Question 1033076
{{{y=xlnx-x^2}}}  ==> y' = lnx + 1 - 2x  ==> y" = {{{1/x-2}}}

The critical values of the 2nd derivative are x = 0 and x = 1/2, but the function {{{y=xlnx-x^2}}} is undefined at x = 0, so there is no inflection point there.

Now in ({{{-infinity}}}, 0), y" < 0.
In (0, 1/2), y" > 0, and 
at (1/2, {{{infinity}}}), y" < 0.
(At x = 1/2, y" = 0.)

Hence there is only one inflection point at (1/2, {{{-ln2/2-1/4}}}).  (Even though there is a change in the direction of concavity from the interval ({{{-infinity}}}, 0) to (0, 1/2).)