Posted by 009 on .
How can I find the asymptotes of [f(x) = 1/ (1+e^x)]?

Math 
Reiny,
f(x) = 1/(1 + e^x)
check for vertical asymptotes,
to have those, the denominator has to be zero, so
1+ e^x = 0
e^x = 1
or 1/e^x = 1/1
e^x = 1
no solution , > if we take ln of both sides to solve for x, we would have to take ln(1) which is undefined
Therefore, no vertical asymptotes
f(0) = 1/(1+1) = 1/2, so we have a yintercept of 1/2
let's look at far right and far left.
as x > +∞
e^x > 0
and 1/(1+e^x) > 1
as x > ∞
e^x > ∞
and 1/(1+e^x)  1/very large > 0
as x> +∞ , f(x) >1
as x > ∞ , f(x) > 0