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

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

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

extreme points

\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, y) = e^{x^{2} + y^{2} - 2 x}

4 x^{3} - 12 x^{2}

3 x y - x^{2} y - x y^{2}

f (x) = x^{3} - 4 x^{2} + 4 x - 1

f (x, y) = \frac{1}{3} x^{3} + \frac{1}{3} y^{3} - x - y + 10

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