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

**Boo** · January 16th 2013, 04:23 AM

Dear All!
please, if we have

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

How do we get local masimums and points of inflection?

I tried:

$e^{lnx}=x$

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

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

what now??? many thanks!

**Barioth** · January 16th 2013, 08:15 AM

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

**topsquark** · January 16th 2013, 12:11 PM

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

Thread: f(x)=e^((lnx)(1-lnx))

LinkBack

Thread Tools

Display

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

Re: f(x)=e^((lnx)(1-lnx))

Re: f(x)=e^((lnx)(1-lnx))

Search Tags