# local maxima and minima

• Oct 16th 2009, 07:17 PM
fvaras89
local maxima and minima
How would you find the local maxima and minima for y= xe^-x

Im not sure how to do it and would appreciate very much if someone could help me.
• Oct 16th 2009, 07:21 PM
TKHunny
Generally, you would examine it for continuity, then find the first derivative. Where the first derivative exists and is zero, this is likely to be a local max or min. That's only part of the story, of course. You tell me the rest.

Can you find dy/dx? It will take the Product Rule.
• Oct 16th 2009, 07:24 PM
fvaras89
so y'= (e^-x)(-x e^-x)
= e^-x(1-x)

so do u let y'= 0?

if so, i put e^-x(1-x)=0 and thats when i got stuck

so what do i do next?
• Oct 16th 2009, 09:17 PM
mr fantastic
Quote:

Originally Posted by fvaras89
so y'= (e^-x)(-x e^-x)
= e^-x(1-x)

so do u let y'= 0?

if so, i put e^-x(1-x)=0 and thats when i got stuck

so what do i do next?

Well, obviously either \$\displaystyle e^{-x} = 0\$ or \$\displaystyle (1 - x) = 0\$ .... (Hint: \$\displaystyle e^{-x} = 0\$ has no real solution ....)
• Oct 17th 2009, 07:19 AM
TKHunny
Personally, I'm a little curious how how got from here

Quote:

Originally Posted by fvaras89
so y'= (e^-x)(-x e^-x)

to here

Quote:

= e^-x(1-x)
Seems to be some notational difficulty requiring magic to get to the right result. Please be more careful. You WILL unconfuse yourself.
• Oct 17th 2009, 08:15 PM
fvaras89
I just thought u could take out the common factor like i did before or would there be another way to present it?
• Oct 18th 2009, 05:50 PM
TKHunny
I think it would help if the addition actually appeared in the first equation. Thus my encouragement to be more careful. I have little doubt that you had the correct thing in your mind, you just didn't write it.