Calculus Examples

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

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

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

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

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 $e^{y} + 1$ .

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Step 3.3.1.1.1

Cancel the common factor.

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

Step 3.3.1.1.2

Rewrite the expression.

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

Step 3.3.2

Simplify the right side.

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Step 3.3.2.1

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

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Step 3.3.2.1.1

Apply the distributive property.

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

Step 3.3.2.1.2

Multiply $x$ by $1$ .

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

Step 3.4

Solve for $y$ .

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Step 3.4.1

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

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

Step 3.4.2

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

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

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

Step 5.2.3

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

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Step 5.2.3.1

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

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

Step 5.2.3.2

Combine.

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

Step 5.2.4

Apply the distributive property.

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

Step 5.2.5

Simplify by cancelling.

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Step 5.2.5.1

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

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Step 5.2.5.1.1

Cancel the common factor.

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

Step 5.2.5.1.2

Rewrite the expression.

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

Step 5.2.5.2

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

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Step 5.2.5.2.1

Move the leading negative in $- \frac{e^{x} - 1}{e^{x} + 1}$ into the numerator.

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

Step 5.2.5.2.2

Cancel the common factor.

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

Step 5.2.5.2.3

Rewrite the expression.

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

Step 5.2.6

Simplify the numerator.

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Step 5.2.6.1

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

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

Step 5.2.6.2

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

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

Step 5.2.6.3

Add $- 1$ and $1$ .

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

Step 5.2.6.4

Add $2 e^{x}$ and $0$ .

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

Step 5.2.7

Simplify the denominator.

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Step 5.2.7.1

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

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

Step 5.2.7.2

Apply the distributive property.

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

Step 5.2.7.3

Multiply $- 1$ by $- 1$ .

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

Step 5.2.7.4

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

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

Step 5.2.7.5

Add $0$ and $1$ .

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

Step 5.2.7.6

Add $1$ and $1$ .

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

Step 5.2.8

Cancel the common factor of $2$ .

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Step 5.2.8.1

Cancel the common factor.

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

Step 5.2.8.2

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

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

Step 5.2.9

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

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

Step 5.2.10

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

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

Step 5.2.11

Multiply $x$ by $1$ .

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

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 $f$ .

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

Step 5.3.3

Simplify the numerator.

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Step 5.3.3.1

Exponentiation and log are inverse functions.

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

Step 5.3.3.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{1 - x}{1 - x}$ .

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

Step 5.3.3.3

Combine $- 1$ and $\frac{1 - x}{1 - x}$ .

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

Step 5.3.3.4

Combine the numerators over the common denominator.

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

Step 5.3.3.5

Reorder terms.

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

Step 5.3.3.6

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

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Step 5.3.3.6.1

Apply the distributive property.

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

Step 5.3.3.6.2

Multiply $- 1$ by $1$ .

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

Step 5.3.3.6.3

Multiply $- - x$ .

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Step 5.3.3.6.3.1

Multiply $- 1$ by $- 1$ .

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

Step 5.3.3.6.3.2

Multiply $x$ by $1$ .

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

Step 5.3.3.6.4

Add $x$ and $x$ .

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

Step 5.3.3.6.5

Subtract $1$ from $1$ .

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

Step 5.3.3.6.6

Add $2 x$ and $0$ .

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

Step 5.3.4

Simplify the denominator.

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Step 5.3.4.1

Exponentiation and log are inverse functions.

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

Step 5.3.4.2

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

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

Step 5.3.4.3

Combine the numerators over the common denominator.

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

Step 5.3.4.4

Reorder terms.

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

Step 5.3.4.5

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

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Step 5.3.4.5.1

Subtract $x$ from $x$ .

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

Step 5.3.4.5.2

Add $0$ and $1$ .

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

Step 5.3.4.5.3

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

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

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

Step 5.3.8

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

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Step 5.3.8.1

Cancel the common factor.

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

Step 5.3.8.2

Divide $x$ by $1$ .

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

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

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