Precalculus Examples

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

Step 1

Write $f (x) = \frac{x}{2 x + 1}$ as an equation.

$y = \frac{x}{2 x + 1}$

Step 2

Interchange the variables.

$x = \frac{y}{2 y + 1}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{y}{2 y + 1} = x$ .

$\frac{y}{2 y + 1} = x$

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.

$2 y + 1, 1$

Step 3.2.2

Remove parentheses.

$2 y + 1, 1$

Step 3.2.3

The LCM of one and any expression is the expression.

$2 y + 1$

Step 3.3

Multiply each term in $\frac{y}{2 y + 1} = x$ by $2 y + 1$ to eliminate the fractions.

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Step 3.3.1

Multiply each term in $\frac{y}{2 y + 1} = x$ by $2 y + 1$ .

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

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $2 y + 1$ .

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Step 3.3.2.1.1

Cancel the common factor.

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

Step 3.3.2.1.2

Rewrite the expression.

$y = x (2 y + 1)$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Apply the distributive property.

$y = x (2 y) + x \cdot 1$

Step 3.3.3.2

Simplify the expression.

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Step 3.3.3.2.1

Rewrite using the commutative property of multiplication.

$y = 2 x y + x \cdot 1$

Step 3.3.3.2.2

Multiply $x$ by $1$ .

$y = 2 x y + x$

Step 3.4

Solve the equation.

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Step 3.4.1

Subtract $2 x y$ from both sides of the equation.

$y - 2 x y = x$

Step 3.4.2

Factor $y$ out of $y - 2 x y$ .

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Step 3.4.2.1

Factor $y$ out of $y^{1}$ .

$y \cdot 1 - 2 x y = x$

Step 3.4.2.2

Factor $y$ out of $- 2 x y$ .

$y \cdot 1 + y (- 2 x) = x$

Step 3.4.2.3

Factor $y$ out of $y \cdot 1 + y (- 2 x)$ .

$y (1 - 2 x) = x$

Step 3.4.3

Divide each term in $y (1 - 2 x) = x$ by $1 - 2 x$ and simplify.

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Step 3.4.3.1

Divide each term in $y (1 - 2 x) = x$ by $1 - 2 x$ .

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

Step 3.4.3.2

Simplify the left side.

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Step 3.4.3.2.1

Cancel the common factor of $1 - 2 x$ .

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Step 3.4.3.2.1.1

Cancel the common factor.

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

Step 3.4.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{x}{1 - 2 x}$

Step 4

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

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

Step 5

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

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

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

Step 5.2.3

Multiply the numerator by the reciprocal of the denominator.

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

Step 5.2.4

Simplify the denominator.

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Step 5.2.4.1

Combine $- 2$ and $\frac{x}{2 x + 1}$ .

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

Step 5.2.4.2

Move the negative in front of the fraction.

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

Step 5.2.4.3

Write $1$ as a fraction with a common denominator.

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

Step 5.2.4.4

Combine the numerators over the common denominator.

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

Step 5.2.4.5

Reorder terms.

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

Step 5.2.4.6

Rewrite $\frac{2 x - 2 x + 1}{2 x + 1}$ in a factored form.

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Step 5.2.4.6.1

Subtract $2 x$ from $2 x$ .

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

Step 5.2.4.6.2

Add $0$ and $1$ .

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

Step 5.2.5

Multiply the numerator by the reciprocal of the denominator.

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

Step 5.2.6

Cancel the common factor of $2 x + 1$ .

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Step 5.2.6.1

Factor $2 x + 1$ out of $1 (2 x + 1)$ .

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

Step 5.2.6.2

Cancel the common factor.

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

Step 5.2.6.3

Rewrite the expression.

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

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

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

Step 5.3.3

Multiply the numerator by the reciprocal of the denominator.

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

Step 5.3.4

Simplify the denominator.

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Step 5.3.4.1

Combine $2$ and $\frac{x}{1 - 2 x}$ .

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

Step 5.3.4.2

Write $1$ as a fraction with a common denominator.

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

Step 5.3.4.3

Combine the numerators over the common denominator.

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

Step 5.3.4.4

Reorder terms.

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

Step 5.3.4.5

Rewrite $\frac{2 x - 2 x + 1}{- 2 x + 1}$ in a factored form.

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Step 5.3.4.5.1

Subtract $2 x$ from $2 x$ .

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

Step 5.3.4.5.2

Add $0$ and $1$ .

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

Step 5.3.5

Multiply the numerator by the reciprocal of the denominator.

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

Step 5.3.6

Multiply $- 2 x + 1$ by $1$ .

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

Step 5.3.7

Multiply $\frac{x}{1 - 2 x}$ by $- 2 x + 1$ .

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

Step 5.3.8

Cancel the common factor of $- 2 x + 1$ and $1 - 2 x$ .

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Step 5.3.8.1

Reorder terms.

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

Step 5.3.8.2

Cancel the common factor.

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

Step 5.3.8.3

Divide $x$ by $1$ .

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

Step 5.4

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

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