What is the range of (e^{-x})/((1+e^{-x))^2} ?

The range of (e^{-x})/((1+e^{-x))^2} is 0<f(x)<= 1/4

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range of (e^{-x})/((1+e^{-x))^2}

range

\frac{e ^{- x}}{( 1 + e ^{- x} ) ^{2}}

0 < f (x) \leq \frac{1}{4}

Interval Notation

(0, \frac{1}{4}]

Solution steps

The function range is the combined domain of the inverse functions

Find the inverse function(s) of:

\frac{e ^{- x}}{( 1 + e ^{- x} ) ^{2}}

Inverse of

\frac{e ^{- x}}{( 1 + e ^{- x} ) ^{2}} : - ln (\frac{- 2 x + 1 + - 4 x + 1}{2 x}), - ln (\frac{- 2 x + 1 - - 4 x + 1}{2 x})

- ln (\frac{- 2 x + 1 + - 4 x + 1}{2 x}), - ln (\frac{- 2 x + 1 - - 4 x + 1}{2 x})

Find the domain of each inverse function

Domain of

- ln (\frac{- 2 x + 1 + - 4 x + 1}{2 x}) : 0 < x \leq \frac{1}{4}

Domain of

- ln (\frac{- 2 x + 1 - - 4 x + 1}{2 x}) : 0 < x \leq \frac{1}{4}

Combine the intervals

0 < f (x) \leq \frac{1}{4}

2 x + 24

f (x) = \frac{6 - x}{x + 7}

f (x) = (7 - 2 x) e^{x}

5 x - 4 y = 12

f (x) = 5 x - 1

What is the range of (e^{-x})/((1+e^{-x))^2} ?
The range of (e^{-x})/((1+e^{-x))^2} is 0<f(x)<= 1/4