What is the range of the function $f (x) = \frac{1}{1 + e^{- x}}$ $f(x)=1/(1+e^-x)$ ?

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Answer 1 · 2017-08-10T13:18:54

The range is $(0, 1)$ $(0,1)$

The function is $f (x) = \frac{1}{1 + e^{- x}}$ $f(x)=1/(1+e^-x)$

The domain of the functions $e^{x}$ $e^x$ and $e^{- x}$ $e^-x$ 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]}

What is the range of the function f(x)=1/(1+e^-x)f(x)=11+e−xf(x)=1/(1+e^-x)?