How do you find the Vertical, Horizontal, and Oblique Asymptote given (e^x)/(1+e^x)?

1 Answer
Jul 17, 2016

The function has horizontal asymptotes at f(x)=1 and f(x)=0 and a point of inflection at the point (0,1/2)

Explanation:

f(x)=e^x/(1+e^x)

This can also be written as:
f(x)=1/(1/e^x+1)

Now consider f(x) as x→+∞ and x→−∞

lim_"x->+oo" f(x)=1/(0+1)=1

lim_"x->-oo" f(x)= lim_"x->+oo" 1/(e^x+1)=0

Consider the graph of f(x) below. By inspection we can see that limiting values arrived at above form horizontal asymptotes.

Evaluating f(0) as a point of interest.

f(0) = e^0/(1+e^0) = 1/(1+1) = 1/2

Notice that the curvature of f(x) changes at (0,1/2) from concave downward to concave upward. This indicates a point of inflection.

graph{e^x/(1+e^x) [-3.08, 3.08, -1.538, 1.54]}

Hence:

The function has horizontal asymptotes at f(x)=1 and f(x)=0 and a point of inflection at the point (0,1/2)