Chapter 3 - The Derivative In Graphing And Applications - 3.4 Absolute Maxima And Minima - Exercises Set 3.4 - Page 222: 7

$$\eqalign{ & {\text{max}} = 2{\text{ at }}x = 1,2\,\,\, \cr & {\text{min}} = 1{\text{ at }}x = \frac{3}{2}\,\,\,\,\, \cr} $$

$$\eqalign{ & f\left( x \right) = 4{x^2} - 12x + 10;\,\,\left[ {1,2} \right] \cr & f\left( x \right){\text{ is }}\,{\text{continuous }}\,{\text{and}}\,{\text{differentiable everywhere}}{\text{, the }} \cr & {\text{absolute}}\,{\text{ extrema }}\,{\text{must }}\,{\text{occur}}\,{\text{ }}\,{\text{either }}\,{\text{at }}\,{\text{endpoints }}\,{\text{of }} \cr & {\text{the interval or at solutions to the equation }}f'\left( x \right) = 0{\text{ in}} \cr & {\text{the open interval }}\left( {1,2} \right){\text{.}} \cr & \cr & {\text{Differentiating }}f\left( x \right) \cr & f'\left( x \right) = 8x - 12 \cr & {\text{setting }}f'\left( x \right) = 0 \cr & 8x - 12 = 0 \cr & 8x = 12 \cr & x = \frac{3}{2} \cr & {\text{Thus}}{\text{, there is a stationary point at }}x = \frac{3}{2}. \cr & {\text{Evaluating }}f\left( x \right){\text{ at the endpoints}}{\text{, }}x = 1,{\text{ }}x = 2{\text{ and }}x = \frac{3}{2} \cr & f\left( 1 \right) = 4{\left( 1 \right)^2} - 12\left( 1 \right) + 10 = 2 \cr & f\left( {3/2} \right) = 4{\left( {3/2} \right)^2} - 12\left( {3/2} \right) + 10 = 1 \cr & f\left( 2 \right) = 4{\left( 2 \right)^2} - 12\left( 2 \right) + 10 = 2 \cr & {\text{Then we can conclude that the absolute minimum of }}f\left( x \right){\text{ on}} \cr & {\text{the interval }}\left[ {1,2} \right]{\text{ is 1 at }}x = \frac{3}{2}.{\text{ And the absolute maximum is}} \cr & 2{\text{ at }}x = 1{\text{ and }}x = 2\,\,\left( {{\text{Endpoints}}} \right). \cr & \cr & {\text{max}} = 2{\text{ at }}x = 1,2\,\,\, \cr & {\text{min}} = 1{\text{ at }}x = \frac{3}{2}\,\,\,\,\,\,\, \cr} $$