Precalculus Examples

$f (x) = \frac{5 x - 6}{7}$

Step 1

Write $f (x) = \frac{5 x - 6}{7}$ as an equation.

$y = \frac{5 x - 6}{7}$

Step 2

Interchange the variables.

$x = \frac{5 y - 6}{7}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{5 y - 6}{7} = x$ .

$\frac{5 y - 6}{7} = x$

Step 3.2

Multiply both sides by $7$ .

$\frac{5 y - 6}{7} \cdot 7 = x \cdot 7$

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 $7$ .

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Step 3.3.1.1.1

Cancel the common factor.

$\frac{5 y - 6}{7} \cdot 7 = x \cdot 7$

Step 3.3.1.1.2

Rewrite the expression.

$5 y - 6 = x \cdot 7$

Step 3.3.2

Simplify the right side.

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Step 3.3.2.1

Move $7$ to the left of $x$ .

$5 y - 6 = 7 x$

Step 3.4

Solve for $y$ .

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Step 3.4.1

Add $6$ to both sides of the equation.

$5 y = 7 x + 6$

Step 3.4.2

Divide each term in $5 y = 7 x + 6$ by $5$ and simplify.

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Step 3.4.2.1

Divide each term in $5 y = 7 x + 6$ by $5$ .

$\frac{5 y}{5} = \frac{7 x}{5} + \frac{6}{5}$

Step 3.4.2.2

Simplify the left side.

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Step 3.4.2.2.1

Cancel the common factor of $5$ .

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Step 3.4.2.2.1.1

Cancel the common factor.

$\frac{5 y}{5} = \frac{7 x}{5} + \frac{6}{5}$

Step 3.4.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{7 x}{5} + \frac{6}{5}$

Step 4

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

$f^{- 1} (x) = \frac{7 x}{5} + \frac{6}{5}$

Step 5

Verify if $f^{- 1} (x) = \frac{7 x}{5} + \frac{6}{5}$ is the inverse of $f (x) = \frac{5 x - 6}{7}$ .

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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 $f^{- 1} (\frac{5 x - 6}{7})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{5 x - 6}{7}) = \frac{7 (\frac{5 x - 6}{7})}{5} + \frac{6}{5}$

Step 5.2.3

Combine the numerators over the common denominator.

$f^{- 1} (\frac{5 x - 6}{7}) = \frac{7 (\frac{5 x - 6}{7}) + 6}{5}$

Step 5.2.4

Cancel the common factor of $7$ .

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Step 5.2.4.1

Cancel the common factor.

$f^{- 1} (\frac{5 x - 6}{7}) = \frac{7 (\frac{5 x - 6}{7}) + 6}{5}$

Step 5.2.4.2

Rewrite the expression.

$f^{- 1} (\frac{5 x - 6}{7}) = \frac{5 x - 6 + 6}{5}$

Step 5.2.5

Combine the opposite terms in $5 x - 6 + 6$ .

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Step 5.2.5.1

Add $- 6$ and $6$ .

$f^{- 1} (\frac{5 x - 6}{7}) = \frac{5 x + 0}{5}$

Step 5.2.5.2

Add $5 x$ and $0$ .

$f^{- 1} (\frac{5 x - 6}{7}) = \frac{5 x}{5}$

Step 5.2.6

Cancel the common factor of $5$ .

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Step 5.2.6.1

Cancel the common factor.

$f^{- 1} (\frac{5 x - 6}{7}) = \frac{5 x}{5}$

Step 5.2.6.2

Divide $x$ by $1$ .

$f^{- 1} (\frac{5 x - 6}{7}) = x$

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 $f (\frac{7 x}{5} + \frac{6}{5})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{7 x}{5} + \frac{6}{5}) = \frac{5 (\frac{7 x}{5} + \frac{6}{5}) - 6}{7}$

Step 5.3.3

Simplify the numerator.

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Step 5.3.3.1

Apply the distributive property.

$f (\frac{7 x}{5} + \frac{6}{5}) = \frac{5 (\frac{7 x}{5}) + 5 (\frac{6}{5}) - 6}{7}$

Step 5.3.3.2

Cancel the common factor of $5$ .

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Step 5.3.3.2.1

Cancel the common factor.

$f (\frac{7 x}{5} + \frac{6}{5}) = \frac{5 (\frac{7 x}{5}) + 5 (\frac{6}{5}) - 6}{7}$

Step 5.3.3.2.2

Rewrite the expression.

$f (\frac{7 x}{5} + \frac{6}{5}) = \frac{7 x + 5 (\frac{6}{5}) - 6}{7}$

Step 5.3.3.3

Cancel the common factor of $5$ .

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Step 5.3.3.3.1

Cancel the common factor.

$f (\frac{7 x}{5} + \frac{6}{5}) = \frac{7 x + 5 (\frac{6}{5}) - 6}{7}$

Step 5.3.3.3.2

Rewrite the expression.

$f (\frac{7 x}{5} + \frac{6}{5}) = \frac{7 x + 6 - 6}{7}$

Step 5.3.3.4

Subtract $6$ from $6$ .

$f (\frac{7 x}{5} + \frac{6}{5}) = \frac{7 x + 0}{7}$

Step 5.3.3.5

Add $7 x$ and $0$ .

$f (\frac{7 x}{5} + \frac{6}{5}) = \frac{7 x}{7}$

Step 5.3.4

Cancel the common factor of $7$ .

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Step 5.3.4.1

Cancel the common factor.

$f (\frac{7 x}{5} + \frac{6}{5}) = \frac{7 x}{7}$

Step 5.3.4.2

Divide $x$ by $1$ .

$f (\frac{7 x}{5} + \frac{6}{5}) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{7 x}{5} + \frac{6}{5}$ is the inverse of $f (x) = \frac{5 x - 6}{7}$ .

$f^{- 1} (x) = \frac{7 x}{5} + \frac{6}{5}$