What's are the critical numbers the function f(x) = x虏lnx?

I got it down to the where the derivative is 0=x+2xlnx; then 陆=lnx.





Any help would be muchly appreciated.What's are the critical numbers the function f(x) = x虏lnx?
f(x) = x虏lnx


f'(x) = 2xlnx + x虏/x


f'(x) = 2xlnx + x


f'(x) = x(2lnx + 1)





Critical values are at f'(x) = 0. The impled solutions are:





0 = x(2lnx + 1)


x = 0 or 2lnx + 1 = 0


x = 0 or 2lnx = 鈥?


x = 0 or lnx = 鈥韭?br>

x = 0 or x = e鈥韭?





Since lnx is undefined for x=0, only the x = e鈥韭?root is valid.What's are the critical numbers the function f(x) = x虏lnx?
I think you mean lnx=-1/2. This means that x=e^(-1/2).