Chapter 3 - Applications of Differentiation - Review Exercises - Page 239: 72

$$\eqalign{ & y{\text{ - intercept }}\left( {0, - 24} \right) \cr & x{\text{ - intercepts: }}\left( { - 2,0} \right){\text{ and }}\left( {3,0} \right) \cr & {\text{Relative minimum at }}\left( {\frac{7}{4}, - 65.918} \right) \cr & {\text{Inflection points: }}\left( { - 2,0} \right){\text{ and }}\left( {\frac{1}{2}, - 39.0625} \right) \cr & {\text{No vertical asymptotes}} \cr & {\text{No horizontal asymptotes}} \cr} $$

$$\eqalign{ & f\left( x \right) = \left( {x - 3} \right){\left( {x + 2} \right)^3} \cr & {\text{Find the }}y{\text{ intercept, let }}x = 0 \cr & f\left( 0 \right) = \left( {0 - 3} \right){\left( {0 + 2} \right)^3} \cr & f\left( 0 \right) = - 24 \cr & y{\text{ - intercept }}\left( {0, - 24} \right) \cr & {\text{Find the }}x{\text{ intercept, let }}y = 0 \cr & \left( {x - 3} \right){\left( {x + 2} \right)^3} = 0 \cr & x = - 2,{\text{ }}x = 3 \cr & x{\text{ - intercepts: }}\left( { - 2,0} \right){\text{ and }}\left( {3,0} \right) \cr & \cr & {\text{*Differentiate}} \cr & y' = \frac{d}{{dx}}\left[ {\left( {x - 3} \right){{\left( {x + 2} \right)}^3}} \right] \cr & y' = 3\left( {x - 3} \right){\left( {x + 2} \right)^2} + {\left( {x + 2} \right)^3} \cr & {\text{Let }}y' = 0{\text{ to find critical points}} \cr & 3\left( {x - 3} \right){\left( {x + 2} \right)^2} + {\left( {x + 2} \right)^3} = 0 \cr & {\left( {x + 2} \right)^2}\left[ {3\left( {x - 3} \right) + \left( {x + 2} \right)} \right] = 0 \cr & {\left( {x + 2} \right)^2}\left( {3x - 9 + x + 2} \right) = 0 \cr & {\left( {x + 2} \right)^2}\left( {4x - 7} \right) = 0 \cr & x = - 2,{\text{ }}x = \frac{7}{4} \cr & \cr & *{\text{Find the second derivative}} \cr & y'' = \frac{d}{{dx}}\left[ {{{\left( {x + 2} \right)}^2}\left( {4x - 7} \right)} \right] \cr & y'' = 2\left( {x + 2} \right)\left( {4x - 7} \right) + 4{\left( {x + 2} \right)^2} \cr & \cr & {\text{*Evaluate }}y''{\text{ at the critical point }}x = - 2 \cr & {\text{ }}y''\left( { - 2} \right) = 0 \cr & {\text{Using the first derivative test}} \cr & {\text{ }}y'\left( { - 3} \right) < 0{\text{ and }}y'\left( { - 1} \right) < 0 \cr & {\text{The derivative does not change at }}x = - 2,{\text{ then}} \cr & {\text{No extrema at }}x = - 2. \cr & \cr & {\text{*Evaluate }}y''{\text{ at the critical point }}x = \frac{7}{4} \cr & {\text{ }}y''\left( {\frac{7}{4}} \right) > 0,{\text{ there is a relative minimum at }}f\left( {\frac{7}{4}} \right) \cr & {\text{ }}f\left( {\frac{7}{4}} \right) \approx - 65.918 \cr & {\text{Relative minimum at }}\left( {\frac{7}{4}, - 65.918} \right) \cr & \cr & {\text{Let }}y''\left( x \right) = 0 \cr & 2\left( {x + 2} \right)\left( {4x - 7} \right) + 4{\left( {x + 2} \right)^2} = 0 \cr & 2\left( {x + 2} \right)\left[ {\left( {4x - 7} \right) + 2\left( {x + 2} \right)} \right] = 0 \cr & 2\left( {x + 2} \right)\left( {6x - 3} \right) = 0 \cr & x = - 2,{\text{ }}x = \frac{1}{2} \cr & f\left( { - 2} \right) = 0{\text{ and }}f\left( {\frac{1}{2}} \right) = - 39.0625 \cr & {\text{The inflection points are:}} \cr & \left( { - 2,0} \right){\text{ and }}\left( {\frac{1}{2}, - 39.0625} \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 - 3} \right){\left( {x + 2} \right)^3} = \infty \cr & \mathop {\lim }\limits_{x \to - \infty } \left( {x - 3} \right){\left( {x + 2} \right)^3} = \infty \cr & {\text{No horizontal asymptotes}} \cr & \cr } $$