Chapter 4 - Section 4.1 - Maximum and Minimum Values - 4.1 Exercises - Page 287: 47

$x = \frac{1}{{\sqrt e }}$

$$\eqalign{ & g\left( x \right) = {x^2}\ln x \cr & {\text{The domain of the function is }}\left( {0,\infty } \right) \cr & {\text{Differentiate with respect to }}x \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^2}\ln x} \right] \cr & {\text{Using the product rule for derivatives}} \cr & f'\left( x \right) = {x^2}\frac{d}{{dx}}\left[ {\ln x} \right] + \ln x\frac{d}{{dx}}\left[ {{x^2}} \right] \cr & {\text{Computing derivatives and simplify}} \cr & f'\left( x \right) = {x^2}\left( {\frac{1}{x}} \right) + \ln x\left( {2x} \right) \cr & f'\left( x \right) = x + 2x\ln x \cr & {\text{Factoring}} \cr & f'\left( x \right) = x\left( {1 + 2\ln x} \right) \cr & {\text{The derivative does not exist for }}x \leq 0 \cr & {\text{Find the values of }}x{\text{ where }}f'\left( x \right) = 0 \cr & x\left( {1 + 2\ln x} \right) = 0 \cr & x = 0, \cr & or \cr & 1 + 2\ln x = 0 \cr & \ln x = - \frac{1}{2} \cr & x = {e^{ - 1/2}} \cr & x = \frac{1}{{\sqrt e }} \cr & x = 0{\text{ is not in the domain of the function}}{\text{, so the only }} \cr & {\text{critical point is}} \cr & x = \frac{1}{{\sqrt e }} \cr} $$