Finite Math Examples

$f (x) = (\frac{e^{3 x}}{e^{3 x} + 1})$

Step 1

Write $f (x) = (\frac{e^{3 x}}{e^{3 x} + 1})$ as an equation.

$y = (\frac{e^{3 x}}{e^{3 x} + 1})$

Step 2

Interchange the variables.

$x = \frac{e^{3 y}}{e^{3 y} + 1}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{e^{3 y}}{e^{3 y} + 1} = x$ .

$\frac{e^{3 y}}{e^{3 y} + 1} = x$

Step 3.2

Multiply both sides by $e^{3 y} + 1$ .

$\frac{e^{3 y}}{e^{3 y} + 1} (e^{3 y} + 1) = x (e^{3 y} + 1)$

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 $e^{3 y} + 1$ .

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Step 3.3.1.1.1

Cancel the common factor.

$\frac{e^{3 y}}{e^{3 y} + 1} (e^{3 y} + 1) = x (e^{3 y} + 1)$

Step 3.3.1.1.2

Rewrite the expression.

$e^{3 y} = x (e^{3 y} + 1)$

Step 3.3.2

Simplify the right side.

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Step 3.3.2.1

Simplify $x (e^{3 y} + 1)$ .

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Step 3.3.2.1.1

Apply the distributive property.

$e^{3 y} = x e^{3 y} + x \cdot 1$

Step 3.3.2.1.2

Multiply $x$ by $1$ .

$e^{3 y} = x e^{3 y} + x$

Step 3.4

Solve for $y$ .

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Step 3.4.1

Subtract $x e^{3 y}$ from both sides of the equation.

$e^{3 y} - x e^{3 y} = x$

Step 3.4.2

Factor $e^{3 y}$ out of $e^{3 y} - x e^{3 y}$ .

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Step 3.4.2.1

Multiply by $1$ .

$e^{3 y} \cdot 1 - x e^{3 y} = x$

Step 3.4.2.2

Factor $e^{3 y}$ out of $- x e^{3 y}$ .

$e^{3 y} \cdot 1 + e^{3 y} (- x) = x$

Step 3.4.2.3

Factor $e^{3 y}$ out of $e^{3 y} \cdot 1 + e^{3 y} (- x)$ .

$e^{3 y} (1 - x) = x$

Step 3.4.3

Divide each term in $e^{3 y} (1 - x) = x$ by $1 - x$ and simplify.

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Step 3.4.3.1

Divide each term in $e^{3 y} (1 - x) = x$ by $1 - x$ .

$\frac{e^{3 y} (1 - x)}{1 - x} = \frac{x}{1 - x}$

Step 3.4.3.2

Simplify the left side.

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Step 3.4.3.2.1

Cancel the common factor of $1 - x$ .

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Step 3.4.3.2.1.1

Cancel the common factor.

$\frac{e^{3 y} (1 - x)}{1 - x} = \frac{x}{1 - x}$

Step 3.4.3.2.1.2

Divide $e^{3 y}$ by $1$ .

$e^{3 y} = \frac{x}{1 - x}$

Step 3.4.4

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

$\ln (e^{3 y}) = \ln (\frac{x}{1 - x})$

Step 3.4.5

Expand the left side.

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Step 3.4.5.1

Expand $\ln (e^{3 y})$ by moving $3 y$ outside the logarithm.

$3 y \ln (e) = \ln (\frac{x}{1 - x})$

Step 3.4.5.2

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

$3 y \cdot 1 = \ln (\frac{x}{1 - x})$

Step 3.4.5.3

Multiply $3$ by $1$ .

$3 y = \ln (\frac{x}{1 - x})$

Step 3.4.6

Divide each term in $3 y = \ln (\frac{x}{1 - x})$ by $3$ and simplify.

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Step 3.4.6.1

Divide each term in $3 y = \ln (\frac{x}{1 - x})$ by $3$ .

$\frac{3 y}{3} = \frac{\ln (\frac{x}{1 - x})}{3}$

Step 3.4.6.2

Simplify the left side.

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Step 3.4.6.2.1

Cancel the common factor of $3$ .

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Step 3.4.6.2.1.1

Cancel the common factor.

$\frac{3 y}{3} = \frac{\ln (\frac{x}{1 - x})}{3}$

Step 3.4.6.2.1.2

Divide $y$ by $1$ .

$y = \frac{\ln (\frac{x}{1 - x})}{3}$

Step 4

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

$f^{- 1} (x) = \frac{\ln (\frac{x}{1 - x})}{3}$

Step 5

Verify if $f^{- 1} (x) = \frac{\ln (\frac{x}{1 - x})}{3}$ is the inverse of $f (x) = \frac{e^{3 x}}{e^{3 x} + 1}$ .

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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 $f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \frac{\ln (\frac{\frac{e^{3 x}}{e^{3 x} + 1}}{1 - (\frac{e^{3 x}}{e^{3 x} + 1})})}{3}$

Step 5.2.3

Simplify $\frac{1}{3} \ln (\frac{\frac{e^{3 x}}{e^{3 x} + 1}}{1 - \frac{e^{3 x}}{e^{3 x} + 1}})$ by moving $\frac{1}{3}$ inside the logarithm.

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{\frac{e^{3 x}}{e^{3 x} + 1}}{1 - \frac{e^{3 x}}{e^{3 x} + 1}})}^{\frac{1}{3}})$

Step 5.2.4

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{e^{3 x} + 1} \cdot \frac{1}{1 - \frac{e^{3 x}}{e^{3 x} + 1}})}^{\frac{1}{3}})$

Step 5.2.5

Simplify the denominator.

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Step 5.2.5.1

Rewrite $e^{3 x}$ as ${(e^{x})}^{3}$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{{(e^{x})}^{3} + 1} \cdot \frac{1}{1 - \frac{e^{3 x}}{e^{3 x} + 1}})}^{\frac{1}{3}})$

Step 5.2.5.2

Rewrite $1$ as $1^{3}$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{{(e^{x})}^{3} + 1^{3}} \cdot \frac{1}{1 - \frac{e^{3 x}}{e^{3 x} + 1}})}^{\frac{1}{3}})$

Step 5.2.5.3

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = e^{x}$ and $b = 1$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) ({(e^{x})}^{2} - e^{x} \cdot 1 + 1^{2})} \cdot \frac{1}{1 - \frac{e^{3 x}}{e^{3 x} + 1}})}^{\frac{1}{3}})$

Step 5.2.5.4

Simplify.

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Step 5.2.5.4.1

Multiply the exponents in ${(e^{x})}^{2}$ .

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Step 5.2.5.4.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{x \cdot 2} - e^{x} \cdot 1 + 1^{2})} \cdot \frac{1}{1 - \frac{e^{3 x}}{e^{3 x} + 1}})}^{\frac{1}{3}})$

Step 5.2.5.4.1.2

Move $2$ to the left of $x$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} - e^{x} \cdot 1 + 1^{2})} \cdot \frac{1}{1 - \frac{e^{3 x}}{e^{3 x} + 1}})}^{\frac{1}{3}})$

Step 5.2.5.4.2

Multiply $- 1$ by $1$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} - e^{x} + 1^{2})} \cdot \frac{1}{1 - \frac{e^{3 x}}{e^{3 x} + 1}})}^{\frac{1}{3}})$

Step 5.2.5.4.3

One to any power is one.

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} - e^{x} + 1)} \cdot \frac{1}{1 - \frac{e^{3 x}}{e^{3 x} + 1}})}^{\frac{1}{3}})$

Step 5.2.5.4.4

Reorder terms.

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{1 - \frac{e^{3 x}}{e^{3 x} + 1}})}^{\frac{1}{3}})$

Step 5.2.6

Simplify the denominator.

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Step 5.2.6.1

Rewrite $e^{3 x}$ as ${(e^{x})}^{3}$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{1 - \frac{e^{3 x}}{{(e^{x})}^{3} + 1}})}^{\frac{1}{3}})$

Step 5.2.6.2

Rewrite $1$ as $1^{3}$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{1 - \frac{e^{3 x}}{{(e^{x})}^{3} + 1^{3}}})}^{\frac{1}{3}})$

Step 5.2.6.3

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = e^{x}$ and $b = 1$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{1 - \frac{e^{3 x}}{(e^{x} + 1) ({(e^{x})}^{2} - e^{x} \cdot 1 + 1^{2})}})}^{\frac{1}{3}})$

Step 5.2.6.4

Simplify.

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Step 5.2.6.4.1

Multiply the exponents in ${(e^{x})}^{2}$ .

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Step 5.2.6.4.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{1 - \frac{e^{3 x}}{(e^{x} + 1) (e^{x \cdot 2} - e^{x} \cdot 1 + 1^{2})}})}^{\frac{1}{3}})$

Step 5.2.6.4.1.2

Move $2$ to the left of $x$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{1 - \frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} - e^{x} \cdot 1 + 1^{2})}})}^{\frac{1}{3}})$

Step 5.2.6.4.2

Multiply $- 1$ by $1$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{1 - \frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} - e^{x} + 1^{2})}})}^{\frac{1}{3}})$

Step 5.2.6.4.3

One to any power is one.

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{1 - \frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} - e^{x} + 1)}})}^{\frac{1}{3}})$

Step 5.2.6.4.4

Reorder terms.

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{1 - \frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}})}^{\frac{1}{3}})$

Step 5.2.7

Simplify the denominator.

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Step 5.2.7.1

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

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{\frac{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} - \frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}})}^{\frac{1}{3}})$

Step 5.2.7.2

Combine the numerators over the common denominator.

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{\frac{(e^{x} + 1) (e^{2 x} + 1 - e^{x}) - e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}})}^{\frac{1}{3}})$

Step 5.2.7.3

Simplify the numerator.

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Step 5.2.7.3.1

Expand $(e^{x} + 1) (e^{2 x} + 1 - e^{x})$ by multiplying each term in the first expression by each term in the second expression.

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{\frac{e^{x} e^{2 x} + e^{x} \cdot 1 + e^{x} (- e^{x}) + 1 e^{2 x} + 1 \cdot 1 + 1 (- e^{x}) - e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}})}^{\frac{1}{3}})$

Step 5.2.7.3.2

Simplify each term.

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Step 5.2.7.3.2.1

Multiply $e^{x}$ by $e^{2 x}$ by adding the exponents.

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Step 5.2.7.3.2.1.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{\frac{e^{x + 2 x} + e^{x} \cdot 1 + e^{x} (- e^{x}) + 1 e^{2 x} + 1 \cdot 1 + 1 (- e^{x}) - e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}})}^{\frac{1}{3}})$

Step 5.2.7.3.2.1.2

Add $x$ and $2 x$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{\frac{e^{3 x} + e^{x} \cdot 1 + e^{x} (- e^{x}) + 1 e^{2 x} + 1 \cdot 1 + 1 (- e^{x}) - e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}})}^{\frac{1}{3}})$

Step 5.2.7.3.2.2

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

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{\frac{e^{3 x} + e^{x} + e^{x} (- e^{x}) + 1 e^{2 x} + 1 \cdot 1 + 1 (- e^{x}) - e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}})}^{\frac{1}{3}})$

Step 5.2.7.3.2.3

Rewrite using the commutative property of multiplication.

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{\frac{e^{3 x} + e^{x} - e^{x} e^{x} + 1 e^{2 x} + 1 \cdot 1 + 1 (- e^{x}) - e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}})}^{\frac{1}{3}})$

Step 5.2.7.3.2.4

Multiply $e^{x}$ by $e^{x}$ by adding the exponents.

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Step 5.2.7.3.2.4.1

Move $e^{x}$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{\frac{e^{3 x} + e^{x} - (e^{x} e^{x}) + 1 e^{2 x} + 1 \cdot 1 + 1 (- e^{x}) - e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}})}^{\frac{1}{3}})$

Step 5.2.7.3.2.4.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{\frac{e^{3 x} + e^{x} - e^{x + x} + 1 e^{2 x} + 1 \cdot 1 + 1 (- e^{x}) - e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}})}^{\frac{1}{3}})$

Step 5.2.7.3.2.4.3

Add $x$ and $x$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{\frac{e^{3 x} + e^{x} - e^{2 x} + 1 e^{2 x} + 1 \cdot 1 + 1 (- e^{x}) - e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}})}^{\frac{1}{3}})$

Step 5.2.7.3.2.5

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

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{\frac{e^{3 x} + e^{x} - e^{2 x} + e^{2 x} + 1 \cdot 1 + 1 (- e^{x}) - e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}})}^{\frac{1}{3}})$

Step 5.2.7.3.2.6

Multiply $1$ by $1$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{\frac{e^{3 x} + e^{x} - e^{2 x} + e^{2 x} + 1 + 1 (- e^{x}) - e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}})}^{\frac{1}{3}})$

Step 5.2.7.3.2.7

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

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{\frac{e^{3 x} + e^{x} - e^{2 x} + e^{2 x} + 1 - e^{x} - e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}})}^{\frac{1}{3}})$

Step 5.2.7.3.3

Combine the opposite terms in $e^{3 x} + e^{x} - e^{2 x} + e^{2 x} + 1 - e^{x}$ .

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Step 5.2.7.3.3.1

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

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{\frac{e^{3 x} + e^{x} + 0 + 1 - e^{x} - e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}})}^{\frac{1}{3}})$

Step 5.2.7.3.3.2

Add $e^{3 x} + e^{x}$ and $0$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{\frac{e^{3 x} + e^{x} + 1 - e^{x} - e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}})}^{\frac{1}{3}})$

Step 5.2.7.3.3.3

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

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{\frac{e^{3 x} + 0 + 1 - e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}})}^{\frac{1}{3}})$

Step 5.2.7.3.3.4

Add $e^{3 x}$ and $0$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{\frac{e^{3 x} + 1 - e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}})}^{\frac{1}{3}})$

Step 5.2.7.3.4

Subtract $e^{3 x}$ from $e^{3 x}$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{\frac{0 + 1}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}})}^{\frac{1}{3}})$

Step 5.2.7.3.5

Add $0$ and $1$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})} \cdot \frac{1}{\frac{1}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}})}^{\frac{1}{3}})$

Step 5.2.8

Combine fractions.

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Step 5.2.8.1

Combine.

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x} \cdot 1}{(e^{x} + 1) (e^{2 x} + 1 - e^{x}) (\frac{1}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})})})}^{\frac{1}{3}})$

Step 5.2.8.2

Multiply $e^{3 x}$ by $1$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x}) (\frac{1}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})})})}^{\frac{1}{3}})$

Step 5.2.9

Factor $\frac{1}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}$ out of $\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x}) \frac{1}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}}$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({((e^{x} + 1) (e^{2 x} + 1 - e^{x}) (\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}))}^{\frac{1}{3}})$

Step 5.2.10

Cancel the common factor of $(e^{x} + 1) (e^{2 x} + 1 - e^{x})$ .

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Step 5.2.10.1

Cancel the common factor.

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({((e^{x} + 1) (e^{2 x} + 1 - e^{x}) (\frac{e^{3 x}}{(e^{x} + 1) (e^{2 x} + 1 - e^{x})}))}^{\frac{1}{3}})$

Step 5.2.10.2

Rewrite the expression.

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln ({(e^{3 x})}^{\frac{1}{3}})$

Step 5.2.11

Multiply the exponents in ${(e^{3 x})}^{\frac{1}{3}}$ .

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Step 5.2.11.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln (e^{3 x (\frac{1}{3})})$

Step 5.2.11.2

Cancel the common factor of $3$ .

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Step 5.2.11.2.1

Factor $3$ out of $3 x$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln (e^{3 (x) (\frac{1}{3})})$

Step 5.2.11.2.2

Cancel the common factor.

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln (e^{3 x (\frac{1}{3})})$

Step 5.2.11.2.3

Rewrite the expression.

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = \ln (e^{x})$

Step 5.2.12

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

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = x \ln (e)$

Step 5.2.13

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

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = x \cdot 1$

Step 5.2.14

Multiply $x$ by $1$ .

$f^{- 1} (\frac{e^{3 x}}{e^{3 x} + 1}) = x$

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 $f (\frac{\ln (\frac{x}{1 - x})}{3})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{e^{3 (\frac{\ln (\frac{x}{1 - x})}{3})}}{e^{3 (\frac{\ln (\frac{x}{1 - x})}{3})} + 1}$

Step 5.3.3

Remove parentheses.

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{e^{3 (\frac{\ln (\frac{x}{1 - x})}{3})}}{e^{3 (\frac{\ln (\frac{x}{1 - x})}{3})} + 1}$

Step 5.3.4

Simplify the numerator.

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Step 5.3.4.1

Cancel the common factor of $3$ .

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Step 5.3.4.1.1

Cancel the common factor.

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{e^{3 (\frac{\ln (\frac{x}{1 - x})}{3})}}{e^{3 (\frac{\ln (\frac{x}{1 - x})}{3})} + 1}$

Step 5.3.4.1.2

Rewrite the expression.

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{e^{\ln (\frac{x}{1 - x})}}{e^{3 (\frac{\ln (\frac{x}{1 - x})}{3})} + 1}$

Step 5.3.4.2

Exponentiation and log are inverse functions.

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{e^{3 (\frac{\ln (\frac{x}{1 - x})}{3})} + 1}$

Step 5.3.5

Simplify the denominator.

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Step 5.3.5.1

Rewrite $e^{3 \frac{\ln (\frac{x}{1 - x})}{3}}$ as ${(e^{\frac{\ln (\frac{x}{1 - x})}{3}})}^{3}$ .

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{{(e^{\frac{\ln (\frac{x}{1 - x})}{3}})}^{3} + 1}$

Step 5.3.5.2

Rewrite $1$ as $1^{3}$ .

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{{(e^{\frac{\ln (\frac{x}{1 - x})}{3}})}^{3} + 1^{3}}$

Step 5.3.5.3

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = e^{\frac{\ln (\frac{x}{1 - x})}{3}}$ and $b = 1$ .

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{(e^{\frac{\ln (\frac{x}{1 - x})}{3}} + 1) ({(e^{\frac{\ln (\frac{x}{1 - x})}{3}})}^{2} - e^{\frac{\ln (\frac{x}{1 - x})}{3}} \cdot 1 + 1^{2})}$

Step 5.3.5.4

Simplify.

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Step 5.3.5.4.1

Rewrite $\frac{\ln (\frac{x}{1 - x})}{3}$ as $\frac{1}{3} \ln (\frac{x}{1 - x})$ .

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{(e^{\frac{1}{3} \cdot \ln (\frac{x}{1 - x})} + 1) ({(e^{\frac{\ln (\frac{x}{1 - x})}{3}})}^{2} - e^{\frac{\ln (\frac{x}{1 - x})}{3}} \cdot 1 + 1^{2})}$

Step 5.3.5.4.2

Simplify $\frac{1}{3} \ln (\frac{x}{1 - x})$ by moving $\frac{1}{3}$ inside the logarithm.

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{(e^{\ln ({(\frac{x}{1 - x})}^{\frac{1}{3}})} + 1) ({(e^{\frac{\ln (\frac{x}{1 - x})}{3}})}^{2} - e^{\frac{\ln (\frac{x}{1 - x})}{3}} \cdot 1 + 1^{2})}$

Step 5.3.5.4.3

Exponentiation and log are inverse functions.

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{({(\frac{x}{1 - x})}^{\frac{1}{3}} + 1) ({(e^{\frac{\ln (\frac{x}{1 - x})}{3}})}^{2} - e^{\frac{\ln (\frac{x}{1 - x})}{3}} \cdot 1 + 1^{2})}$

Step 5.3.5.4.4

Apply the product rule to $\frac{x}{1 - x}$ .

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{(\frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}} + 1) ({(e^{\frac{\ln (\frac{x}{1 - x})}{3}})}^{2} - e^{\frac{\ln (\frac{x}{1 - x})}{3}} \cdot 1 + 1^{2})}$

Step 5.3.5.4.5

Rewrite $\frac{\ln (\frac{x}{1 - x})}{3}$ as $\frac{1}{3} \ln (\frac{x}{1 - x})$ .

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{(\frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}} + 1) ({(e^{\frac{1}{3} \cdot \ln (\frac{x}{1 - x})})}^{2} - e^{\frac{\ln (\frac{x}{1 - x})}{3}} \cdot 1 + 1^{2})}$

Step 5.3.5.4.6

Simplify $\frac{1}{3} \ln (\frac{x}{1 - x})$ by moving $\frac{1}{3}$ inside the logarithm.

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{(\frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}} + 1) ({(e^{\ln ({(\frac{x}{1 - x})}^{\frac{1}{3}})})}^{2} - e^{\frac{\ln (\frac{x}{1 - x})}{3}} \cdot 1 + 1^{2})}$

Step 5.3.5.4.7

Exponentiation and log are inverse functions.

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{(\frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}} + 1) ({({(\frac{x}{1 - x})}^{\frac{1}{3}})}^{2} - e^{\frac{\ln (\frac{x}{1 - x})}{3}} \cdot 1 + 1^{2})}$

Step 5.3.5.4.8

Multiply the exponents in ${({(\frac{x}{1 - x})}^{\frac{1}{3}})}^{2}$ .

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Step 5.3.5.4.8.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{(\frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}} + 1) ({(\frac{x}{1 - x})}^{\frac{1}{3} \cdot 2} - e^{\frac{\ln (\frac{x}{1 - x})}{3}} \cdot 1 + 1^{2})}$

Step 5.3.5.4.8.2

Combine $\frac{1}{3}$ and $2$ .

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{(\frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}} + 1) ({(\frac{x}{1 - x})}^{\frac{2}{3}} - e^{\frac{\ln (\frac{x}{1 - x})}{3}} \cdot 1 + 1^{2})}$

Step 5.3.5.4.9

Apply the product rule to $\frac{x}{1 - x}$ .

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{(\frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}} + 1) (\frac{x^{\frac{2}{3}}}{{(1 - x)}^{\frac{2}{3}}} - e^{\frac{\ln (\frac{x}{1 - x})}{3}} \cdot 1 + 1^{2})}$

Step 5.3.5.4.10

Rewrite $\frac{\ln (\frac{x}{1 - x})}{3}$ as $\frac{1}{3} \ln (\frac{x}{1 - x})$ .

Step 5.3.5.4.11

Simplify $\frac{1}{3} \ln (\frac{x}{1 - x})$ by moving $\frac{1}{3}$ inside the logarithm.

Step 5.3.5.4.12

Exponentiation and log are inverse functions.

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{(\frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}} + 1) (\frac{x^{\frac{2}{3}}}{{(1 - x)}^{\frac{2}{3}}} - {(\frac{x}{1 - x})}^{\frac{1}{3}} \cdot 1 + 1^{2})}$

Step 5.3.5.4.13

Apply the product rule to $\frac{x}{1 - x}$ .

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{(\frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}} + 1) (\frac{x^{\frac{2}{3}}}{{(1 - x)}^{\frac{2}{3}}} - \frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}} \cdot 1 + 1^{2})}$

Step 5.3.5.4.14

Multiply $- 1$ by $1$ .

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{(\frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}} + 1) (\frac{x^{\frac{2}{3}}}{{(1 - x)}^{\frac{2}{3}}} - \frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}} + 1^{2})}$

Step 5.3.5.4.15

One to any power is one.

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{(\frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}} + 1) (\frac{x^{\frac{2}{3}}}{{(1 - x)}^{\frac{2}{3}}} - \frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}} + 1)}$

Step 5.3.5.4.16

Reorder terms.

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{(\frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}} + 1) (\frac{x^{\frac{2}{3}}}{{(1 - x)}^{\frac{2}{3}}} + 1 - \frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}})}$

Step 5.3.5.5

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

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{(\frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}} + \frac{{(1 - x)}^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}}) (\frac{x^{\frac{2}{3}}}{{(1 - x)}^{\frac{2}{3}}} + 1 - \frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}})}$

Step 5.3.5.6

Combine the numerators over the common denominator.

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{\frac{x^{\frac{1}{3}} + {(1 - x)}^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}} \cdot (\frac{x^{\frac{2}{3}}}{{(1 - x)}^{\frac{2}{3}}} + 1 - \frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}})}$

Step 5.3.5.7

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

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{\frac{x^{\frac{1}{3}} + {(1 - x)}^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}} \cdot (\frac{x^{\frac{2}{3}}}{{(1 - x)}^{\frac{2}{3}}} + \frac{{(1 - x)}^{\frac{2}{3}}}{{(1 - x)}^{\frac{2}{3}}} - \frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}})}$

Step 5.3.5.8

Combine the numerators over the common denominator.

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{\frac{x^{\frac{1}{3}} + {(1 - x)}^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}} \cdot (\frac{x^{\frac{2}{3}} + {(1 - x)}^{\frac{2}{3}}}{{(1 - x)}^{\frac{2}{3}}} - \frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}})}$

Step 5.3.5.9

To write $- \frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}}$ as a fraction with a common denominator, multiply by $\frac{{(1 - x)}^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}}$ .

Step 5.3.5.10

Write each expression with a common denominator of ${(1 - x)}^{\frac{2}{3}}$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.3.5.10.1

Multiply $\frac{x^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}}$ by $\frac{{(1 - x)}^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}}$ .

Step 5.3.5.10.2

Multiply ${(1 - x)}^{\frac{1}{3}}$ by ${(1 - x)}^{\frac{1}{3}}$ by adding the exponents.

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Step 5.3.5.10.2.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

Step 5.3.5.10.2.2

Combine the numerators over the common denominator.

Step 5.3.5.10.2.3

Add $1$ and $1$ .

Step 5.3.5.11

Combine the numerators over the common denominator.

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{\frac{x^{\frac{1}{3}} + {(1 - x)}^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}} \cdot \frac{x^{\frac{2}{3}} + {(1 - x)}^{\frac{2}{3}} - x^{\frac{1}{3}} {(1 - x)}^{\frac{1}{3}}}{{(1 - x)}^{\frac{2}{3}}}}$

Step 5.3.5.12

Reorder terms.

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{\frac{x^{\frac{1}{3}} + {(1 - x)}^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}} \cdot \frac{- x^{\frac{1}{3}} {(1 - x)}^{\frac{1}{3}} + x^{\frac{2}{3}} + {(1 - x)}^{\frac{2}{3}}}{{(1 - x)}^{\frac{2}{3}}}}$

Step 5.3.6

Multiply $\frac{x^{\frac{1}{3}} + {(1 - x)}^{\frac{1}{3}}}{{(1 - x)}^{\frac{1}{3}}}$ by $\frac{- x^{\frac{1}{3}} {(1 - x)}^{\frac{1}{3}} + x^{\frac{2}{3}} + {(1 - x)}^{\frac{2}{3}}}{{(1 - x)}^{\frac{2}{3}}}$ .

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{\frac{x}{1 - x}}{\frac{(x^{\frac{1}{3}} + {(1 - x)}^{\frac{1}{3}}) (- x^{\frac{1}{3}} {(1 - x)}^{\frac{1}{3}} + x^{\frac{2}{3}} + {(1 - x)}^{\frac{2}{3}})}{{(1 - x)}^{\frac{1}{3}} {(1 - x)}^{\frac{2}{3}}}}$

Step 5.3.7

Simplify the denominator.

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Step 5.3.7.1

Multiply ${(1 - x)}^{\frac{1}{3}}$ by ${(1 - x)}^{\frac{2}{3}}$ by adding the exponents.

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Step 5.3.7.1.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

Step 5.3.7.1.2

Combine the numerators over the common denominator.

Step 5.3.7.1.3

Add $1$ and $2$ .

Step 5.3.7.1.4

Divide $3$ by $3$ .

Step 5.3.7.2

Simplify ${(1 - x)}^{1}$ .

Step 5.3.8

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{x}{1 - x} \cdot \frac{1 - x}{(x^{\frac{1}{3}} + {(1 - x)}^{\frac{1}{3}}) (- x^{\frac{1}{3}} {(1 - x)}^{\frac{1}{3}} + x^{\frac{2}{3}} + {(1 - x)}^{\frac{2}{3}})}$

Step 5.3.9

Cancel the common factor of $1 - x$ .

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Step 5.3.9.1

Cancel the common factor.

Step 5.3.9.2

Rewrite the expression.

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = x (\frac{1}{(x^{\frac{1}{3}} + {(1 - x)}^{\frac{1}{3}}) (- x^{\frac{1}{3}} {(1 - x)}^{\frac{1}{3}} + x^{\frac{2}{3}} + {(1 - x)}^{\frac{2}{3}})})$

Step 5.3.10

Combine $x$ and $\frac{1}{(x^{\frac{1}{3}} + {(1 - x)}^{\frac{1}{3}}) (- x^{\frac{1}{3}} {(1 - x)}^{\frac{1}{3}} + x^{\frac{2}{3}} + {(1 - x)}^{\frac{2}{3}})}$ .

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{x}{(x^{\frac{1}{3}} + {(1 - x)}^{\frac{1}{3}}) (- x^{\frac{1}{3}} {(1 - x)}^{\frac{1}{3}} + x^{\frac{2}{3}} + {(1 - x)}^{\frac{2}{3}})}$

Step 5.3.11

Factor $- 1$ out of $- x^{\frac{1}{3}} {(1 - x)}^{\frac{1}{3}}$ .

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{x}{(x^{\frac{1}{3}} + {(1 - x)}^{\frac{1}{3}}) (- (x^{\frac{1}{3}} {(1 - x)}^{\frac{1}{3}}) + x^{\frac{2}{3}} + {(1 - x)}^{\frac{2}{3}})}$

Step 5.3.12

Factor $- 1$ out of $x^{\frac{2}{3}}$ .

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{x}{(x^{\frac{1}{3}} + {(1 - x)}^{\frac{1}{3}}) (- (x^{\frac{1}{3}} {(1 - x)}^{\frac{1}{3}}) - 1 (- x^{\frac{2}{3}}) + {(1 - x)}^{\frac{2}{3}})}$

Step 5.3.13

Factor $- 1$ out of $- (x^{\frac{1}{3}} {(1 - x)}^{\frac{1}{3}}) - 1 (- x^{\frac{2}{3}})$ .

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{x}{(x^{\frac{1}{3}} + {(1 - x)}^{\frac{1}{3}}) (- (x^{\frac{1}{3}} {(1 - x)}^{\frac{1}{3}} - x^{\frac{2}{3}}) + {(1 - x)}^{\frac{2}{3}})}$

Step 5.3.14

Factor $- 1$ out of ${(1 - x)}^{\frac{2}{3}}$ .

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{x}{(x^{\frac{1}{3}} + {(1 - x)}^{\frac{1}{3}}) (- (x^{\frac{1}{3}} {(1 - x)}^{\frac{1}{3}} - x^{\frac{2}{3}}) - 1 (- {(1 - x)}^{\frac{2}{3}}))}$

Step 5.3.15

Factor $- 1$ out of $- (x^{\frac{1}{3}} {(1 - x)}^{\frac{1}{3}} - x^{\frac{2}{3}}) - 1 (- {(1 - x)}^{\frac{2}{3}})$ .

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{x}{(x^{\frac{1}{3}} + {(1 - x)}^{\frac{1}{3}}) (- (x^{\frac{1}{3}} {(1 - x)}^{\frac{1}{3}} - x^{\frac{2}{3}} - {(1 - x)}^{\frac{2}{3}}))}$

Step 5.3.16

Rewrite negatives.

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Step 5.3.16.1

Rewrite $- (x^{\frac{1}{3}} {(1 - x)}^{\frac{1}{3}} - x^{\frac{2}{3}} - {(1 - x)}^{\frac{2}{3}})$ as $- 1 (x^{\frac{1}{3}} {(1 - x)}^{\frac{1}{3}} - x^{\frac{2}{3}} - {(1 - x)}^{\frac{2}{3}})$ .

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = \frac{x}{(x^{\frac{1}{3}} + {(1 - x)}^{\frac{1}{3}}) (- 1 (x^{\frac{1}{3}} {(1 - x)}^{\frac{1}{3}} - x^{\frac{2}{3}} - {(1 - x)}^{\frac{2}{3}}))}$

Step 5.3.16.2

Move the negative in front of the fraction.

$f (\frac{\ln (\frac{x}{1 - x})}{3}) = - \frac{x}{(x^{\frac{1}{3}} + {(1 - x)}^{\frac{1}{3}}) (x^{\frac{1}{3}} {(1 - x)}^{\frac{1}{3}} - x^{\frac{2}{3}} - {(1 - x)}^{\frac{2}{3}})}$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\ln (\frac{x}{1 - x})}{3}$ is the inverse of $f (x) = \frac{e^{3 x}}{e^{3 x} + 1}$ .

$f^{- 1} (x) = \frac{\ln (\frac{x}{1 - x})}{3}$