Chapter 4 - Applications of the Derivative - 4.2 What Derivatives Tell Us - 4.2 Exercises - Page 257: 44

$$\eqalign{ & \left( a \right){\text{Critical points }}x = 0 \cr & \left( b \right){\text{No local min}}{\text{, no local max}}{\text{.}} \cr & \left( c \right){\text{Absolute min:}}\, - \,3964{\text{ at }}x = 4 \cr & \,\,\,\,\,\,{\text{Absolute max:}}\,\,68{\text{ at }}x = - 2 \cr} $$

$$\eqalign{ & f\left( x \right) = - 2{x^5} - 5{x^4} - 10{x^3}{\text{ + 4 on the interval }}\left[ { - 2,4} \right] \cr & {\text{Derivative}} \cr & f'\left( x \right) = - 10{x^4} - 20{x^3} - 30{x^2} \cr & {\text{Set the derivative to 0}} \cr & - 10{x^4} - 20{x^3} - 30{x^2} = 0 \cr & - 10{x^2}\left( {{x^2} + 2x + 3} \right) = 0 \cr & - 10{x^2}\left( {{x^2} + 2x + 3} \right) = 0 \cr & {\text{Solving the equation we obtain}} \cr & x = 0\,\, \cr & \cr & {\text{Evaluating that point and the endpoints we obtain}} \cr & f\left( { - 2} \right) = - 2{\left( { - 2} \right)^5} - 5{\left( { - 2} \right)^4} - 10{\left( { - 2} \right)^3} + 4 = 68 \cr & f\left( 0 \right) = - 2{\left( 0 \right)^5} - 5{\left( 0 \right)^4} - 10{\left( 0 \right)^3} + 4 = 4 \cr & f\left( { - 2} \right) = - 2{\left( { - 2} \right)^5} - 5{\left( { - 2} \right)^4} - 10{\left( { - 2} \right)^3} + 4 = - 3964 \cr & \cr & {\text{Then}}{\text{, }} \cr & {\text{The largest result is 68 at }}x = - 2\,\,\,\left( {{\text{Absolute maximum}}} \right) \cr & {\text{The smallest result is }} - 3964{\text{ at }}x = 4\,\,\left( {{\text{Absolute minimum}}} \right) \cr & {\text{There are no Local min or max}} \cr & \cr & {\text{We can conclude that}} \cr & \left( a \right){\text{Critical points }}x = 0 \cr & \left( b \right){\text{No local min}}{\text{, no local max}}{\text{.}} \cr & \left( c \right){\text{Absolute min:}}\, - \,3964{\text{ at }}x = 4 \cr & \,\,\,\,\,\,{\text{Absolute max:}}\,\,68{\text{ at }}x = - 2 \cr} $$