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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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