Show that $\displaystyle x\ln x$ has only one stationary value, and find it.

I differentiated the equation:

$\displaystyle \frac{d}{dx}(x\ln x) = x(\frac{1}{x})+ \ln x(1) = 1+\ln x$

Then

$\displaystyle 1+\ln x=0$

$\displaystyle \ln x=-1$

$\displaystyle x=e^{-1}=\frac{1}{e}$

Am i correct? Because the answer is supposed to be $\displaystyle -\frac{1}{e}$