Precalculus Examples

$f (x) = - \frac{1}{3} x + 5$

Step 1

Write $f (x) = - \frac{1}{3} x + 5$ as an equation.

$y = - \frac{1}{3} x + 5$

Step 2

Interchange the variables.

$x = - \frac{1}{3} y + 5$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $- \frac{1}{3} y + 5 = x$ .

$- \frac{1}{3} y + 5 = x$

Step 3.2

Combine $y$ and $\frac{1}{3}$ .

$- \frac{y}{3} + 5 = x$

Step 3.3

Subtract $5$ from both sides of the equation.

$- \frac{y}{3} = x - 5$

Step 3.4

Multiply both sides of the equation by $- 3$ .

$- 3 (- \frac{y}{3}) = - 3 (x - 5)$

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 $- 3 (- \frac{y}{3})$ .

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Step 3.5.1.1.1

Cancel the common factor of $3$ .

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Step 3.5.1.1.1.1

Move the leading negative in $- \frac{y}{3}$ into the numerator.

$- 3 \frac{- y}{3} = - 3 (x - 5)$

Step 3.5.1.1.1.2

Factor $3$ out of $- 3$ .

$3 (- 1) \frac{- y}{3} = - 3 (x - 5)$

Step 3.5.1.1.1.3

Cancel the common factor.

$3 \cdot - 1 \frac{- y}{3} = - 3 (x - 5)$

Step 3.5.1.1.1.4

Rewrite the expression.

$- - y = - 3 (x - 5)$

Step 3.5.1.1.2

Multiply.

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Step 3.5.1.1.2.1

Multiply $- 1$ by $- 1$ .

$1 y = - 3 (x - 5)$

Step 3.5.1.1.2.2

Multiply $y$ by $1$ .

$y = - 3 (x - 5)$

Step 3.5.2

Simplify the right side.

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Step 3.5.2.1

Simplify $- 3 (x - 5)$ .

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Step 3.5.2.1.1

Apply the distributive property.

$y = - 3 x - 3 \cdot - 5$

Step 3.5.2.1.2

Multiply $- 3$ by $- 5$ .

$y = - 3 x + 15$

Step 4

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

$f^{- 1} (x) = - 3 x + 15$

Step 5

Verify if $f^{- 1} (x) = - 3 x + 15$ is the inverse of $f (x) = - \frac{1}{3} x + 5$ .

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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 $f^{- 1} (- \frac{1}{3} x + 5)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (- \frac{1}{3} x + 5) = - 3 (- \frac{1}{3} x + 5) + 15$

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Combine $x$ and $\frac{1}{3}$ .

$f^{- 1} (- \frac{1}{3} x + 5) = - 3 (- \frac{x}{3} + 5) + 15$

Step 5.2.3.2

Apply the distributive property.

$f^{- 1} (- \frac{1}{3} x + 5) = - 3 (- \frac{x}{3}) - 3 \cdot 5 + 15$

Step 5.2.3.3

Cancel the common factor of $3$ .

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Step 5.2.3.3.1

Move the leading negative in $- \frac{x}{3}$ into the numerator.

$f^{- 1} (- \frac{1}{3} x + 5) = - 3 \frac{- x}{3} - 3 \cdot 5 + 15$

Step 5.2.3.3.2

Factor $3$ out of $- 3$ .

$f^{- 1} (- \frac{1}{3} x + 5) = 3 (- 1) (\frac{- x}{3}) - 3 \cdot 5 + 15$

Step 5.2.3.3.3

Cancel the common factor.

$f^{- 1} (- \frac{1}{3} x + 5) = 3 \cdot (- 1 \frac{- x}{3}) - 3 \cdot 5 + 15$

Step 5.2.3.3.4

Rewrite the expression.

$f^{- 1} (- \frac{1}{3} x + 5) = - 1 (- x) - 3 \cdot 5 + 15$

Step 5.2.3.4

Multiply $- 1$ by $- 1$ .

$f^{- 1} (- \frac{1}{3} x + 5) = 1 x - 3 \cdot 5 + 15$

Step 5.2.3.5

Multiply $x$ by $1$ .

$f^{- 1} (- \frac{1}{3} x + 5) = x - 3 \cdot 5 + 15$

Step 5.2.3.6

Multiply $- 3$ by $5$ .

$f^{- 1} (- \frac{1}{3} x + 5) = x - 15 + 15$

Step 5.2.4

Combine the opposite terms in $x - 15 + 15$ .

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Step 5.2.4.1

Add $- 15$ and $15$ .

$f^{- 1} (- \frac{1}{3} x + 5) = x + 0$

Step 5.2.4.2

Add $x$ and $0$ .

$f^{- 1} (- \frac{1}{3} x + 5) = x$

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 $f (- 3 x + 15)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (- 3 x + 15) = - \frac{1}{3} \cdot (- 3 x + 15) + 5$

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Apply the distributive property.

$f (- 3 x + 15) = - \frac{1}{3} \cdot (- 3 x) - \frac{1}{3} \cdot 15 + 5$

Step 5.3.3.2

Cancel the common factor of $3$ .

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Step 5.3.3.2.1

Move the leading negative in $- \frac{1}{3}$ into the numerator.

$f (- 3 x + 15) = \frac{- 1}{3} \cdot (- 3 x) - \frac{1}{3} \cdot 15 + 5$

Step 5.3.3.2.2

Factor $3$ out of $- 3 x$ .

$f (- 3 x + 15) = \frac{- 1}{3} \cdot (3 (- x)) - \frac{1}{3} \cdot 15 + 5$

Step 5.3.3.2.3

Cancel the common factor.

$f (- 3 x + 15) = \frac{- 1}{3} \cdot (3 (- x)) - \frac{1}{3} \cdot 15 + 5$

Step 5.3.3.2.4

Rewrite the expression.

$f (- 3 x + 15) = - 1 (- x) - \frac{1}{3} \cdot 15 + 5$

Step 5.3.3.3

Multiply $- 1$ by $- 1$ .

$f (- 3 x + 15) = 1 x - \frac{1}{3} \cdot 15 + 5$

Step 5.3.3.4

Multiply $x$ by $1$ .

$f (- 3 x + 15) = x - \frac{1}{3} \cdot 15 + 5$

Step 5.3.3.5

Cancel the common factor of $3$ .

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Step 5.3.3.5.1

Move the leading negative in $- \frac{1}{3}$ into the numerator.

$f (- 3 x + 15) = x + \frac{- 1}{3} \cdot 15 + 5$

Step 5.3.3.5.2

Factor $3$ out of $15$ .

$f (- 3 x + 15) = x + \frac{- 1}{3} \cdot (3 (5)) + 5$

Step 5.3.3.5.3

Cancel the common factor.

$f (- 3 x + 15) = x + \frac{- 1}{3} \cdot (3 \cdot 5) + 5$

Step 5.3.3.5.4

Rewrite the expression.

$f (- 3 x + 15) = x - 1 \cdot 5 + 5$

Step 5.3.3.6

Multiply $- 1$ by $5$ .

$f (- 3 x + 15) = x - 5 + 5$

Step 5.3.4

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

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Step 5.3.4.1

Add $- 5$ and $5$ .

$f (- 3 x + 15) = x + 0$

Step 5.3.4.2

Add $x$ and $0$ .

$f (- 3 x + 15) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - 3 x + 15$ is the inverse of $f (x) = - \frac{1}{3} x + 5$ .

$f^{- 1} (x) = - 3 x + 15$