Finite Math Examples

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

Step 1

Write $f (x) = \frac{x - 1}{8}$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{y - 1}{8} = x$ .

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

Step 3.2

Multiply both sides of the equation by $8$ .

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

Step 3.3

Simplify the left side.

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Step 3.3.1

Cancel the common factor of $8$ .

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Step 3.3.1.1

Cancel the common factor.

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

Step 3.3.1.2

Rewrite the expression.

$y - 1 = 8 x$

Step 3.4

Add $1$ to both sides of the equation.

$y = 8 x + 1$

Step 4

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

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

Step 5

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

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

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

Step 5.2.3

Cancel the common factor of $8$ .

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Step 5.2.3.1

Cancel the common factor.

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

Step 5.2.3.2

Rewrite the expression.

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

Step 5.2.4

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

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Step 5.2.4.1

Add $- 1$ and $1$ .

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

Step 5.2.4.2

Add $x$ and $0$ .

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

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

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

Step 5.3.3

Simplify the numerator.

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Step 5.3.3.1

Subtract $1$ from $1$ .

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

Step 5.3.3.2

Add $8 x$ and $0$ .

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

Step 5.3.4

Cancel the common factor of $8$ .

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Step 5.3.4.1

Cancel the common factor.

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

Step 5.3.4.2

Divide $x$ by $1$ .

$f (8 x + 1) = x$

Step 5.4

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

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