What is the range of the function #f(x)=1/(1+e^-x)#?

1 Answer
Aug 10, 2017

Answer:

The range is #(0,1)#

Explanation:

The function is #f(x)=1/(1+e^-x)#

The domain of the functions #e^x# and #e^-x# are #RR#

The domain of #f(x)# is #D_f(x)=RR#

We calculate the limits to #+-oo# of #f(x)#

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

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

Therefore,

The range is #(0,1)#

graph{1/(1+e^-x) [-7.9, 7.904, -3.95, 3.95]}