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

1 Answer
Jul 17, 2016

Answer:

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)#