# f(x)=e^((lnx)(1-lnx))

• Jan 16th 2013, 04:23 AM
Boo
f(x)=e^((lnx)(1-lnx))
Dear All!

$\displaystyle f(x)=e^{lnx(1-lnx)}$

How do we get local masimums and points of inflection?

I tried:

$\displaystyle e^{lnx}=x$

$\displaystyle f(x)=x^{(1-lnx)}$...

I got $\displaystyle f(x)=\frac{x}{x^{lnx}}$

what now??? many thanks!
• Jan 16th 2013, 08:15 AM
Barioth
Re: f(x)=e^((lnx)(1-lnx))
Do find the maximums and points of inflection I would derivate it now, have you seen derivate yet?

edit: It would been ezier (for me..) to derivate the first equation that you gave. but thats just how I leanred my derivate :)
• Jan 16th 2013, 12:11 PM
topsquark
Re: f(x)=e^((lnx)(1-lnx))
Quote:

Originally Posted by Barioth
Do find the maximums and points of inflection I would derivate it now, have you seen derivate yet?

edit: It would been ezier (for me..) to derivate the first equation that you gave. but thats just how I leanred my derivate :)

I agree totally with Barioth. It's easier in this case to take the derivative of the function before "simplifying" it.

-Dan