What is the extreme f(x)=(e^{-x})/((1+e^{-x))^2} ?

The extreme f(x)=(e^{-x})/((1+e^{-x))^2} is Maximum(0, 1/4)

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extreme f(x)=(e^{-x})/((1+e^{-x))^2}

extreme points

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

Maximum (0, \frac{1}{4})

Solution steps

f^{'} (x) = \frac{e ^{- 2 x} ( - e ^{x} + 1 )}{( 1 + e ^{- x} ) ^{3}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = 0

into

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

Maximum (0, \frac{1}{4})

f (x) = \frac{( ln ( x ) ) ^{2}}{x}

f (x) = \frac{x ^{4}}{4} - x^{3}

f (x) = - 4 x^{2} - 16 x

f (x) = x + y + 1

\frac{x ^{2} - 1}{x}

What is the extreme f(x)=(e^{-x})/((1+e^{-x))^2} ?
The extreme f(x)=(e^{-x})/((1+e^{-x))^2} is Maximum(0, 1/4)