Precalculus Examples

$f (x) = 4 x - \frac{1}{3}$

Step 1

Write $f (x) = 4 x - \frac{1}{3}$ as an equation.

$y = 4 x - \frac{1}{3}$

Step 2

Interchange the variables.

$x = 4 y - \frac{1}{3}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $4 y - \frac{1}{3} = x$ .

$4 y - \frac{1}{3} = x$

Step 3.2

Add $\frac{1}{3}$ to both sides of the equation.

$4 y = x + \frac{1}{3}$

Step 3.3

Divide each term in $4 y = x + \frac{1}{3}$ by $4$ and simplify.

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Step 3.3.1

Divide each term in $4 y = x + \frac{1}{3}$ by $4$ .

$\frac{4 y}{4} = \frac{x}{4} + \frac{\frac{1}{3}}{4}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $4$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{4 y}{4} = \frac{x}{4} + \frac{\frac{1}{3}}{4}$

Step 3.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{x}{4} + \frac{\frac{1}{3}}{4}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Simplify each term.

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Step 3.3.3.1.1

Multiply the numerator by the reciprocal of the denominator.

$y = \frac{x}{4} + \frac{1}{3} \cdot \frac{1}{4}$

Step 3.3.3.1.2

Multiply $\frac{1}{3} \cdot \frac{1}{4}$ .

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Step 3.3.3.1.2.1

Multiply $\frac{1}{3}$ by $\frac{1}{4}$ .

$y = \frac{x}{4} + \frac{1}{3 \cdot 4}$

Step 3.3.3.1.2.2

Multiply $3$ by $4$ .

$y = \frac{x}{4} + \frac{1}{12}$

Step 4

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

$f^{- 1} (x) = \frac{x}{4} + \frac{1}{12}$

Step 5

Verify if $f^{- 1} (x) = \frac{x}{4} + \frac{1}{12}$ is the inverse of $f (x) = 4 x - \frac{1}{3}$ .

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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 $f^{- 1} (4 x - \frac{1}{3})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (4 x - \frac{1}{3}) = \frac{4 x - \frac{1}{3}}{4} + \frac{1}{12}$

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Simplify the numerator.

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Step 5.2.3.1.1

To write $4 x$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$f^{- 1} (4 x - \frac{1}{3}) = \frac{4 x \cdot \frac{3}{3} - \frac{1}{3}}{4} + \frac{1}{12}$

Step 5.2.3.1.2

Combine $4 x$ and $\frac{3}{3}$ .

$f^{- 1} (4 x - \frac{1}{3}) = \frac{\frac{4 x \cdot 3}{3} - \frac{1}{3}}{4} + \frac{1}{12}$

Step 5.2.3.1.3

Combine the numerators over the common denominator.

$f^{- 1} (4 x - \frac{1}{3}) = \frac{\frac{4 x \cdot 3 - 1}{3}}{4} + \frac{1}{12}$

Step 5.2.3.1.4

Multiply $3$ by $4$ .

$f^{- 1} (4 x - \frac{1}{3}) = \frac{\frac{12 x - 1}{3}}{4} + \frac{1}{12}$

Step 5.2.3.2

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (4 x - \frac{1}{3}) = \frac{12 x - 1}{3} \cdot \frac{1}{4} + \frac{1}{12}$

Step 5.2.3.3

Multiply $\frac{12 x - 1}{3} \cdot \frac{1}{4}$ .

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Step 5.2.3.3.1

Multiply $\frac{12 x - 1}{3}$ by $\frac{1}{4}$ .

$f^{- 1} (4 x - \frac{1}{3}) = \frac{12 x - 1}{3 \cdot 4} + \frac{1}{12}$

Step 5.2.3.3.2

Multiply $3$ by $4$ .

$f^{- 1} (4 x - \frac{1}{3}) = \frac{12 x - 1}{12} + \frac{1}{12}$

Step 5.2.4

Simplify terms.

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Step 5.2.4.1

Combine the numerators over the common denominator.

$f^{- 1} (4 x - \frac{1}{3}) = \frac{12 x - 1 + 1}{12}$

Step 5.2.4.2

Combine the opposite terms in $12 x - 1 + 1$ .

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Step 5.2.4.2.1

Add $- 1$ and $1$ .

$f^{- 1} (4 x - \frac{1}{3}) = \frac{12 x + 0}{12}$

Step 5.2.4.2.2

Add $12 x$ and $0$ .

$f^{- 1} (4 x - \frac{1}{3}) = \frac{12 x}{12}$

Step 5.2.4.3

Cancel the common factor of $12$ .

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Step 5.2.4.3.1

Cancel the common factor.

$f^{- 1} (4 x - \frac{1}{3}) = \frac{12 x}{12}$

Step 5.2.4.3.2

Divide $x$ by $1$ .

$f^{- 1} (4 x - \frac{1}{3}) = x$

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 $f (\frac{x}{4} + \frac{1}{12})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x}{4} + \frac{1}{12}) = 4 (\frac{x}{4} + \frac{1}{12}) - \frac{1}{3}$

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Apply the distributive property.

$f (\frac{x}{4} + \frac{1}{12}) = 4 (\frac{x}{4}) + 4 (\frac{1}{12}) - \frac{1}{3}$

Step 5.3.3.2

Cancel the common factor of $4$ .

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Step 5.3.3.2.1

Cancel the common factor.

$f (\frac{x}{4} + \frac{1}{12}) = 4 (\frac{x}{4}) + 4 (\frac{1}{12}) - \frac{1}{3}$

Step 5.3.3.2.2

Rewrite the expression.

$f (\frac{x}{4} + \frac{1}{12}) = x + 4 (\frac{1}{12}) - \frac{1}{3}$

Step 5.3.3.3

Cancel the common factor of $4$ .

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Step 5.3.3.3.1

Factor $4$ out of $12$ .

$f (\frac{x}{4} + \frac{1}{12}) = x + 4 (\frac{1}{4 (3)}) - \frac{1}{3}$

Step 5.3.3.3.2

Cancel the common factor.

$f (\frac{x}{4} + \frac{1}{12}) = x + 4 (\frac{1}{4 \cdot 3}) - \frac{1}{3}$

Step 5.3.3.3.3

Rewrite the expression.

$f (\frac{x}{4} + \frac{1}{12}) = x + \frac{1}{3} - \frac{1}{3}$

Step 5.3.4

Combine the opposite terms in $x + \frac{1}{3} - \frac{1}{3}$ .

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Step 5.3.4.1

Combine the numerators over the common denominator.

$f (\frac{x}{4} + \frac{1}{12}) = x + \frac{1 - 1}{3}$

Step 5.3.4.2

Subtract $1$ from $1$ .

$f (\frac{x}{4} + \frac{1}{12}) = x + \frac{0}{3}$

Step 5.3.4.3

Divide $0$ by $3$ .

$f (\frac{x}{4} + \frac{1}{12}) = x + 0$

Step 5.3.4.4

Add $x$ and $0$ .

$f (\frac{x}{4} + \frac{1}{12}) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x}{4} + \frac{1}{12}$ is the inverse of $f (x) = 4 x - \frac{1}{3}$ .

$f^{- 1} (x) = \frac{x}{4} + \frac{1}{12}$