Precalculus Examples

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

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

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

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Step 3.3.1

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

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

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.

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

Step 3.3.2.1.2

Rewrite the expression.

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

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Apply the distributive property.

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

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.

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

Step 3.3.3.2.2

Multiply $x$ by $1$ .

$y + 1 = 2 x y + x$

Step 3.4

Solve the equation.

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Step 3.4.1

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

$y + 1 - 2 x y = x$

Step 3.4.2

Subtract $1$ from both sides of the equation.

$y - 2 x y = x - 1$

Step 3.4.3

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

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Step 3.4.3.1

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

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

Step 3.4.3.2

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

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

Step 3.4.3.3

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

$y (1 - 2 x) = x - 1$

Step 3.4.4

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

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Step 3.4.4.1

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

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

Step 3.4.4.2

Simplify the left side.

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Step 3.4.4.2.1

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

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Step 3.4.4.2.1.1

Cancel the common factor.

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

Step 3.4.4.2.1.2

Divide $y$ by $1$ .

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

Step 3.4.4.3

Simplify the right side.

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Step 3.4.4.3.1

Combine the numerators over the common denominator.

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

Step 4

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

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

Step 5

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

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

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

Step 5.2.3

Multiply the numerator and denominator of the fraction by $2 x + 1$ .

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Step 5.2.3.1

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

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

Step 5.2.3.2

Combine.

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

Step 5.2.4

Apply the distributive property.

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

Step 5.2.5

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

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Step 5.2.5.1

Cancel the common factor.

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

Step 5.2.5.2

Rewrite the expression.

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

Step 5.2.6

Simplify the numerator.

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Step 5.2.6.1

Apply the distributive property.

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

Step 5.2.6.2

Multiply $- 1$ by $2$ .

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

Step 5.2.6.3

Multiply $- 1$ by $1$ .

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

Step 5.2.6.4

Subtract $2 x$ from $x$ .

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

Step 5.2.6.5

Subtract $1$ from $1$ .

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

Step 5.2.6.6

Add $- x$ and $0$ .

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

Step 5.2.7

Simplify the denominator.

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Step 5.2.7.1

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

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Step 5.2.7.1.1

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

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

Step 5.2.7.1.2

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

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

Step 5.2.7.2

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

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

Step 5.2.7.3

Move the negative in front of the fraction.

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

Step 5.2.7.4

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

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

Step 5.2.7.5

Combine the numerators over the common denominator.

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

Step 5.2.7.6

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

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Step 5.2.7.6.1

Apply the distributive property.

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

Step 5.2.7.6.2

Multiply $- 2$ by $1$ .

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

Step 5.2.7.6.3

Subtract $2 x$ from $2 x$ .

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

Step 5.2.7.6.4

Add $0$ and $1$ .

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

Step 5.2.7.6.5

Subtract $2$ from $1$ .

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

Step 5.2.7.7

Move the negative in front of the fraction.

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

Step 5.2.7.8

Factor out negative.

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

Step 5.2.8

Dividing two negative values results in a positive value.

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

Step 5.2.9

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

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

Step 5.2.10

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

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Step 5.2.10.1

Cancel the common factor.

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

Step 5.2.10.2

Rewrite the expression.

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

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

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

Step 5.3.3

Multiply the numerator and denominator of the fraction by $1 - 2 x$ .

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Step 5.3.3.1

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

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

Step 5.3.3.2

Combine.

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

Step 5.3.4

Apply the distributive property.

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

Step 5.3.5

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

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Step 5.3.5.1

Cancel the common factor.

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

Step 5.3.5.2

Rewrite the expression.

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

Step 5.3.6

Simplify the numerator.

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Step 5.3.6.1

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

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

Step 5.3.6.2

Subtract $2 x$ from $x$ .

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

Step 5.3.6.3

Add $- 1$ and $1$ .

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

Step 5.3.6.4

Add $- x$ and $0$ .

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

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

Step 5.3.7

Simplify the denominator.

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Step 5.3.7.1

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

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Step 5.3.7.1.1

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

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

Step 5.3.7.1.2

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

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

Step 5.3.7.2

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

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

Step 5.3.7.3

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

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

Step 5.3.7.4

Combine the numerators over the common denominator.

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

Step 5.3.7.5

Reorder terms.

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

Step 5.3.7.6

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

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Step 5.3.7.6.1

Apply the distributive property.

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

Step 5.3.7.6.2

Multiply $2$ by $- 1$ .

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

Step 5.3.7.6.3

Add $- 2 x$ and $2 x$ .

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

Step 5.3.7.6.4

Subtract $2$ from $0$ .

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

Step 5.3.7.6.5

Add $- 2$ and $1$ .

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

Step 5.3.7.7

Move the negative in front of the fraction.

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

Step 5.3.7.8

Factor out negative.

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

Step 5.3.8

Dividing two negative values results in a positive value.

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

Step 5.3.9

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

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

Step 5.3.10

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

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

Step 5.3.11

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

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Step 5.3.11.1

Reorder terms.

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

Step 5.3.11.2

Cancel the common factor.

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

Step 5.3.11.3

Divide $x$ by $1$ .

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

Step 5.4

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

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