JustAnotherUser
Nov 2nd, 2005, 05:32 PM
Title pretty much said it. Maybe I'm just being a complete fool and its obvious.
For the precise problem:
Find stationary points: f(x) = (x^3)/3 + e^(-x)
f ' (x) = 0 = x^2 - e^(-x)
e^(-x) = x^2
I could use a numerical method (eg iteration), but surely theres a way to get the answer algebraically?
Thanks
For the precise problem:
Find stationary points: f(x) = (x^3)/3 + e^(-x)
f ' (x) = 0 = x^2 - e^(-x)
e^(-x) = x^2
I could use a numerical method (eg iteration), but surely theres a way to get the answer algebraically?
Thanks