Chapter 4 - Applications of the Derivative - 4.2 What Derivatives Tell Us - 4.2 Exercises - Page 257: 75

$$\eqalign{ & {\text{Critical points: }}x = 6 \cr & {\text{local minimum at }}x = 6 \cr} $$

$$\eqalign{ & f\left( x \right) = {e^x}\left( {x - 7} \right) \cr & {\text{Calculate the first derivative }} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{e^x}\left( {x - 7} \right)} \right] \cr & f'\left( x \right) = {e^x} + {e^x}\left( {x - 7} \right) \cr & f'\left( x \right) = {e^x}\left( {1 + x - 7} \right) \cr & f'\left( x \right) = {e^x}\left( {x - 6} \right) \cr & {\text{Find the critical point}}{\text{, set }}f'\left( x \right) = 0 \cr & {e^x}\left( {x - 6} \right) = 0 \cr & {\text{The critical points is }}x = 6 \cr & \cr & {\text{Calculate the second derivative }} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {{e^x}\left( {x - 6} \right)} \right] \cr & f''\left( x \right) = {e^x} + {e^x}\left( {x - 6} \right) \cr & f''\left( x \right) = {e^x}\left( {1 + x - 6} \right) \cr & f''\left( x \right) = {e^x}\left( {x - 5} \right) \cr & \cr & {\text{Using the second test derivative into the critical points}} \cr & {\text{Evaluate }}f''\left( 6 \right) \cr & f''\left( 6 \right) = {e^6}\left( {6 - 5} \right) \cr & f''\left( 6 \right) = {e^6} \cr & f''\left( 6 \right) > 0,{\text{ then }}f\left( x \right){\text{ has a local minimun at }}x = 6 \cr & \cr & {\text{Critical points: }}x = 6 \cr & {\text{local minimum at }}x = 6 \cr} $$