Calculus 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

Simplify the expression.

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Step 3.3.2.1.2.1

Multiply $x$ by $1$ .

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

Step 3.3.2.1.2.2

Rewrite using the commutative property of multiplication.

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

Step 3.4

Solve for $y$ .

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Step 3.4.1

Add $x e^{y}$ to 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

Multiply by $1$ .

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

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 3.4.5

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 $y$ by $1$ .

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

Step 4

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

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

Step 5

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

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

Step 5.2.3

Remove parentheses.

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

Step 5.2.4

Multiply the numerator and denominator of the fraction by $1 - e^{x}$ .

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Step 5.2.4.1

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

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

Step 5.2.4.2

Combine.

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

Step 5.2.5

Apply the distributive property.

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

Step 5.2.6

Simplify by cancelling.

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Step 5.2.6.1

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

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Step 5.2.6.1.1

Cancel the common factor.

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

Step 5.2.6.1.2

Rewrite the expression.

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

Step 5.2.6.2

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

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Step 5.2.6.2.1

Cancel the common factor.

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

Step 5.2.6.2.2

Rewrite the expression.

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

Step 5.2.7

Simplify the numerator.

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Step 5.2.7.1

Apply the distributive property.

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

Step 5.2.7.2

Multiply $- 1$ by $1$ .

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

Step 5.2.7.3

Multiply $- e^{x} \cdot - 1$ .

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Step 5.2.7.3.1

Multiply $- 1$ by $- 1$ .

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

Step 5.2.7.3.2

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

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

Step 5.2.7.4

Subtract $1$ from $1$ .

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

Step 5.2.7.5

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

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

Step 5.2.7.6

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

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

Step 5.2.8

Simplify the denominator.

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Step 5.2.8.1

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

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

Step 5.2.8.2

Add $1$ and $1$ .

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

Step 5.2.8.3

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

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

Step 5.2.8.4

Add $2$ and $0$ .

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

Step 5.2.9

Cancel the common factor of $2$ .

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Step 5.2.9.1

Cancel the common factor.

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

Step 5.2.9.2

Divide $e^{x}$ by $1$ .

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

Step 5.2.10

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

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

Step 5.2.11

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

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

Step 5.2.12

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

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

Step 5.3.3

Simplify the numerator.

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Step 5.3.3.1

Exponentiation and log are inverse functions.

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

Step 5.3.3.2

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

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

Step 5.3.3.3

Combine the numerators over the common denominator.

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

Step 5.3.3.4

Reorder terms.

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

Step 5.3.3.5

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

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Step 5.3.3.5.1

Add $x$ and $x$ .

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

Step 5.3.3.5.2

Subtract $1$ from $1$ .

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

Step 5.3.3.5.3

Add $2 x$ and $0$ .

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

Step 5.3.4

Simplify the denominator.

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Step 5.3.4.1

Exponentiation and log are inverse functions.

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

Step 5.3.4.2

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

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

Step 5.3.4.3

Combine the numerators over the common denominator.

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

Step 5.3.4.4

Reorder terms.

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

Step 5.3.4.5

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

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Step 5.3.4.5.1

Apply the distributive property.

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

Step 5.3.4.5.2

Multiply $- 1$ by $- 1$ .

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

Step 5.3.4.5.3

Subtract $x$ from $x$ .

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

Step 5.3.4.5.4

Add $0$ and $1$ .

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

Step 5.3.4.5.5

Add $1$ and $1$ .

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

Step 5.3.5

Multiply the numerator by the reciprocal of the denominator.

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

Step 5.3.6

Cancel the common factor of $2$ .

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Step 5.3.6.1

Factor $2$ out of $2 x$ .

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

Step 5.3.6.2

Cancel the common factor.

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

Step 5.3.6.3

Rewrite the expression.

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

Step 5.3.7

Cancel the common factor of $x + 1$ .

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Step 5.3.7.1

Cancel the common factor.

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

Step 5.3.7.2

Rewrite the expression.

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

Step 5.4

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

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