Chapter 4 - Applications of the Derivative - 4.2 What Derivatives Tell Us - 4.2 Exercises - Page 257: 76

$$\eqalign{ & {\text{Critical points: }}x = \frac{{5 + \sqrt {101} }}{2}{\text{ and }}x = \frac{{5 - \sqrt {101} }}{2} \cr & {\text{local minimum at }}x = \frac{{5 + \sqrt {101} }}{2} \cr & {\text{local maximum at }}x = \frac{{5 - \sqrt {101} }}{2} \cr} $$

$$\eqalign{ & f\left( x \right) = {e^x}\left( {{x^2} - 7x - 12} \right) \cr & {\text{Calculate the first derivative }} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{e^x}\left( {{x^2} - 7x - 12} \right)} \right] \cr & f'\left( x \right) = {e^x}\left( {2x - 7} \right) + \left( {{x^2} - 7x - 12} \right){e^x} \cr & f'\left( x \right) = {e^x}\left( {2x - 7 + {x^2} - 7x - 12} \right) \cr & f'\left( x \right) = {e^x}\left( {{x^2} - 5x - 19} \right) \cr & {\text{Find the critical point}}{\text{, set }}f'\left( x \right) = 0 \cr & {e^x}\left( {{x^2} - 5x - 19} \right) = 0 \cr & {x^2} - 5x - 19 = 0 \cr & {\text{By the Quadratic Formula we obtain}} \cr & {x_1} = \frac{{5 + \sqrt {101} }}{2}{\text{ and }}{x_2} = \frac{{5 - \sqrt {101} }}{2} \cr & {\text{The critical points are }}x = \frac{{5 + \sqrt {101} }}{2}{\text{ and }}x = \frac{{5 - \sqrt {101} }}{2} \cr & \cr & {\text{Calculate the second derivative }} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {{e^x}\left( {{x^2} - 5x - 19} \right)} \right] \cr & f''\left( x \right) = {e^x}\left( {2x - 5} \right) + {e^x}\left( {{x^2} - 5x - 19} \right) \cr & f''\left( x \right) = {e^x}\left( {2x - 5 + {x^2} - 5x - 19} \right) \cr & f''\left( x \right) = {e^x}\left( {{x^2} - 3x - 24} \right) \cr & \cr & {\text{Using the second test derivative into the critical points}} \cr & {\text{Evaluate }}f''\left( {\frac{{5 + \sqrt {101} }}{2}} \right) \cr & f''\left( {\frac{{5 + \sqrt {101} }}{2}} \right) = {e^{\frac{{5 + \sqrt {101} }}{2}}}\left( {{{\left( {\frac{{5 + \sqrt {101} }}{2}} \right)}^2} - 3\left( {\frac{{5 + \sqrt {101} }}{2}} \right) - 24} \right) \cr & {\text{By using a calculator we obtain}} \cr & f''\left( {\frac{{5 + \sqrt {101} }}{2}} \right) \approx 18629.433 \cr & f''\left( {\frac{{5 + \sqrt {101} }}{2}} \right) > 0,{\text{ then }}f\left( x \right){\text{ has a local minimun at }}x = \frac{{5 + \sqrt {101} }}{2} \cr & \cr & {\text{Evaluate }}f''\left( {\frac{{5 - \sqrt {101} }}{2}} \right) \cr & f''\left( {\frac{{5 - \sqrt {101} }}{2}} \right) = {e^{\frac{{5 + \sqrt {101} }}{2}}}\left( {{{\left( {\frac{{5 - \sqrt {101} }}{2}} \right)}^2} - 3\left( {\frac{{5 - \sqrt {101} }}{2}} \right) - 24} \right) \cr & {\text{By using a calculator we obtain}} \cr & f''\left( {\frac{{5 - \sqrt {101} }}{2}} \right) \approx - 0.8046 \cr & f''\left( {\frac{{5 - \sqrt {101} }}{2}} \right) < 0,{\text{ then }}f\left( x \right){\text{ has a local maximun at }}x = \frac{{5 - \sqrt {101} }}{2} \cr & \cr & {\text{Critical points: }}x = \frac{{5 + \sqrt {101} }}{2}{\text{ and }}x = \frac{{5 - \sqrt {101} }}{2} \cr & {\text{local minimum at }}x = \frac{{5 + \sqrt {101} }}{2} \cr & {\text{local maximum at }}x = \frac{{5 - \sqrt {101} }}{2} \cr} $$