Algebra Examples

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

Step 1

Write $f (x) = \frac{1 - e^{- x}}{1 + e^{- x}}$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

Simplify.

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Step 3.3.1

Simplify the left side.

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Step 3.3.1.1

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

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Step 3.3.1.1.1

Cancel the common factor.

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

Step 3.3.1.1.2

Rewrite the expression.

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

Step 3.3.2

Simplify the right side.

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Step 3.3.2.1

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

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Step 3.3.2.1.1

Apply the distributive property.

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

Step 3.3.2.1.2

Multiply $x$ by $1$ .

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

Step 3.4

Solve for $y$ .

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Step 3.4.1

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

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

Step 3.4.2

Subtract $1$ from both sides of the equation.

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

Step 3.4.3

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

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Step 3.4.3.1

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

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

Step 3.4.3.2

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

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

Step 3.4.3.3

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

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

Step 3.4.4

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

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Step 3.4.4.1

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

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

Step 3.4.4.2

Simplify the left side.

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Step 3.4.4.2.1

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

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Step 3.4.4.2.1.1

Cancel the common factor.

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

Step 3.4.4.2.1.2

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

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

Step 3.4.4.3

Simplify the right side.

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Step 3.4.4.3.1

Combine the numerators over the common denominator.

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

Step 3.4.4.3.2

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

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

Step 3.4.4.3.3

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

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

Step 3.4.4.3.4

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

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

Step 3.4.4.3.5

Move the negative in front of the fraction.

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

Step 3.4.5

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

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

Step 3.4.6

Expand the left side.

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Step 3.4.6.1

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

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

Step 3.4.6.2

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

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

Step 3.4.6.3

Multiply $- 1$ by $1$ .

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

Step 3.4.7

Simplify the right side.

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Step 3.4.7.1

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

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Step 3.4.7.1.1

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

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

Step 3.4.7.1.2

Move the negative in front of the fraction.

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

Step 3.4.7.1.3

Apply the distributive property.

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

Step 3.4.7.1.4

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

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Step 3.4.7.1.4.1

Multiply $- 1$ by $- 1$ .

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

Step 3.4.7.1.4.2

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

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

Step 3.4.8

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

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Step 3.4.8.1

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

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

Step 3.4.8.2

Simplify the left side.

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Step 3.4.8.2.1

Dividing two negative values results in a positive value.

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

Step 3.4.8.2.2

Divide $y$ by $1$ .

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

Step 3.4.8.3

Simplify the right side.

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Step 3.4.8.3.1

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

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

Step 3.4.8.3.2

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

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

Step 4

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

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

Step 5

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

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Step 5.1

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

Step 5.2

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

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Step 5.2.1

Set up the composite result function.

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

Step 5.2.2

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

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

Step 5.2.3

Simplify terms.

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Step 5.2.3.1

Remove parentheses.

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

Step 5.2.3.2

Combine the numerators over the common denominator.

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

Step 5.2.3.3

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

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

Step 5.2.3.4

Combine the numerators over the common denominator.

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

Step 5.2.4

Simplify the numerator.

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Step 5.2.4.1

Apply the distributive property.

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

Step 5.2.4.2

Multiply $- 1$ by $1$ .

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

Step 5.2.4.3

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

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Step 5.2.4.3.1

Multiply $- 1$ by $- 1$ .

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

Step 5.2.4.3.2

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

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

Step 5.2.4.4

Add $- 1$ and $1$ .

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

Step 5.2.4.5

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

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

Step 5.2.4.6

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

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

Step 5.2.5

Multiply the numerator by the reciprocal of the denominator.

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

Step 5.2.6

Simplify the denominator.

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Step 5.2.6.1

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

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

Step 5.2.6.2

Combine the numerators over the common denominator.

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

Step 5.2.6.3

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

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Step 5.2.6.3.1

Add $1$ and $1$ .

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

Step 5.2.6.3.2

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

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

Step 5.2.6.3.3

Add $2$ and $0$ .

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

Step 5.2.7

Multiply the numerator by the reciprocal of the denominator.

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

Step 5.2.8

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

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

Step 5.2.9

Cancel the common factor of $2$ .

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Step 5.2.9.1

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

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

Step 5.2.9.2

Cancel the common factor.

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

Step 5.2.9.3

Rewrite the expression.

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

Step 5.2.10

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

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Step 5.2.10.1

Cancel the common factor.

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

Step 5.2.10.2

Rewrite the expression.

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

Step 5.2.11

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

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

Step 5.2.12

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

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

Step 5.2.13

Multiply $- 1$ by $1$ .

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

Step 5.3

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

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Step 5.3.1

Set up the composite result function.

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

Step 5.3.2

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

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

Step 5.3.3

Simplify the numerator.

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Step 5.3.3.1

Combine the numerators over the common denominator.

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

Step 5.3.3.2

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

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Step 5.3.3.2.1

Multiply $- 1$ by $- 1$ .

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

Step 5.3.3.2.2

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

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

Step 5.3.3.3

Exponentiation and log are inverse functions.

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

Step 5.3.3.4

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

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

Step 5.3.3.5

Combine the numerators over the common denominator.

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

Step 5.3.3.6

Reorder terms.

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

Step 5.3.3.7

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

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Step 5.3.3.7.1

Apply the distributive property.

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

Step 5.3.3.7.2

Multiply $- - x$ .

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Step 5.3.3.7.2.1

Multiply $- 1$ by $- 1$ .

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

Step 5.3.3.7.2.2

Multiply $x$ by $1$ .

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

Step 5.3.3.7.3

Multiply $- 1$ by $1$ .

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

Step 5.3.3.7.4

Add $x$ and $x$ .

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

Step 5.3.3.7.5

Subtract $1$ from $1$ .

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

Step 5.3.3.7.6

Add $2 x$ and $0$ .

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

Step 5.3.4

Simplify the denominator.

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Step 5.3.4.1

Combine the numerators over the common denominator.

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

Step 5.3.4.2

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

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Step 5.3.4.2.1

Multiply $- 1$ by $- 1$ .

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

Step 5.3.4.2.2

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

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

Step 5.3.4.3

Exponentiation and log are inverse functions.

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

Step 5.3.4.4

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

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

Step 5.3.4.5

Combine the numerators over the common denominator.

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

Step 5.3.4.6

Reorder terms.

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

Step 5.3.4.7

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

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Step 5.3.4.7.1

Subtract $x$ from $x$ .

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

Step 5.3.4.7.2

Add $0$ and $1$ .

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

Step 5.3.4.7.3

Add $1$ and $1$ .

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

Step 5.3.5

Multiply the numerator by the reciprocal of the denominator.

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

Step 5.3.6

Cancel the common factor of $2$ .

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Step 5.3.6.1

Factor $2$ out of $2 x$ .

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

Step 5.3.6.2

Cancel the common factor.

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

Step 5.3.6.3

Rewrite the expression.

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

Step 5.3.7

Cancel the common factor of $x + 1$ .

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Step 5.3.7.1

Cancel the common factor.

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

Step 5.3.7.2

Rewrite the expression.

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

Step 5.4

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

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