Inflexion point for y = x loge x-x^2 (1 Viewer)

[Thylacine]

Having problems with this one.
Find the point of inflexion on the curve y = x log_e x-x²

Can anyone help?

 

[Drongoski]

Pt of inflexion:

 

[funnytomato]

y=xlnx-x^2
y'=lnx+1-2x
y''=1/x-2 =0 at pt of inf, so x=1/2

and sub into y , what you get should be the same as Drongoski's answer

of course you should verify the change of concavity

 

[Thylacine]

Thankyou both for your help.
I realise now I was treating the x - x^2 as though it was in brackets.
ie loge(x-x^2).
This caused major problems when trying to find the second derivative.
Thanks again.