Precalculus Examples

$f (x) = - 1.5 x + 250$

Step 1

Write $f (x) = - 1.5 x + 250$ as an equation.

$y = - 1.5 x + 250$

Step 2

Interchange the variables.

$x = - 1.5 y + 250$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $- 1.5 y + 250 = x$ .

$- 1.5 y + 250 = x$

Step 3.2

Subtract $250$ from both sides of the equation.

$- 1.5 y = x - 250$

Step 3.3

Divide each term in $- 1.5 y = x - 250$ by $- 1.5$ and simplify.

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Step 3.3.1

Divide each term in $- 1.5 y = x - 250$ by $- 1.5$ .

$\frac{- 1.5 y}{- 1.5} = \frac{x}{- 1.5} + \frac{- 250}{- 1.5}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $- 1.5$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{- 1.5 y}{- 1.5} = \frac{x}{- 1.5} + \frac{- 250}{- 1.5}$

Step 3.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{x}{- 1.5} + \frac{- 250}{- 1.5}$

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

Move the negative in front of the fraction.

$y = - \frac{x}{1.5} + \frac{- 250}{- 1.5}$

Step 3.3.3.1.2

Multiply by $1$ .

$y = - \frac{1 x}{1.5} + \frac{- 250}{- 1.5}$

Step 3.3.3.1.3

Factor $1.5$ out of $1.5$ .

$y = - \frac{1 x}{1.5 (1)} + \frac{- 250}{- 1.5}$

Step 3.3.3.1.4

Separate fractions.

$y = - (\frac{1}{1.5} \cdot \frac{x}{1}) + \frac{- 250}{- 1.5}$

Step 3.3.3.1.5

Divide $1$ by $1.5$ .

$y = - (0.\overline{6} \frac{x}{1}) + \frac{- 250}{- 1.5}$

Step 3.3.3.1.6

Divide $x$ by $1$ .

$y = - (0.\overline{6} x) + \frac{- 250}{- 1.5}$

Step 3.3.3.1.7

Multiply $0.\overline{6}$ by $- 1$ .

$y = - 0.\overline{6} x + \frac{- 250}{- 1.5}$

Step 3.3.3.1.8

Divide $- 250$ by $- 1.5$ .

$y = - 0.\overline{6} x + 166.\overline{6}$

Step 4

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

$f^{- 1} (x) = - 0.\overline{6} x + 166.\overline{6}$

Step 5

Verify if $f^{- 1} (x) = - 0.\overline{6} x + 166.\overline{6}$ is the inverse of $f (x) = - 1.5 x + 250$ .

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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 $f^{- 1} (- 1.5 x + 250)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (- 1.5 x + 250) = - 0.\overline{6} (- 1.5 x + 250) + 166.\overline{6}$

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Apply the distributive property.

$f^{- 1} (- 1.5 x + 250) = - 0.\overline{6} (- 1.5 x) - 0.\overline{6} \cdot 250 + 166.\overline{6}$

Step 5.2.3.2

Multiply $- 0.\overline{6} (- 1.5 x)$ .

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Step 5.2.3.2.1

Multiply $- 1.5$ by $- 0.\overline{6}$ .

$f^{- 1} (- 1.5 x + 250) = 1 x - 0.\overline{6} \cdot 250 + 166.\overline{6}$

Step 5.2.3.2.2

Multiply $x$ by $1$ .

$f^{- 1} (- 1.5 x + 250) = x - 0.\overline{6} \cdot 250 + 166.\overline{6}$

Step 5.2.3.3

Multiply $- 0.\overline{6}$ by $250$ .

$f^{- 1} (- 1.5 x + 250) = x - 166.\overline{6} + 166.\overline{6}$

Step 5.2.4

Add $- 166.\overline{6}$ and $166.\overline{6}$ .

$f^{- 1} (- 1.5 x + 250) = x 0$

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 $f (- 0.\overline{6} x + 166.\overline{6})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (- 0.\overline{6} x + 166.\overline{6}) = - 1.5 (- 0.\overline{6} x + 166.\overline{6}) + 250$

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Apply the distributive property.

$f (- 0.\overline{6} x + 166.\overline{6}) = - 1.5 (- 0.\overline{6} x) - 1.5 \cdot 166.\overline{6} + 250$

Step 5.3.3.2

Multiply $- 1.5 (- 0.\overline{6} x)$ .

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Step 5.3.3.2.1

Multiply $- 0.\overline{6}$ by $- 1.5$ .

$f (- 0.\overline{6} x + 166.\overline{6}) = 1 x - 1.5 \cdot 166.\overline{6} + 250$

Step 5.3.3.2.2

Multiply $x$ by $1$ .

$f (- 0.\overline{6} x + 166.\overline{6}) = x - 1.5 \cdot 166.\overline{6} + 250$

Step 5.3.3.3

Multiply $- 1.5$ by $166.\overline{6}$ .

$f (- 0.\overline{6} x + 166.\overline{6}) = x - 250 + 250$

Step 5.3.4

Simplify by adding numbers.

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Step 5.3.4.1

Add $- 250$ and $250$ .

$f (- 0.\overline{6} x + 166.\overline{6}) = x + 0$

Step 5.3.4.2

Add $x$ and $0$ .

$f (- 0.\overline{6} x + 166.\overline{6}) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - 0.\overline{6} x + 166.\overline{6}$ is the inverse of $f (x) = - 1.5 x + 250$ .

$f^{- 1} (x) = - 0.\overline{6} x + 166.\overline{6}$