Calculus Examples

$y = \frac{e^{x}}{1 + 2 e^{x}}$

Step 1

Interchange the variables.

$x = \frac{e^{y}}{1 + 2 e^{y}}$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $\frac{e^{y}}{1 + 2 e^{y}} = x$ .

$\frac{e^{y}}{1 + 2 e^{y}} = x$

Step 2.2

Multiply both sides by $1 + 2 e^{y}$ .

$\frac{e^{y}}{1 + 2 e^{y}} (1 + 2 e^{y}) = x (1 + 2 e^{y})$

Step 2.3

Simplify.

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Step 2.3.1

Simplify the left side.

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Step 2.3.1.1

Cancel the common factor of $1 + 2 e^{y}$ .

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Step 2.3.1.1.1

Cancel the common factor.

$\frac{e^{y}}{1 + 2 e^{y}} (1 + 2 e^{y}) = x (1 + 2 e^{y})$

Step 2.3.1.1.2

Rewrite the expression.

$e^{y} = x (1 + 2 e^{y})$

Step 2.3.2

Simplify the right side.

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Step 2.3.2.1

Simplify $x (1 + 2 e^{y})$ .

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Step 2.3.2.1.1

Apply the distributive property.

$e^{y} = x \cdot 1 + x (2 e^{y})$

Step 2.3.2.1.2

Simplify the expression.

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Step 2.3.2.1.2.1

Multiply $x$ by $1$ .

$e^{y} = x + x (2 e^{y})$

Step 2.3.2.1.2.2

Rewrite using the commutative property of multiplication.

$e^{y} = x + 2 x e^{y}$

Step 2.4

Solve for $y$ .

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Step 2.4.1

Subtract $2 x e^{y}$ from both sides of the equation.

$e^{y} - 2 x e^{y} = x$

Step 2.4.2

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

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Step 2.4.2.1

Multiply by $1$ .

$e^{y} \cdot 1 - 2 x e^{y} = x$

Step 2.4.2.2

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

$e^{y} \cdot 1 + e^{y} (- 2 x) = x$

Step 2.4.2.3

Factor $e^{y}$ out of $e^{y} \cdot 1 + e^{y} (- 2 x)$ .

$e^{y} (1 - 2 x) = x$

Step 2.4.3

Divide each term in $e^{y} (1 - 2 x) = x$ by $1 - 2 x$ and simplify.

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Step 2.4.3.1

Divide each term in $e^{y} (1 - 2 x) = x$ by $1 - 2 x$ .

$\frac{e^{y} (1 - 2 x)}{1 - 2 x} = \frac{x}{1 - 2 x}$

Step 2.4.3.2

Simplify the left side.

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Step 2.4.3.2.1

Cancel the common factor of $1 - 2 x$ .

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Step 2.4.3.2.1.1

Cancel the common factor.

$\frac{e^{y} (1 - 2 x)}{1 - 2 x} = \frac{x}{1 - 2 x}$

Step 2.4.3.2.1.2

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

$e^{y} = \frac{x}{1 - 2 x}$

Step 2.4.4

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

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

Step 2.4.5

Expand the left side.

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Step 2.4.5.1

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

$y \ln (e) = \ln (\frac{x}{1 - 2 x})$

Step 2.4.5.2

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

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

Step 2.4.5.3

Multiply $y$ by $1$ .

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

Step 3

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

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

Step 4

Verify if $f^{- 1} (x) = \ln (\frac{x}{1 - 2 x})$ is the inverse of $f (x) = \frac{e^{x}}{1 + 2 e^{x}}$ .

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Step 4.1

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

Step 4.2

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

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Step 4.2.1

Set up the composite result function.

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

Step 4.2.2

Evaluate $f^{- 1} (\frac{e^{x}}{1 + 2 e^{x}})$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Step 4.2.3

Multiply the numerator by the reciprocal of the denominator.

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

Step 4.2.4

Simplify the denominator.

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Step 4.2.4.1

Combine $- 2$ and $\frac{e^{x}}{1 + 2 e^{x}}$ .

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

Step 4.2.4.2

Move the negative in front of the fraction.

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

Step 4.2.4.3

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

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

Step 4.2.4.4

Combine the numerators over the common denominator.

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

Step 4.2.4.5

Reorder terms.

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

Step 4.2.4.6

Rewrite $\frac{2 e^{x} + 1 - 2 e^{x}}{2 e^{x} + 1}$ in a factored form.

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Step 4.2.4.6.1

Subtract $2 e^{x}$ from $2 e^{x}$ .

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

Step 4.2.4.6.2

Add $0$ and $1$ .

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

Step 4.2.5

Multiply the numerator by the reciprocal of the denominator.

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

Step 4.2.6

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

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

Step 4.2.7

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

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

Step 4.2.8

Cancel the common factor of $2 e^{x} + 1$ and $1 + 2 e^{x}$ .

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Step 4.2.8.1

Reorder terms.

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

Step 4.2.8.2

Cancel the common factor.

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

Step 4.2.8.3

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

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

Step 4.2.9

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

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

Step 4.2.10

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

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

Step 4.2.11

Multiply $x$ by $1$ .

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

Step 4.3

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

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Step 4.3.1

Set up the composite result function.

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

Step 4.3.2

Evaluate $f (\ln (\frac{x}{1 - 2 x}))$ by substituting in the value of $f^{- 1}$ into $f$ .

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

Step 4.3.3

Exponentiation and log are inverse functions.

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

Step 4.3.4

Simplify the denominator.

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Step 4.3.4.1

Exponentiation and log are inverse functions.

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

Step 4.3.4.2

Combine $2$ and $\frac{x}{1 - 2 x}$ .

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

Step 4.3.4.3

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

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

Step 4.3.4.4

Combine the numerators over the common denominator.

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

Step 4.3.4.5

Rewrite $\frac{1 - 2 x + 2 x}{1 - 2 x}$ in a factored form.

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Step 4.3.4.5.1

Add $- 2 x$ and $2 x$ .

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

Step 4.3.4.5.2

Add $1$ and $0$ .

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

Step 4.3.5

Multiply the numerator by the reciprocal of the denominator.

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

Step 4.3.6

Cancel the common factor of $1 - 2 x$ .

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Step 4.3.6.1

Cancel the common factor.

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

Step 4.3.6.2

Rewrite the expression.

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

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \ln (\frac{x}{1 - 2 x})$ is the inverse of $f (x) = \frac{e^{x}}{1 + 2 e^{x}}$ .

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