Chapter 4 - Applications of the Derivative - 4.1 Maxima and Minima - 4.1 Exercises - Page 243: 28

$$\eqalign{ & \left( a \right){\text{Critical points }}x = - 2,\,\,x = - \frac{3}{2},\,\,x = 0,\,\,x = \frac{3}{2},\,\,x = 2 \cr & \left( b \right)f\left( { - \frac{3}{2}} \right) = \frac{{181}}{{20}}{\text{ Is the absolute }}\left( {{\text{and local}}} \right)\,{\text{maximum}} \cr & \,\,\,\,\,\,f\left( {\frac{3}{2}} \right) = \frac{{19}}{{20}}{\text{is the absolute }}\left( {{\text{and local}}} \right)\,{\text{minimum}} \cr} $$

$$\eqalign{ & f\left( x \right) = \frac{{4{x^5}}}{5} - 3{x^3} + 5,{\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) = \frac{{4\left( {5{x^4}} \right)}}{5} - 3\left( {3{x^2}} \right) \cr & f'\left( x \right) = 4{x^4} - 9{x^2} \cr & {\text{Setting the derivative equal to zero}}{\text{, we have}} \cr & 4{x^4} - 9{x^2} = 0 \cr & {x^2}\left( {4{x^2} - 9} \right) = 0 \cr & {x^2}\left( {2x + 3} \right)\left( {2x - 3} \right) = 0 \cr & {\text{Solving this equation gives the critical points}} \cr & x = 0,\,\,x = - \frac{3}{2}\,{\text{ and }}x = \frac{3}{2} \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 - \frac{3}{2},0,\frac{3}{2},2} \right\} \cr & \cr & {\text{Evaluating these points}} \cr & f\left( { - 2} \right) = \frac{{4{{\left( { - 2} \right)}^5}}}{5} - 3{\left( { - 2} \right)^3} + 5 = \frac{{17}}{5} \cr & f\left( { - 3/2} \right) = \frac{{4{{\left( { - 3/2} \right)}^5}}}{5} - 3{\left( { - 3/2} \right)^3} + 5 = \frac{{181}}{{20}} \cr & f\left( 0 \right) = \frac{{4{{\left( 0 \right)}^5}}}{5} - 3{\left( 0 \right)^3} + 5 = 5 \cr & f\left( {3/2} \right) = \frac{{4{{\left( {3/2} \right)}^5}}}{5} - 3{\left( {3/2} \right)^3} + 5 = \frac{{19}}{{20}} \cr & f\left( 2 \right) = \frac{{4{{\left( 2 \right)}^5}}}{5} - 3{\left( 2 \right)^3} + 5 = \frac{{33}}{5} \cr & \cr & {\text{The largest of these function values is }}f\left( { - 3/2} \right) = \frac{{181}}{{20}},{\text{ which is the}} \cr & {\text{absolute }}\left( {{\text{and local}}} \right)\,{\text{maximum on the interval }}\left[ { - 2,2} \right]. \cr & \cr & {\text{The smallest of these values is }}f\left( {\frac{3}{2}} \right) = \frac{{19}}{{20}}{\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} $$