Find the Inverse y=(1-e^(-x))/(1+e^(-x))

$y = \frac{1 - e^{- x}}{1 + e^{- x}}$

Interchange the variables.

$x = \frac{1 - e^{- y}}{1 + e^{- y}}$

Solve for $y$ .

Rewrite the equation as $\frac{1 - e^{- y}}{1 + e^{- y}} = x$ .

$\frac{1 - e^{- y}}{1 + e^{- y}} = x$

Multiply both sides by $1 + e^{- y}$ .

$\frac{1 - e^{- y}}{1 + e^{- y}} (1 + e^{- y}) = x (1 + e^{- y})$

Simplify.

Simplify the left side.

Cancel the common factor of $1 + e^{- y}$ .

Cancel the common factor.

$\frac{1 - e^{- y}}{1 + e^{- y}} (1 + e^{- y}) = x (1 + e^{- y})$

Rewrite the expression.

$1 - e^{- y} = x (1 + e^{- y})$

Simplify the right side.

Simplify $x (1 + e^{- y})$ .

Apply the distributive property.

$1 - e^{- y} = x \cdot 1 + x e^{- y}$

Multiply $x$ by $1$ .

$1 - e^{- y} = x + x e^{- y}$

Solve for $y$ .

Subtract $x e^{- y}$ from both sides of the equation.

$1 - e^{- y} - x e^{- y} = x$

Subtract $1$ from both sides of the equation.

$- e^{- y} - x e^{- y} = x - 1$

Factor $e^{- y}$ out of $- e^{- y} - x e^{- y}$ .

Factor $e^{- y}$ out of $- e^{- y}$ .

$e^{- y} \cdot - 1 - x e^{- y} = x - 1$

Factor $e^{- y}$ out of $- x e^{- y}$ .

$e^{- y} \cdot - 1 + e^{- y} (- x) = x - 1$

Factor $e^{- y}$ out of $e^{- y} \cdot - 1 + e^{- y} (- x)$ .

$e^{- y} (- 1 - x) = x - 1$

Divide each term in $e^{- y} (- 1 - x) = x - 1$ by $- 1 - x$ and simplify.

Divide each term in $e^{- y} (- 1 - x) = x - 1$ by $- 1 - x$ .

$\frac{e^{- y} (- 1 - x)}{- 1 - x} = \frac{x}{- 1 - x} + \frac{- 1}{- 1 - x}$

Simplify the left side.

Cancel the common factor of $- 1 - x$ .

Cancel the common factor.

$\frac{e^{- y} (- 1 - x)}{- 1 - x} = \frac{x}{- 1 - x} + \frac{- 1}{- 1 - x}$

Divide $e^{- y}$ by $1$ .

$e^{- y} = \frac{x}{- 1 - x} + \frac{- 1}{- 1 - x}$

Simplify the right side.

Combine the numerators over the common denominator.

$e^{- y} = \frac{x - 1}{- 1 - x}$

Rewrite $- 1$ as $- 1 (1)$ .

$e^{- y} = \frac{x - 1}{- 1 (1) - x}$

Factor $- 1$ out of $- x$ .

$e^{- y} = \frac{x - 1}{- 1 (1) - (x)}$

Factor $- 1$ out of $- 1 (1) - (x)$ .

$e^{- y} = \frac{x - 1}{- 1 (1 + x)}$

Move the negative in front of the fraction.

$e^{- y} = - \frac{x - 1}{1 + x}$

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

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

Expand the left side.

Expand $\ln (e^{- y})$ by moving $- y$ outside the logarithm.

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

The natural logarithm of $e$ is $1$ .

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

Multiply $- 1$ by $1$ .

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

Simplify the right side.

Simplify $\ln (- \frac{x - 1}{1 + x})$ .

Split the fraction $\frac{x - 1}{1 + x}$ into two fractions.

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

Move the negative in front of the fraction.

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

Apply the distributive property.

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

Multiply $- - \frac{1}{1 + x}$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $\frac{1}{1 + x}$ by $1$ .

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

Divide each term in $- y = \ln (- \frac{x}{1 + x} + \frac{1}{1 + x})$ by $- 1$ and simplify.

Divide each term in $- y = \ln (- \frac{x}{1 + x} + \frac{1}{1 + x})$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Move the negative one from the denominator of $\frac{\ln (- \frac{x}{1 + x} + \frac{1}{1 + x})}{- 1}$ .

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

Rewrite $- 1 \cdot \ln (- \frac{x}{1 + x} + \frac{1}{1 + x})$ as $- \ln (- \frac{x}{1 + x} + \frac{1}{1 + x})$ .

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

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

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

Verify if $f^{- 1} (x) = - \ln (- \frac{x}{1 + x} + \frac{1}{1 + x})$ is the inverse of $f (x) = \frac{1 - e^{- x}}{1 + e^{- x}}$ .

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Evaluate $f^{- 1} (f (x))$ .

Set up the composite result function.

$f^{- 1} (f (x))$

Evaluate $f^{- 1} (\frac{1 - e^{- x}}{1 + e^{- x}})$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Simplify terms.

Remove parentheses.

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

Combine the numerators over the common denominator.

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

Write $1$ as a fraction with a common denominator.

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Apply the distributive property.

$f^{- 1} (\frac{1 - e^{- x}}{1 + e^{- x}}) = - \ln (\frac{\frac{- 1 \cdot 1 + e^{- x} + 1 + e^{- x}}{1 + e^{- x}}}{1 + \frac{1 - e^{- x}}{1 + e^{- x}}})$

Multiply $- 1$ by $1$ .

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

Multiply $- - e^{- x}$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $e^{- x}$ by $1$ .

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

Add $- 1$ and $1$ .

$f^{- 1} (\frac{1 - e^{- x}}{1 + e^{- x}}) = - \ln (\frac{\frac{0 + e^{- x} + e^{- x}}{1 + e^{- x}}}{1 + \frac{1 - e^{- x}}{1 + e^{- x}}})$

Add $0$ and $e^{- x}$ .

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

Add $e^{- x}$ and $e^{- x}$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Simplify the denominator.

Write $1$ as a fraction with a common denominator.

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

Combine the numerators over the common denominator.

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

Rewrite $\frac{1 + e^{- x} + 1 - e^{- x}}{1 + e^{- x}}$ in a factored form.

Add $1$ and $1$ .

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

Subtract $e^{- x}$ from $e^{- x}$ .

$f^{- 1} (\frac{1 - e^{- x}}{1 + e^{- x}}) = - \ln (\frac{2 e^{- x}}{1 + e^{- x}} \cdot \frac{1}{\frac{2 + 0}{1 + e^{- x}}})$

Add $2$ and $0$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $\frac{1 + e^{- x}}{2}$ by $1$ .

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

Cancel the common factor of $2$ .

Factor $2$ out of $2 e^{- x}$ .

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

Cancel the common factor.

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

Rewrite the expression.

$f^{- 1} (\frac{1 - e^{- x}}{1 + e^{- x}}) = - \ln (\frac{e^{- x}}{1 + e^{- x}} \cdot (1 + e^{- x}))$

Cancel the common factor of $1 + e^{- x}$ .

Cancel the common factor.

$f^{- 1} (\frac{1 - e^{- x}}{1 + e^{- x}}) = - \ln (\frac{e^{- x}}{1 + e^{- x}} \cdot (1 + e^{- x}))$

Rewrite the expression.

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

Use logarithm rules to move $- x$ out of the exponent.

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

The natural logarithm of $e$ is $1$ .

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

Multiply $- 1$ by $1$ .

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

Evaluate $f (f^{- 1} (x))$ .

Set up the composite result function.

$f (f^{- 1} (x))$

Evaluate $f (- \ln (- \frac{x}{1 + x} + \frac{1}{1 + x}))$ by substituting in the value of $f^{- 1}$ into $f$ .

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

Simplify the numerator.

Combine the numerators over the common denominator.

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

Multiply $- - \ln (\frac{- x + 1}{1 + x})$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $\ln (\frac{- x + 1}{1 + x})$ by $1$ .

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

Exponentiation and log are inverse functions.

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

Write $1$ as a fraction with a common denominator.

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

Rewrite $\frac{x + 1 - (- x + 1)}{x + 1}$ in a factored form.

Apply the distributive property.

$f (- \ln (- \frac{x}{1 + x} + \frac{1}{1 + x})) = \frac{\frac{x + 1 + x - 1 \cdot 1}{x + 1}}{1 + e^{\ln (- \frac{x}{1 + x} + \frac{1}{1 + x})}}$

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

$f (- \ln (- \frac{x}{1 + x} + \frac{1}{1 + x})) = \frac{\frac{x + 1 + 1 x - 1 \cdot 1}{x + 1}}{1 + e^{\ln (- \frac{x}{1 + x} + \frac{1}{1 + x})}}$

Multiply $x$ by $1$ .

$f (- \ln (- \frac{x}{1 + x} + \frac{1}{1 + x})) = \frac{\frac{x + 1 + x - 1 \cdot 1}{x + 1}}{1 + e^{\ln (- \frac{x}{1 + x} + \frac{1}{1 + x})}}$

Multiply $- 1$ by $1$ .

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

Add $x$ and $x$ .

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

Subtract $1$ from $1$ .

$f (- \ln (- \frac{x}{1 + x} + \frac{1}{1 + x})) = \frac{\frac{2 x + 0}{x + 1}}{1 + e^{\ln (- \frac{x}{1 + x} + \frac{1}{1 + x})}}$

Add $2 x$ and $0$ .

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

Simplify the denominator.

Combine the numerators over the common denominator.

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

Multiply $- - \ln (\frac{- x + 1}{1 + x})$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $\ln (\frac{- x + 1}{1 + x})$ by $1$ .

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

Exponentiation and log are inverse functions.

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

Write $1$ as a fraction with a common denominator.

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

Rewrite $\frac{x - x + 1 + 1}{x + 1}$ in a factored form.

Subtract $x$ from $x$ .

$f (- \ln (- \frac{x}{1 + x} + \frac{1}{1 + x})) = \frac{\frac{2 x}{x + 1}}{\frac{0 + 1 + 1}{x + 1}}$

Add $0$ and $1$ .

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

Add $1$ and $1$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Cancel the common factor of $2$ .

Factor $2$ out of $2 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x + 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - \ln (- \frac{x}{1 + x} + \frac{1}{1 + x})$ is the inverse of $f (x) = \frac{1 - e^{- x}}{1 + e^{- x}}$ .

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

Find the Inverse y=(1-e^(-x))/(1+e^(-x))

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