Chapter 4 - Applications of the Derivative - 4.1 Maxima and Minima - 4.1 Exercises - Page 243: 49

$$\eqalign{ & \left( {\text{a}} \right)x = 1{\text{ and }}x = 4 \cr & \left( {\text{b}} \right){\text{absolute maximum of }}11{\text{ at }}x = 1 \cr & {\text{absolute minimum of }} - 16{\text{ at }}x = 4 \cr & \left( {\text{c}} \right){\text{ graph}} \cr} $$

$$\eqalign{ & f\left( x \right) = 2{x^3} - 15{x^2} + 24x{\text{ on }}\left[ {0,5} \right] \cr & \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {2{x^3} - 15{x^2} + 24x} \right] \cr & f'\left( x \right) = 6{x^2} - 30x + 24 \cr & \cr & \left( a \right){\text{ Set the derivative equal to 0}} \cr & f'\left( x \right) = 0 \cr & 6{x^2} - 30x + 24 = 0 \cr & {x^2} - 5x + 4 = 0 \cr & {\text{Factoring}} \cr & \left( {x - 4} \right)\left( {x - 1} \right) = 0 \cr & {\text{The critical points are:}} \cr & x = 1{\text{ and }}x = 4 \cr & \cr & \left( {\text{b}} \right){\text{Evaluate }}f\left( x \right){\text{ at the critical points and the endpoints }} \cr & f\left( 0 \right) = 2{\left( 0 \right)^3} - 15{\left( 0 \right)^2} + 24\left( 0 \right) = 0 \cr & f\left( 1 \right) = 2{\left( 1 \right)^3} - 15{\left( 1 \right)^2} + 24\left( 1 \right) = 11,{\text{ }}\left( {{\text{Largest}}} \right) \cr & f\left( 4 \right) = 2{\left( 4 \right)^3} - 15{\left( 4 \right)^2} + 24\left( 4 \right) = - 16,{\text{ }}\left( {{\text{Smallest}}} \right) \cr & f\left( 5 \right) = 2{\left( 5 \right)^3} - 15{\left( 5 \right)^2} + 24\left( 5 \right) = - 5 \cr & {\text{*The largest of these function values is }}f\left( 1 \right) = 11,{\text{ then}} \cr & {\text{the absolute maximum of }}f\left( x \right){\text{ on }}\left[ {0,5} \right]{\text{ is 11}} \cr & {\text{*The smallest of these function values is }}f\left( 4 \right) = - 16,{\text{ then}} \cr & {\text{the absolute minimum of }}f\left( x \right){\text{ on }}\left[ {0,5} \right]{\text{ is }} - {\text{16}} \cr & \cr & \left( {\text{c}} \right){\text{Graph}} \cr} $$