Finite Math Examples

$f (x) = \frac{x}{x + 8}$

Step 1

Write $f (x) = \frac{x}{x + 8}$ as an equation.

$y = \frac{x}{x + 8}$

Step 2

Interchange the variables.

$x = \frac{y}{y + 8}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{y}{y + 8} = x$ .

$\frac{y}{y + 8} = x$

Step 3.2

Find the LCD of the terms in the equation.

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Step 3.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$y + 8, 1$

Step 3.2.2

Remove parentheses.

$y + 8, 1$

Step 3.2.3

The LCM of one and any expression is the expression.

$y + 8$

Step 3.3

Multiply each term in $\frac{y}{y + 8} = x$ by $y + 8$ to eliminate the fractions.

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Step 3.3.1

Multiply each term in $\frac{y}{y + 8} = x$ by $y + 8$ .

$\frac{y}{y + 8} (y + 8) = x (y + 8)$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $y + 8$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{y}{y + 8} (y + 8) = x (y + 8)$

Step 3.3.2.1.2

Rewrite the expression.

$y = x (y + 8)$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Apply the distributive property.

$y = x y + x \cdot 8$

Step 3.3.3.2

Move $8$ to the left of $x$ .

$y = x y + 8 x$

Step 3.4

Solve the equation.

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Step 3.4.1

Subtract $x y$ from both sides of the equation.

$y - x y = 8 x$

Step 3.4.2

Factor $y$ out of $y - x y$ .

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Step 3.4.2.1

Factor $y$ out of $y^{1}$ .

$y \cdot 1 - x y = 8 x$

Step 3.4.2.2

Factor $y$ out of $- x y$ .

$y \cdot 1 + y (- x) = 8 x$

Step 3.4.2.3

Factor $y$ out of $y \cdot 1 + y (- x)$ .

$y (1 - x) = 8 x$

Step 3.4.3

Divide each term in $y (1 - x) = 8 x$ by $1 - x$ and simplify.

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Step 3.4.3.1

Divide each term in $y (1 - x) = 8 x$ by $1 - x$ .

$\frac{y (1 - x)}{1 - x} = \frac{8 x}{1 - x}$

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.

$\frac{y (1 - x)}{1 - x} = \frac{8 x}{1 - x}$

Step 3.4.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{8 x}{1 - x}$

Step 4

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

$f^{- 1} (x) = \frac{8 x}{1 - x}$

Step 5

Verify if $f^{- 1} (x) = \frac{8 x}{1 - x}$ is the inverse of $f (x) = \frac{x}{x + 8}$ .

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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 $f^{- 1} (\frac{x}{x + 8})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{x}{x + 8}) = \frac{8 (\frac{x}{x + 8})}{1 - (\frac{x}{x + 8})}$

Step 5.2.3

Combine $8$ and $\frac{x}{x + 8}$ .

$f^{- 1} (\frac{x}{x + 8}) = \frac{\frac{8 x}{x + 8}}{1 - \frac{x}{x + 8}}$

Step 5.2.4

Simplify the denominator.

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Step 5.2.4.1

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

$f^{- 1} (\frac{x}{x + 8}) = \frac{\frac{8 x}{x + 8}}{\frac{x + 8}{x + 8} - \frac{x}{x + 8}}$

Step 5.2.4.2

Combine the numerators over the common denominator.

$f^{- 1} (\frac{x}{x + 8}) = \frac{\frac{8 x}{x + 8}}{\frac{x + 8 - x}{x + 8}}$

Step 5.2.4.3

Reorder terms.

$f^{- 1} (\frac{x}{x + 8}) = \frac{\frac{8 x}{x + 8}}{\frac{x - x + 8}{x + 8}}$

Step 5.2.4.4

Rewrite $\frac{x - x + 8}{x + 8}$ in a factored form.

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Step 5.2.4.4.1

Subtract $x$ from $x$ .

$f^{- 1} (\frac{x}{x + 8}) = \frac{\frac{8 x}{x + 8}}{\frac{0 + 8}{x + 8}}$

Step 5.2.4.4.2

Add $0$ and $8$ .

$f^{- 1} (\frac{x}{x + 8}) = \frac{\frac{8 x}{x + 8}}{\frac{8}{x + 8}}$

Step 5.2.5

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (\frac{x}{x + 8}) = \frac{8 x}{x + 8} \cdot \frac{x + 8}{8}$

Step 5.2.6

Cancel the common factor of $8$ .

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Step 5.2.6.1

Factor $8$ out of $8 x$ .

$f^{- 1} (\frac{x}{x + 8}) = \frac{8 (x)}{x + 8} \cdot \frac{x + 8}{8}$

Step 5.2.6.2

Cancel the common factor.

$f^{- 1} (\frac{x}{x + 8}) = \frac{8 x}{x + 8} \cdot \frac{x + 8}{8}$

Step 5.2.6.3

Rewrite the expression.

$f^{- 1} (\frac{x}{x + 8}) = \frac{x}{x + 8} \cdot (x + 8)$

Step 5.2.7

Cancel the common factor of $x + 8$ .

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Step 5.2.7.1

Cancel the common factor.

$f^{- 1} (\frac{x}{x + 8}) = \frac{x}{x + 8} \cdot (x + 8)$

Step 5.2.7.2

Rewrite the expression.

$f^{- 1} (\frac{x}{x + 8}) = x$

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 $f (\frac{8 x}{1 - x})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{8 x}{1 - x}) = \frac{\frac{8 x}{1 - x}}{(\frac{8 x}{1 - x}) + 8}$

Step 5.3.3

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{8 x}{1 - x}) = \frac{8 x}{1 - x} \cdot \frac{1}{\frac{8 x}{1 - x} + 8}$

Step 5.3.4

Simplify the denominator.

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Step 5.3.4.1

To write $8$ as a fraction with a common denominator, multiply by $\frac{1 - x}{1 - x}$ .

$f (\frac{8 x}{1 - x}) = \frac{8 x}{1 - x} \cdot \frac{1}{\frac{8 x}{1 - x} + \frac{8 (1 - x)}{1 - x}}$

Step 5.3.4.2

Combine the numerators over the common denominator.

$f (\frac{8 x}{1 - x}) = \frac{8 x}{1 - x} \cdot \frac{1}{\frac{8 x + 8 (1 - x)}{1 - x}}$

Step 5.3.4.3

Reorder terms.

$f (\frac{8 x}{1 - x}) = \frac{8 x}{1 - x} \cdot \frac{1}{\frac{8 x + 8 (1 - x)}{- x + 1}}$

Step 5.3.4.4

Rewrite $\frac{8 x + 8 (1 - x)}{- x + 1}$ in a factored form.

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Step 5.3.4.4.1

Factor $8$ out of $8 x + 8 (1 - x)$ .

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Step 5.3.4.4.1.1

Factor $8$ out of $8 x$ .

$f (\frac{8 x}{1 - x}) = \frac{8 x}{1 - x} \cdot \frac{1}{\frac{8 (x) + 8 (1 - x)}{- x + 1}}$

Step 5.3.4.4.1.2

Factor $8$ out of $8 (x) + 8 (1 - x)$ .

$f (\frac{8 x}{1 - x}) = \frac{8 x}{1 - x} \cdot \frac{1}{\frac{8 (x + 1 - x)}{- x + 1}}$

Step 5.3.4.4.2

Subtract $x$ from $x$ .

$f (\frac{8 x}{1 - x}) = \frac{8 x}{1 - x} \cdot \frac{1}{\frac{8 (0 + 1)}{- x + 1}}$

Step 5.3.4.4.3

Add $0$ and $1$ .

$f (\frac{8 x}{1 - x}) = \frac{8 x}{1 - x} \cdot \frac{1}{\frac{8 \cdot 1}{- x + 1}}$

Step 5.3.4.5

Multiply $8$ by $1$ .

$f (\frac{8 x}{1 - x}) = \frac{8 x}{1 - x} \cdot \frac{1}{\frac{8}{- x + 1}}$

Step 5.3.5

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{8 x}{1 - x}) = \frac{8 x}{1 - x} \cdot (1 (\frac{- x + 1}{8}))$

Step 5.3.6

Multiply $\frac{- x + 1}{8}$ by $1$ .

$f (\frac{8 x}{1 - x}) = \frac{8 x}{1 - x} \cdot \frac{- x + 1}{8}$

Step 5.3.7

Cancel the common factor of $8$ .

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Step 5.3.7.1

Factor $8$ out of $8 x$ .

$f (\frac{8 x}{1 - x}) = \frac{8 (x)}{1 - x} \cdot \frac{- x + 1}{8}$

Step 5.3.7.2

Cancel the common factor.

$f (\frac{8 x}{1 - x}) = \frac{8 x}{1 - x} \cdot \frac{- x + 1}{8}$

Step 5.3.7.3

Rewrite the expression.

$f (\frac{8 x}{1 - x}) = \frac{x}{1 - x} \cdot (- x + 1)$

Step 5.3.8

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

$f (\frac{8 x}{1 - x}) = \frac{x (- x + 1)}{1 - x}$

Step 5.3.9

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

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Step 5.3.9.1

Reorder terms.

$f (\frac{8 x}{1 - x}) = \frac{x (- x + 1)}{- x + 1}$

Step 5.3.9.2

Cancel the common factor.

$f (\frac{8 x}{1 - x}) = \frac{x (- x + 1)}{- x + 1}$

Step 5.3.9.3

Divide $x$ by $1$ .

$f (\frac{8 x}{1 - x}) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{8 x}{1 - x}$ is the inverse of $f (x) = \frac{x}{x + 8}$ .

$f^{- 1} (x) = \frac{8 x}{1 - x}$