Chapter 4 - Applications of the Derivative - 4.2 What Derivatives Tell Us - 4.2 Exercises - Page 257: 81

$$\eqalign{ & {\text{Critical points: }}x = {e^5} \cr & {\text{local minimum at }}x = {e^5} \cr} $$

$$\eqalign{ & f\left( x \right) = 2{x^2}\ln x - 11{x^2} \cr & {\text{Calculate the first derivative }} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {2{x^2}\ln x - 11{x^2}} \right] \cr & f'\left( x \right) = 2{x^2}\left( {\frac{1}{x}} \right) + 4x\ln x - 22x \cr & f'\left( x \right) = 2x + 4x\ln x - 22x \cr & f'\left( x \right) = 4x\ln x - 20x \cr & {\text{Find the critical point}}{\text{, set }}f'\left( x \right) = 0 \cr & 4x\ln x - 20x = 0 \cr & 4x\left( {\ln x - 5} \right) = 0 \cr & 4x = 0{\text{ or }}\ln x - 5 = 0 \cr & x = 0,\,\,\,\,\,x = {e^5} \cr & {\text{The critical points are }}x = 0{\text{ and }}x = {e^5} \cr & x = 0{\text{ is not in the domain of }}f\left( x \right) = 2{x^2}\ln x - 11{x^2} \cr & {\text{Then the critical point is }}x = {e^5} \cr & \cr & {\text{Calculate the second derivative }} \cr & f''\left( x \right) = \frac{d}{{dx}}\left[ {4x\ln x - 20x} \right] \cr & f''\left( x \right) = 4x\left( {\frac{1}{x}} \right) + 4\ln x - 20 \cr & f''\left( x \right) = 4\ln x - 16 \cr & \cr & {\text{Using the second test derivative into the critical point}} \cr & {\text{Evaluate }}f''\left( {{e^5}} \right) \cr & f''\left( {{e^5}} \right) = 4\ln {e^5} - 16 \cr & f''\left( {{e^5}} \right) = 20 - 16 \cr & f''\left( {{e^5}} \right) = 4 \cr & f''\left( {{e^5}} \right) > 0,{\text{ then }}f\left( x \right){\text{ has a local minimun at }}x = {e^5} \cr & \cr & {\text{Critical points: }}x = {e^5} \cr & {\text{local minimum at }}x = {e^5} \cr} $$