Chapter 4 - Applications of the Derivative - 4.2 What Derivatives Tell Us - 4.2 Exercises - Page 257: 72

$$\eqalign{ & {\text{Critical points: }}x = 0{\text{ and}}\,\,\,x = 4 \cr & {\text{local maximum at }}x = 4 \cr & {\text{local minimum at }}x = 0 \cr} $$

$$\eqalign{ & f\left( x \right) = 6{x^2} - {x^3} \cr & {\text{Calculate the first derivative }} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {6{x^2} - {x^3}} \right] \cr & f'\left( x \right) = 12x - 3{x^2} \cr & {\text{Find the critical point}}{\text{, set }}f'\left( x \right) = 0 \cr & 12x - 3{x^2} = 0 \cr & 3x\left( {4 - x} \right) = 0 \cr & {\text{The critical points are }}x = 0,\,\,\,x = 4 \cr & \cr & {\text{Calculate the second derivative }} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {12x - 3{x^2}} \right] \cr & f''\left( x \right) = 12 - 6x \cr & {\text{Using the second test derivative into the critical points}} \cr & {\text{Evaluate }}f''\left( 0 \right) \cr & f''\left( 0 \right) = 12 - 6\left( 0 \right) \cr & f''\left( 0 \right) = 12 \cr & f''\left( 0 \right) > 0,{\text{ then }}f\left( x \right){\text{ has a local minimum at }}x = 0 \cr & \cr & {\text{Evaluate }}f''\left( 4 \right) \cr & f''\left( 4 \right) = 12 - 6\left( 4 \right) \cr & f''\left( 4 \right) = - 12 \cr & f''\left( 4 \right) < 0,{\text{ then }}f\left( x \right){\text{ has a local maximum at }}x = 4 \cr & \cr & {\text{Critical points: }}x = 0{\text{ and}}\,\,\,x = 4 \cr & {\text{local maximum at }}x = 4 \cr & {\text{local minimum at }}x = 0 \cr} $$