Chapter 4 - Applications of the Derivative - 4.1 Maxima and Minima - 4.1 Exercises - Page 243: 30

$$\eqalign{ & \left( a \right){\text{Critical points }}x = - 2,\,\,x = - 1,\,\,x = 0,\,\,x = 1,{\text{ }}x = 2 \cr & \left( b \right)f\left( 2 \right) = 224{\text{ Is the absolute maximum}} \cr & \,\,\,\,\,\,f\left( { - 2} \right) = - 224{\text{ Is the absolute minimum}} \cr & \,\,\,\,\,\,\,f\left( { - 1} \right) = 8{\text{ is a local maximum}} \cr & \,\,\,\,\,\,\,f\left( 1 \right) = - 8{\text{ is a local minimum}} \cr & \,\,\,\,\,\,\,f\left( 0 \right) = 0{\text{ No local extremum}} \cr} $$

$$\eqalign{ & f\left( x \right) = 12{x^5} - 20{x^3},{\text{ on the interval }}\left[ { - 2,2} \right] \cr & \cr & {\text{Because }}f{\text{ is a polynomial}}{\text{, its derivative exists everywhere}} \cr & f'\left( x \right) = 60{x^4} - 60{x^2} \cr & {\text{Setting the derivative equal to zero}}{\text{, we have}} \cr & 60{x^4} - 60{x^2} = 0 \cr & 60{x^2}\left( {{x^2} - 1} \right) = 0 \cr & {\text{Solving this equation gives the critical points}} \cr & x = 0,\,\,\,\,x = 1\,{\text{ and }}x = - 1 \cr & {\text{All of them lies in the given interval }}\left[ { - 2,2} \right]. \cr & {\text{These points and the endpoints are candidates for the location}} \cr & {\text{of absolute extrema:}} \cr & {\text{Points }}x = \left\{ { - 2, - 1,0,1,2} \right\} \cr & \cr & {\text{Evaluating these points}} \cr & f\left( { - 2} \right) = 12{\left( { - 2} \right)^5} - 20{\left( { - 2} \right)^3} = - 224 \cr & f\left( { - 1} \right) = 12{\left( { - 1} \right)^5} - 20{\left( { - 1} \right)^3} = 8 \cr & f\left( 0 \right) = 12{\left( 0 \right)^5} - 20{\left( 0 \right)^3} = 0 \cr & f\left( 1 \right) = 12{\left( 1 \right)^5} - 20{\left( 1 \right)^3} = - 8 \cr & f\left( 2 \right) = 12{\left( 2 \right)^5} - 20{\left( 2 \right)^3} = 224 \cr & \cr & {\text{The largest of these function values is }}f\left( 2 \right) = 224,{\text{ which is the}} \cr & {\text{absolute maximum on the interval }}\left[ { - 2,2} \right]. \cr & \cr & {\text{The smallest of these values is }}f\left( { - 2} \right) = - 224\,{\text{which is the absolute}} \cr & \left( {{\text{and local}}} \right){\text{minimum on the interval }}\left[ { - 2,2} \right]. \cr & \cr & {\text{The next graph shows that the critical points corresponds to neither}} \cr & {\text{a local maximum nor a local minimum}}{\text{.}} \cr} $$