If I interpret your problem correctly you have
for x=0.5
and you want
for x=a
so choose:
or
Hi!
I've been at this all day and can't seem to find it...
I've got a function f(x) = 1/x + 1/(1-x)
Its minimum is x = 0.5 and f(x) = 4
The f(x) value doesn't really matter, but the function shape does, as it reaches infinity around 0 and 1. Hence, x is in ]0;1[
However, i'm trying to change the function so as to accept an "a" parameter, "a" being in ]0;1[ as well, such that the minimum isn't in x=0.5 but in x=a.
i.e.
inf(f(x,a)) = f(a,a)
I've been going for something like f(x,a) = a/x + 1/(1-x) but here "a" goes from 0+ to infinity...
Any ideas?
Thanks in advance!