Precalculus Examples

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

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

Combine $\frac{1}{4}$ and $y$ .

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

Step 3.3

Subtract $3$ from both sides of the equation.

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

Step 3.4

Multiply both sides of the equation by $4$ .

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

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

Cancel the common factor of $4$ .

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Step 3.5.1.1.1

Cancel the common factor.

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

Step 3.5.1.1.2

Rewrite the expression.

$y = 4 (x - 3)$

Step 3.5.2

Simplify the right side.

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Step 3.5.2.1

Simplify $4 (x - 3)$ .

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Step 3.5.2.1.1

Apply the distributive property.

$y = 4 x + 4 \cdot - 3$

Step 3.5.2.1.2

Multiply $4$ by $- 3$ .

$y = 4 x - 12$

Step 4

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

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

Step 5

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

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

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

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Combine $\frac{1}{4}$ and $x$ .

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

Step 5.2.3.2

Apply the distributive property.

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

Step 5.2.3.3

Cancel the common factor of $4$ .

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Step 5.2.3.3.1

Cancel the common factor.

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

Step 5.2.3.3.2

Rewrite the expression.

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

Step 5.2.3.4

Multiply $4$ by $3$ .

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

Step 5.2.4

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

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Step 5.2.4.1

Subtract $12$ from $12$ .

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

Step 5.2.4.2

Add $x$ and $0$ .

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

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

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

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Apply the distributive property.

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

Step 5.3.3.2

Cancel the common factor of $4$ .

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Step 5.3.3.2.1

Factor $4$ out of $4 x$ .

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

Step 5.3.3.2.2

Cancel the common factor.

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

Step 5.3.3.2.3

Rewrite the expression.

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

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 $- 12$ .

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

Step 5.3.3.3.2

Cancel the common factor.

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

Step 5.3.3.3.3

Rewrite the expression.

$f (4 x - 12) = x - 3 + 3$

Step 5.3.4

Combine the opposite terms in $x - 3 + 3$ .

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Step 5.3.4.1

Add $- 3$ and $3$ .

$f (4 x - 12) = x + 0$

Step 5.3.4.2

Add $x$ and $0$ .

$f (4 x - 12) = x$

Step 5.4

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

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