}\left( {0,0} \right){\text{ and }}\left( {4,0} \right) \cr & {\text{Relative maximum at }}\left( {3,27} \right) \cr & {\text{Inflection points: }}\left( {0,0} \right){\text{ and }}\left( {1,3} \right) \cr & {\text{No vertical asymptotes}} \cr & {\text{No horizontal asymptotes}} \cr} $$

$$\eqalign{ & f\left( x \right) = 4{x^3} - {x^4} \cr & {\text{Find the }}y{\text{ intercept, let }}x = 0 \cr & y = 4{\left( 0 \right)^3} - {\left( 0 \right)^4} \cr & y = 0 \cr & y{\text{ - intercept }}\left( {0,0} \right) \cr & {\text{Find the }}x{\text{ intercept, let }}y = 0 \cr & 0 = 4{x^3} - {x^4} \cr & {x^3}\left( {4 - x} \right) = 0 \cr & x = 0,{\text{ }}x = 4 \cr & x{\text{ - intercepts: }}\left( {0,0} \right){\text{ and }}\left( {4,0} \right) \cr & \cr & *{\text{Find the extrema}} \cr & {\text{Differentiate}} \cr & y' = \frac{d}{{dx}}\left[ {4{x^3} - {x^4}} \right] \cr & y' = 12{x^2} - 4{x^3} \cr & {\text{Let }}y' = 0{\text{ to find critical points}} \cr & 12{x^2} - 4{x^3} = 0 \cr & 4{x^2}\left( {3 - x} \right) = 0 \cr & x = 0,{\text{ }}x = 3 \cr & \cr & *{\text{Find the second derivative}} \cr & y'' = \frac{d}{{dx}}\left[ {12{x^2} - 4{x^3}} \right] \cr & y'' = 24x - 12{x^2} \cr & {\text{Evaluate }}y''{\text{ at the critical point }}x = 3{\text{ and }}x = 0 \cr & y''\left( 3 \right) = - 36 < 0,{\text{ there are a relative maximum at }}f\left( 3 \right) \cr & f\left( 3 \right) = 27 \cr & {\text{Relative maximum at }}\left( {3,27} \right) \cr & y''\left( 0 \right) = 0 \cr & {\text{Using the first derivative test:}} \cr & f'\left( { - 1} \right) = 16{\text{ and }}f'\left( 1 \right) = 8 \cr & {\text{The sign of the derivative does not change at }}x = 0, \cr & {\text{There are no relative extrema at }}x = 0 \cr & \cr & {\text{Let }}y''\left( x \right) = 0 \cr & 24x - 12{x^2} = 0 \cr & 12x\left( {1 - x} \right) = 0 \cr & x = 0,{\text{ }}x = 1 \cr & f\left( 0 \right) = 0 \cr & f\left( 1 \right) = 3 \cr & {\text{Inflection points at: }}\left( {0,0} \right){\text{ and }}\left( {1,3} \right) \cr & \cr & {\text{*Calculate the asymptotes}} \cr & {\text{No vertical asymptotes, the denominator is 1}}{\text{.}} \cr & \mathop {\lim }\limits_{x \to \infty } \left( {4{x^3} - {x^4}} \right) = - \infty \cr & \mathop {\lim }\limits_{x \to - \infty } \left( {4{x^3} - {x^4}} \right) = - \infty \cr & {\text{No horizontal asymptotes}} \cr & \cr } $$