Precalculus Examples

$f (x) = \frac{2}{x - 2}$

Step 1

Write $f (x) = \frac{2}{x - 2}$ as an equation.

$y = \frac{2}{x - 2}$

Step 2

Interchange the variables.

$x = \frac{2}{y - 2}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{2}{y - 2} = x$ .

$\frac{2}{y - 2} = x$

Step 3.2

Find the LCD of the terms in the equation.

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Step 3.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$y - 2, 1$

Step 3.2.2

Remove parentheses.

$y - 2, 1$

Step 3.2.3

The LCM of one and any expression is the expression.

$y - 2$

Step 3.3

Multiply each term in $\frac{2}{y - 2} = x$ by $y - 2$ to eliminate the fractions.

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Step 3.3.1

Multiply each term in $\frac{2}{y - 2} = x$ by $y - 2$ .

$\frac{2}{y - 2} (y - 2) = x (y - 2)$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $y - 2$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{2}{y - 2} (y - 2) = x (y - 2)$

Step 3.3.2.1.2

Rewrite the expression.

$2 = x (y - 2)$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Apply the distributive property.

$2 = x y + x \cdot - 2$

Step 3.3.3.2

Move $- 2$ to the left of $x$ .

$2 = x y - 2 x$

Step 3.4

Solve the equation.

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Step 3.4.1

Rewrite the equation as $x y - 2 x = 2$ .

$x y - 2 x = 2$

Step 3.4.2

Add $2 x$ to both sides of the equation.

$x y = 2 + 2 x$

Step 3.4.3

Divide each term in $x y = 2 + 2 x$ by $x$ and simplify.

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Step 3.4.3.1

Divide each term in $x y = 2 + 2 x$ by $x$ .

$\frac{x y}{x} = \frac{2}{x} + \frac{2 x}{x}$

Step 3.4.3.2

Simplify the left side.

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Step 3.4.3.2.1

Cancel the common factor of $x$ .

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Step 3.4.3.2.1.1

Cancel the common factor.

$\frac{x y}{x} = \frac{2}{x} + \frac{2 x}{x}$

Step 3.4.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{2}{x} + \frac{2 x}{x}$

Step 3.4.3.3

Simplify the right side.

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Step 3.4.3.3.1

Cancel the common factor of $x$ .

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Step 3.4.3.3.1.1

Cancel the common factor.

$y = \frac{2}{x} + \frac{2 x}{x}$

Step 3.4.3.3.1.2

Divide $2$ by $1$ .

$y = \frac{2}{x} + 2$

Step 4

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

$f^{- 1} (x) = \frac{2}{x} + 2$

Step 5

Verify if $f^{- 1} (x) = \frac{2}{x} + 2$ is the inverse of $f (x) = \frac{2}{x - 2}$ .

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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 $f^{- 1} (\frac{2}{x - 2})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{2}{x - 2}) = \frac{2}{\frac{2}{x - 2}} + 2$

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (\frac{2}{x - 2}) = 2 (\frac{x - 2}{2}) + 2$

Step 5.2.3.2

Cancel the common factor of $2$ .

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Step 5.2.3.2.1

Cancel the common factor.

$f^{- 1} (\frac{2}{x - 2}) = 2 (\frac{x - 2}{2}) + 2$

Step 5.2.3.2.2

Rewrite the expression.

$f^{- 1} (\frac{2}{x - 2}) = x - 2 + 2$

Step 5.2.4

Combine the opposite terms in $x - 2 + 2$ .

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Step 5.2.4.1

Add $- 2$ and $2$ .

$f^{- 1} (\frac{2}{x - 2}) = x + 0$

Step 5.2.4.2

Add $x$ and $0$ .

$f^{- 1} (\frac{2}{x - 2}) = x$

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 $f (\frac{2}{x} + 2)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{2}{x} + 2) = \frac{2}{(\frac{2}{x} + 2) - 2}$

Step 5.3.3

Cancel the common factor of $2$ and $(\frac{2}{x} + 2) - 2$ .

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Step 5.3.3.1

Factor $2$ out of $2$ .

$f (\frac{2}{x} + 2) = \frac{2 (1)}{(\frac{2}{x} + 2) - 2}$

Step 5.3.3.2

Cancel the common factors.

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Step 5.3.3.2.1

Factor $2$ out of $\frac{2}{x}$ .

$f (\frac{2}{x} + 2) = \frac{2 \cdot 1}{2 (\frac{1}{x}) + 2 - 2}$

Step 5.3.3.2.2

Factor $2$ out of $2$ .

$f (\frac{2}{x} + 2) = \frac{2 (1)}{2 (\frac{1}{x}) + 2 \cdot 1 - 2}$

Step 5.3.3.2.3

Factor $2$ out of $2 \frac{1}{x} + 2 \cdot 1$ .

$f (\frac{2}{x} + 2) = \frac{2 (1)}{2 (\frac{1}{x} + 1) - 2}$

Step 5.3.3.2.4

Factor $2$ out of $- 2$ .

$f (\frac{2}{x} + 2) = \frac{2 \cdot 1}{2 (\frac{1}{x} + 1) + 2 \cdot - 1}$

Step 5.3.3.2.5

Factor $2$ out of $2 (\frac{1}{x} + 1) + 2 (- 1)$ .

$f (\frac{2}{x} + 2) = \frac{2 (1)}{2 (\frac{1}{x} + 1 - 1)}$

Step 5.3.3.2.6

Cancel the common factor.

$f (\frac{2}{x} + 2) = \frac{2 \cdot 1}{2 (\frac{1}{x} + 1 - 1)}$

Step 5.3.3.2.7

Rewrite the expression.

$f (\frac{2}{x} + 2) = \frac{1}{\frac{1}{x} + 1 - 1}$

Step 5.3.4

Simplify the denominator.

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Step 5.3.4.1

Subtract $1$ from $1$ .

$f (\frac{2}{x} + 2) = \frac{1}{\frac{1}{x} + 0}$

Step 5.3.4.2

Add $\frac{1}{x}$ and $0$ .

$f (\frac{2}{x} + 2) = \frac{1}{\frac{1}{x}}$

Step 5.3.5

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{2}{x} + 2) = 1 x$

Step 5.3.6

Multiply $x$ by $1$ .

$f (\frac{2}{x} + 2) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{2}{x} + 2$ is the inverse of $f (x) = \frac{2}{x - 2}$ .

$f^{- 1} (x) = \frac{2}{x} + 2$