Finite Math Examples

$f (x) = (\frac{2}{9}) x - \frac{1}{9}$

Step 1

Write $f (x) = (\frac{2}{9}) x - \frac{1}{9}$ as an equation.

$y = (\frac{2}{9}) x - \frac{1}{9}$

Step 2

Interchange the variables.

$x = (\frac{2}{9}) y - \frac{1}{9}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{2}{9} y - \frac{1}{9} = x$ .

$\frac{2}{9} y - \frac{1}{9} = x$

Step 3.2

Combine $\frac{2}{9}$ and $y$ .

$\frac{2 y}{9} - \frac{1}{9} = x$

Step 3.3

Add $\frac{1}{9}$ to both sides of the equation.

$\frac{2 y}{9} = x + \frac{1}{9}$

Step 3.4

Multiply both sides of the equation by $\frac{9}{2}$ .

$\frac{9}{2} \cdot \frac{2 y}{9} = \frac{9}{2} (x + \frac{1}{9})$

Step 3.5

Simplify both sides of the equation.

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Step 3.5.1

Simplify the left side.

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Step 3.5.1.1

Simplify $\frac{9}{2} \cdot \frac{2 y}{9}$ .

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Step 3.5.1.1.1

Cancel the common factor of $9$ .

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Step 3.5.1.1.1.1

Cancel the common factor.

$\frac{9}{2} \cdot \frac{2 y}{9} = \frac{9}{2} (x + \frac{1}{9})$

Step 3.5.1.1.1.2

Rewrite the expression.

$\frac{1}{2} (2 y) = \frac{9}{2} (x + \frac{1}{9})$

Step 3.5.1.1.2

Cancel the common factor of $2$ .

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Step 3.5.1.1.2.1

Factor $2$ out of $2 y$ .

$\frac{1}{2} (2 (y)) = \frac{9}{2} (x + \frac{1}{9})$

Step 3.5.1.1.2.2

Cancel the common factor.

$\frac{1}{2} (2 y) = \frac{9}{2} (x + \frac{1}{9})$

Step 3.5.1.1.2.3

Rewrite the expression.

$y = \frac{9}{2} (x + \frac{1}{9})$

Step 3.5.2

Simplify the right side.

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Step 3.5.2.1

Simplify $\frac{9}{2} (x + \frac{1}{9})$ .

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Step 3.5.2.1.1

Apply the distributive property.

$y = \frac{9}{2} x + \frac{9}{2} \cdot \frac{1}{9}$

Step 3.5.2.1.2

Combine $\frac{9}{2}$ and $x$ .

$y = \frac{9 x}{2} + \frac{9}{2} \cdot \frac{1}{9}$

Step 3.5.2.1.3

Cancel the common factor of $9$ .

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Step 3.5.2.1.3.1

Cancel the common factor.

$y = \frac{9 x}{2} + \frac{9}{2} \cdot \frac{1}{9}$

Step 3.5.2.1.3.2

Rewrite the expression.

$y = \frac{9 x}{2} + \frac{1}{2}$

Step 4

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

$f^{- 1} (x) = \frac{9 x}{2} + \frac{1}{2}$

Step 5

Verify if $f^{- 1} (x) = \frac{9 x}{2} + \frac{1}{2}$ is the inverse of $f (x) = (\frac{2}{9}) x - \frac{1}{9}$ .

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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 $f^{- 1} ((\frac{2}{9}) x - \frac{1}{9})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} ((\frac{2}{9}) x - \frac{1}{9}) = \frac{9 ((\frac{2}{9}) x - \frac{1}{9})}{2} + \frac{1}{2}$

Step 5.2.3

Combine the numerators over the common denominator.

$f^{- 1} ((\frac{2}{9}) x - \frac{1}{9}) = \frac{9 ((\frac{2}{9}) x - \frac{1}{9}) + 1}{2}$

Step 5.2.4

Simplify each term.

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Step 5.2.4.1

Combine $\frac{2}{9}$ and $x$ .

$f^{- 1} ((\frac{2}{9}) x - \frac{1}{9}) = \frac{9 (\frac{2 x}{9} - \frac{1}{9}) + 1}{2}$

Step 5.2.4.2

Apply the distributive property.

$f^{- 1} ((\frac{2}{9}) x - \frac{1}{9}) = \frac{9 (\frac{2 x}{9}) + 9 (- \frac{1}{9}) + 1}{2}$

Step 5.2.4.3

Cancel the common factor of $9$ .

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Step 5.2.4.3.1

Cancel the common factor.

$f^{- 1} ((\frac{2}{9}) x - \frac{1}{9}) = \frac{9 (\frac{2 x}{9}) + 9 (- \frac{1}{9}) + 1}{2}$

Step 5.2.4.3.2

Rewrite the expression.

$f^{- 1} ((\frac{2}{9}) x - \frac{1}{9}) = \frac{2 x + 9 (- \frac{1}{9}) + 1}{2}$

Step 5.2.4.4

Cancel the common factor of $9$ .

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Step 5.2.4.4.1

Move the leading negative in $- \frac{1}{9}$ into the numerator.

$f^{- 1} ((\frac{2}{9}) x - \frac{1}{9}) = \frac{2 x + 9 (\frac{- 1}{9}) + 1}{2}$

Step 5.2.4.4.2

Cancel the common factor.

$f^{- 1} ((\frac{2}{9}) x - \frac{1}{9}) = \frac{2 x + 9 (\frac{- 1}{9}) + 1}{2}$

Step 5.2.4.4.3

Rewrite the expression.

$f^{- 1} ((\frac{2}{9}) x - \frac{1}{9}) = \frac{2 x - 1 + 1}{2}$

Step 5.2.5

Simplify terms.

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Step 5.2.5.1

Combine the opposite terms in $2 x - 1 + 1$ .

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Step 5.2.5.1.1

Add $- 1$ and $1$ .

$f^{- 1} ((\frac{2}{9}) x - \frac{1}{9}) = \frac{2 x + 0}{2}$

Step 5.2.5.1.2

Add $2 x$ and $0$ .

$f^{- 1} ((\frac{2}{9}) x - \frac{1}{9}) = \frac{2 x}{2}$

Step 5.2.5.2

Cancel the common factor of $2$ .

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Step 5.2.5.2.1

Cancel the common factor.

$f^{- 1} ((\frac{2}{9}) x - \frac{1}{9}) = \frac{2 x}{2}$

Step 5.2.5.2.2

Divide $x$ by $1$ .

$f^{- 1} ((\frac{2}{9}) x - \frac{1}{9}) = x$

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 $f (\frac{9 x}{2} + \frac{1}{2})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{9 x}{2} + \frac{1}{2}) = (\frac{2}{9}) (\frac{9 x}{2} + \frac{1}{2}) - \frac{1}{9}$

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Apply the distributive property.

$f (\frac{9 x}{2} + \frac{1}{2}) = \frac{2}{9} \cdot \frac{9 x}{2} + \frac{2}{9} \cdot \frac{1}{2} - \frac{1}{9}$

Step 5.3.3.2

Cancel the common factor of $2$ .

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Step 5.3.3.2.1

Cancel the common factor.

$f (\frac{9 x}{2} + \frac{1}{2}) = \frac{2}{9} \cdot \frac{9 x}{2} + \frac{2}{9} \cdot \frac{1}{2} - \frac{1}{9}$

Step 5.3.3.2.2

Rewrite the expression.

$f (\frac{9 x}{2} + \frac{1}{2}) = \frac{1}{9} \cdot (9 x) + \frac{2}{9} \cdot \frac{1}{2} - \frac{1}{9}$

Step 5.3.3.3

Cancel the common factor of $9$ .

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Step 5.3.3.3.1

Factor $9$ out of $9 x$ .

$f (\frac{9 x}{2} + \frac{1}{2}) = \frac{1}{9} \cdot (9 (x)) + \frac{2}{9} \cdot \frac{1}{2} - \frac{1}{9}$

Step 5.3.3.3.2

Cancel the common factor.

$f (\frac{9 x}{2} + \frac{1}{2}) = \frac{1}{9} \cdot (9 x) + \frac{2}{9} \cdot \frac{1}{2} - \frac{1}{9}$

Step 5.3.3.3.3

Rewrite the expression.

$f (\frac{9 x}{2} + \frac{1}{2}) = x + \frac{2}{9} \cdot \frac{1}{2} - \frac{1}{9}$

Step 5.3.3.4

Cancel the common factor of $2$ .

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Step 5.3.3.4.1

Cancel the common factor.

$f (\frac{9 x}{2} + \frac{1}{2}) = x + \frac{2}{9} \cdot \frac{1}{2} - \frac{1}{9}$

Step 5.3.3.4.2

Rewrite the expression.

$f (\frac{9 x}{2} + \frac{1}{2}) = x + \frac{1}{9} - \frac{1}{9}$

Step 5.3.4

Combine the opposite terms in $x + \frac{1}{9} - \frac{1}{9}$ .

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Step 5.3.4.1

Combine the numerators over the common denominator.

$f (\frac{9 x}{2} + \frac{1}{2}) = x + \frac{1 - 1}{9}$

Step 5.3.4.2

Subtract $1$ from $1$ .

$f (\frac{9 x}{2} + \frac{1}{2}) = x + \frac{0}{9}$

Step 5.3.4.3

Divide $0$ by $9$ .

$f (\frac{9 x}{2} + \frac{1}{2}) = x + 0$

Step 5.3.4.4

Add $x$ and $0$ .

$f (\frac{9 x}{2} + \frac{1}{2}) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{9 x}{2} + \frac{1}{2}$ is the inverse of $f (x) = (\frac{2}{9}) x - \frac{1}{9}$ .

$f^{- 1} (x) = \frac{9 x}{2} + \frac{1}{2}$