# Math

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How can I find the asymptotes of [f(x) = 1/ (1+e^-x)]?

• Math -

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 y-intercept 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

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