What is the f(x)= 1/2 ln((1+x)/(1-x))-2x ?

The f(x)= 1/2 ln((1+x)/(1-x))-2x is

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f(x)= 1/2 ln((1+x)/(1-x))-2x

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

Domain : - 1 < x < 1

Range : - \infty < f (x) < \infty

Y Intercepts : (0, 0)

Asymptotes : Vertical x = - 1, x = 1

Extreme Points : Maximum (- \frac{1}{2}, \frac{1}{2} ln (3 - 22) + 2), Minimum (\frac{1}{2}, \frac{1}{2} ln (3 + 22) - 2)

Interval Notation

Domain : (- 1, 1)

Range : (- \infty, \infty)

Solution steps

Domain of

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

Range of

\frac{1}{2} ln (\frac{1 + x}{1 - x}) - 2 x : - \infty < f (x) < \infty

Axis interception points of

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

Y Intercepts

: (0, 0)

Asymptotes of

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

Vertical

: x = - 1, x = 1

Extreme Points of

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

Maximum

(- \frac{1}{2}, \frac{1}{2} ln (3 - 22) + 2),

Minimum

(\frac{1}{2}, \frac{1}{2} ln (3 + 22) - 2)

f (x) = x^{2} - x - 1

y = 6 x^{2} + x^{- 4}

f (x) = \frac{4}{x ^{2} - 25}

f (s) = e^{- s} - e^{- 2 s}

f (x) = x^{3} - 3 x^{2} + 5 x - 2