Question

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

Answer 1

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]}