What is the f(x)=(e^{-2x}(-e^x+1))/((1+e^{-x))^3} ?

The f(x)=(e^{-2x}(-e^x+1))/((1+e^{-x))^3} is

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

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

Domain : - \infty < x < \infty

Range : - \frac{1}{6 3} \leq f (x) \leq \frac{1}{6 3}

X Intercepts : (0, 0), Y Intercepts : (0, 0)

Asymptotes : Horizontal y = 0

Extreme Points : Maximum (ln (2 - 3), \frac{1}{6 3}), Minimum (ln (2 + 3), - \frac{1}{6 3})

Interval Notation

Domain : (- \infty, \infty)

Range : [- \frac{1}{6 3}, \frac{1}{6 3}]

Solution steps

Domain of

\frac{e ^{- 2 x} ( - e ^{x} + 1 )}{( 1 + e ^{- x} ) ^{3}} : - \infty < x < \infty

Range of

\frac{e ^{- 2 x} ( - e ^{x} + 1 )}{( 1 + e ^{- x} ) ^{3}} : - \frac{1}{6 3} \leq f (x) \leq \frac{1}{6 3}

Axis interception points of

\frac{e ^{- 2 x} ( - e ^{x} + 1 )}{( 1 + e ^{- x} ) ^{3}} :

X Intercepts

: (0, 0),

Y Intercepts

: (0, 0)

Asymptotes of

\frac{e ^{- 2 x} ( - e ^{x} + 1 )}{( 1 + e ^{- x} ) ^{3}} :

Horizontal

: y = 0

Extreme Points of

\frac{e ^{- 2 x} ( - e ^{x} + 1 )}{( 1 + e ^{- x} ) ^{3}} :

Maximum

(ln (2 - 3), \frac{1}{6 3}),

Minimum

(ln (2 + 3), - \frac{1}{6 3})

f (x) = \frac{3 x + 7}{5 x - 2}

f (x) = \frac{2 x + 5}{3}

f (m) = m^{2} - 729

y = ln (\frac{1 + x}{1 - x})

y = 3 tan (π x^{2})

What is the f(x)=(e^{-2x}(-e^x+1))/((1+e^{-x))^3} ?
The f(x)=(e^{-2x}(-e^x+1))/((1+e^{-x))^3} is