Are there inflection points in the function f(x) = x ln(x)?

[Greywolfe1982]

Finding inflection points of x(ln(x)) (or "Why is WolframAlpha telling me this?")

The question is finding the points of inflection for f(x)=xlnx. As far as I know, it has none, and is concave up for the entire function within its domain, x>0. When I look at the graph on WolframAlpha though, it's telling me there's a real part to the function for x<=0. Would somebody care to explain what I'm missing? Or is it a glitch on WolframAlpha's part?

Here's the output:
http://www.wolframalpha.com/input/?i=x(lnx)

 

[Kevin_Axion]

Does this help: http://www.wolframalpha.com/input/?i=x(lnx)+inflection+point.

 

[Kevin_Axion]

Does this help: http://www.wolframalpha.com/input/?i=x(lnx)+inflection+point.

EDIT: I realize now that my answer was pointless.

 

[Kevin_Axion]

It's a complex number. Therefore it has real and imaginary parts when x&lt;0.