## Calculus (3rd Edition)

$x= e^{1/3}$ is a local maximum.
Given $$g(x)=\frac{\ln x}{x^{3}}$$ Since \begin{align*} g^{\prime}(x)&= \frac{\frac{d}{dx}\left(\ln \left(x\right)\right)x^3-\frac{d}{dx}\left(x^3\right)\ln \left(x\right)}{\left(x^3\right)^2}\\ &=\frac{1-3\ln \left(x\right)}{x^4} \end{align*} Then $g(x)$ has critical points when \begin{align*} g'(x)&=0\\ \frac{1-3\ln \left(x\right)}{x^4}&= 0 \\ 1-3\ln \left(x\right)&=0 \end{align*} Then $x= e^{1/3}$ is a critical point. Now, we use the second derivative to check $x= e^{1/3}$ \begin{align*} g''(x) &=\frac{\frac{d}{dx}\left(1-3\ln \left(x\right)\right)x^4-\frac{d}{dx}\left(x^4\right)\left(1-3\ln \left(x\right)\right)}{\left(x^4\right)^2}\\ &= -\frac{-12\ln \left(x\right)+7}{x^5} \end{align*} Hence $$g''(e^{1/3})=\frac{-7+12 \ln e^{1 / 3}}{\left(e^{1 / 3}\right)^{5}}=-0.5666<0$$ Since the second derivative at the critical point is negative, we have a maximum. Thus, $x= e^{1/3}$ is a local maximum.