Chapter 4 - Applications of the Derivative - 4.1 Maxima and Minima - 4.1 Exercises - Page 243: 27

$$\eqalign{ & \left( a \right){\text{Critical points }}x = - 1,\,\,x = - \frac{2}{3},\,\,x = \frac{1}{3},\,\,x = 1 \cr & \left( b \right)f\left( 1 \right) = \frac{5}{2}{\text{ Is the absolute maximum}} \cr & \,\,\,\,\,\,f\left( {\frac{1}{3}} \right) = - \frac{7}{{18}}{\text{is the absolute }}\left( {{\text{and local}}} \right)\,{\text{minimum}} \cr & \,\,\,\,\,\,\,f\left( { - \frac{2}{3}} \right) = \frac{{10}}{3}{\text{ is a local maximum}} \cr} $$

$$\eqalign{ & f\left( x \right) = 3{x^3} + \frac{{3{x^2}}}{2} - 2x,{\text{ on the interval }}\left[ { - 1,1} \right] \cr & \cr & {\text{Because }}f{\text{ is a polynomial}}{\text{, its derivative exists everywhere}} \cr & f'\left( x \right) = 3\left( {3{x^2}} \right) + \frac{{6x}}{2} - 2 \cr & f'\left( x \right) = 9{x^2} + 3x - 2 \cr & {\text{Setting the derivative equal to zero}}{\text{, we have}} \cr & 9{x^2} + 3x - 2 = 0 \cr & \left( {3x - 1} \right)\left( {3x + 2} \right) = 0 \cr & {\text{Solving this equation gives the critical points}} \cr & x = - \frac{2}{3}\,{\text{ and }}x = \frac{1}{3} \cr & {\text{All of them lies in the given interval }}\left[ { - 1,1} \right]. \cr & {\text{These points and the endpoints are candidates for the location}} \cr & {\text{of absolute extrema:}} \cr & {\text{Points }}x = \left\{ { - 1, - \frac{2}{3},\frac{1}{3},1} \right\} \cr & \cr & {\text{Evaluating these points}} \cr & f\left( { - 1} \right) = 3{\left( { - 1} \right)^3} + \frac{{3{{\left( { - 1} \right)}^2}}}{2} - 2\left( { - 1} \right) = \frac{1}{2} \cr & f\left( { - \frac{2}{3}} \right) = 3{\left( { - \frac{2}{3}} \right)^3} + \frac{{3{{\left( { - 2/3} \right)}^2}}}{2} - 2\left( { - \frac{2}{3}} \right) = \frac{{10}}{9} \cr & f\left( {\frac{1}{3}} \right) = 3{\left( {\frac{1}{3}} \right)^3} + \frac{{3{{\left( {1/3} \right)}^2}}}{2} - 2\left( {\frac{1}{3}} \right) = - \frac{7}{{18}} \cr & f\left( 1 \right) = 3{\left( 1 \right)^3} + \frac{{3{{\left( 1 \right)}^2}}}{2} - 2\left( 1 \right) = \frac{5}{2} \cr & \cr & {\text{The largest of these function values is }}f\left( 1 \right) = \frac{5}{2},{\text{ which is the}} \cr & {\text{absolute maximum on the interval }}\left[ { - 1,1} \right]. \cr & \cr & {\text{The smallest of these values is }}f\left( {\frac{1}{3}} \right) = - \frac{7}{{18}}{\text{which is the absolute}} \cr & \left( {{\text{and local}}} \right){\text{minimum on the interval }}\left[ { - 1,1} \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} $$