Answer
$x= 1/e$ is a local minimum
Work Step by Step
Given
$$ g(x)=x\ln x$$
Since
\begin{align*} g^{\prime}(x)&= \frac{d}{dx}\left(x\right)\ln \left(x\right)+\frac{d}{dx}\left(\ln \left(x\right)\right)x\\ &=1+\ln x \end{align*}
Then $g(x)$ has critical points when \begin{align*} g'(x)&=0\\ 1+\ln x&= 0 \end{align*} Then the critical point is $x= 1/e$. Now, we check the sign of the second derivative at $x=1/ e
$ \begin{align*} g''(x) &=\frac{1}{x} \end{align*}
Hence
$$g''(1/e)= e>0 $$
Then $x= 1/e$ is a local minimum.
