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July 1, 2015

July 1, 2015

Posted by **009** on Sunday, February 3, 2013 at 6:47am.

- Math -
**Reiny**, Sunday, February 3, 2013 at 8:18amf(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