Arelative maximum of the function f(x)=(lnx)^2/x occurs
e^2
show work and explain how to come to the answer.

The solution to the answer is as follows:

 f(x) = [lnx]^2 / x 

d[f(x)]/dx = [x d(lnx)/dx-lnxd(x)/dx]/x62 
= (1 -lnx) x^2 

to have a min or max 
du/dxshould be = to 0 

(1-lnx)/x^2 = 0 

to happen that 
1-lnx = 0 
lnx = 1 
log_e x = 1 
e^(1) = x 
x= e