I can't tell if your equation is listed correctly, but here is just a guess.
I most likely am wrong here, but it looks like your equation is a decreasing
function that really does not have a Maximum value, except at points of
discontinuity when you have 1/0.

Your first term is:
1/(1-E^(z1-x))

This is a decreasing function as x tends towards infinity. When x = z1, you
have 1/0 which is a discontinuity (value tends to infinity). Just above x,
the value is a large number. The limit as x goes to infinity is 1.
Each of the other terms (1/(1-E^(z2-x))) have a limit of 0 as x tends
towards infinity. (With jumps as x approaches z2). So, my guess is that it
is hard to tell what you mean by "Maximum" value with the given equation.

--
Dana DeLouis
Using Windows XP & Office XP
= = = = = = = = = = = = = = = = =


"John Hall" wrote in message
m...
I have scoured the newsgroups and web and can't seem to find an anwser
to this.

I need to find the max of an equation. where the equation is in this
form 1/(1-e^(z1-x))*(1-1/(1+e^(z2-x))*.........*(1-1/(1+e^(zn-x)).

Where there are n number of z values that are know. Meaning that z is
not a variable but different constants.

I can calculate the slope at a specific value of x but want to solve
for x where F'(x) = 0.

Any help?

Thanks,
John