Precalculus Examples

$f (x) = \frac{2 - x}{3 + x}$

Step 1

Write $f (x) = \frac{2 - x}{3 + x}$ as an equation.

$y = \frac{2 - x}{3 + x}$

Step 2

Interchange the variables.

$x = \frac{2 - y}{3 + y}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{2 - y}{3 + y} = x$ .

$\frac{2 - y}{3 + y} = x$

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.

$3 + y, 1$

Step 3.2.2

Remove parentheses.

$3 + y, 1$

Step 3.2.3

The LCM of one and any expression is the expression.

$3 + y$

Step 3.3

Multiply each term in $\frac{2 - y}{3 + y} = x$ by $3 + y$ to eliminate the fractions.

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Step 3.3.1

Multiply each term in $\frac{2 - y}{3 + y} = x$ by $3 + y$ .

$\frac{2 - y}{3 + y} (3 + y) = x (3 + y)$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $3 + y$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{2 - y}{3 + y} (3 + y) = x (3 + y)$

Step 3.3.2.1.2

Rewrite the expression.

$2 - y = x (3 + y)$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Apply the distributive property.

$2 - y = x \cdot 3 + x y$

Step 3.3.3.2

Move $3$ to the left of $x$ .

$2 - y = 3 x + x y$

Step 3.4

Solve the equation.

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Step 3.4.1

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

$2 - y - x y = 3 x$

Step 3.4.2

Subtract $2$ from both sides of the equation.

$- y - x y = 3 x - 2$

Step 3.4.3

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

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Step 3.4.3.1

Factor $y$ out of $- y$ .

$y \cdot - 1 - x y = 3 x - 2$

Step 3.4.3.2

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

$y \cdot - 1 + y (- x) = 3 x - 2$

Step 3.4.3.3

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

$y (- 1 - x) = 3 x - 2$

Step 3.4.4

Divide each term in $y (- 1 - x) = 3 x - 2$ by $- 1 - x$ and simplify.

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Step 3.4.4.1

Divide each term in $y (- 1 - x) = 3 x - 2$ by $- 1 - x$ .

$\frac{y (- 1 - x)}{- 1 - x} = \frac{3 x}{- 1 - x} + \frac{- 2}{- 1 - x}$

Step 3.4.4.2

Simplify the left side.

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Step 3.4.4.2.1

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

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Step 3.4.4.2.1.1

Cancel the common factor.

$\frac{y (- 1 - x)}{- 1 - x} = \frac{3 x}{- 1 - x} + \frac{- 2}{- 1 - x}$

Step 3.4.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{3 x}{- 1 - x} + \frac{- 2}{- 1 - x}$

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.

$y = \frac{3 x - 2}{- 1 - x}$

Step 3.4.4.3.2

Rewrite $- 1$ as $- 1 (1)$ .

$y = \frac{3 x - 2}{- 1 (1) - x}$

Step 3.4.4.3.3

Factor $- 1$ out of $- x$ .

$y = \frac{3 x - 2}{- 1 (1) - (x)}$

Step 3.4.4.3.4

Factor $- 1$ out of $- 1 (1) - (x)$ .

$y = \frac{3 x - 2}{- 1 (1 + x)}$

Step 3.4.4.3.5

Move the negative in front of the fraction.

$y = - \frac{3 x - 2}{1 + x}$

Step 4

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

$f^{- 1} (x) = - \frac{3 x - 2}{1 + x}$

Step 5

Verify if $f^{- 1} (x) = - \frac{3 x - 2}{1 + x}$ is the inverse of $f (x) = \frac{2 - x}{3 + x}$ .

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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 $f^{- 1} (\frac{2 - x}{3 + x})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{3 (\frac{2 - x}{3 + x}) - 2}{1 + \frac{2 - x}{3 + x}}$

Step 5.2.3

Remove parentheses.

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{3 (\frac{2 - x}{3 + x}) - 2}{1 + \frac{2 - x}{3 + x}}$

Step 5.2.4

Multiply the numerator and denominator of the fraction by $3 + x$ .

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Step 5.2.4.1

Multiply $\frac{3 \frac{2 - x}{3 + x} - 2}{1 + \frac{2 - x}{3 + x}}$ by $\frac{3 + x}{3 + x}$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - (\frac{3 + x}{3 + x} \cdot \frac{3 (\frac{2 - x}{3 + x}) - 2}{1 + \frac{2 - x}{3 + x}})$

Step 5.2.4.2

Combine.

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) (3 (\frac{2 - x}{3 + x}) - 2)}{(3 + x) (1 + \frac{2 - x}{3 + x})}$

Step 5.2.5

Apply the distributive property.

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) (3 (\frac{2 - x}{3 + x})) + (3 + x) \cdot - 2}{(3 + x) \cdot 1 + (3 + x) (\frac{2 - x}{3 + x})}$

Step 5.2.6

Cancel the common factor of $3 + x$ .

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Step 5.2.6.1

Cancel the common factor.

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) (3 (\frac{2 - x}{3 + x})) + (3 + x) \cdot - 2}{(3 + x) \cdot 1 + (3 + x) (\frac{2 - x}{3 + x})}$

Step 5.2.6.2

Rewrite the expression.

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) (3 (\frac{2 - x}{3 + x})) + (3 + x) \cdot - 2}{(3 + x) \cdot 1 + 2 - x}$

Step 5.2.7

Simplify the numerator.

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Step 5.2.7.1

Factor $3 + x$ out of $(3 + x) \cdot 3 \frac{2 - x}{3 + x} + (3 + x) \cdot - 2$ .

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Step 5.2.7.1.1

Factor $3 + x$ out of $(3 + x) \cdot 3 \frac{2 - x}{3 + x}$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) (3 (\frac{2 - x}{3 + x})) + (3 + x) \cdot - 2}{(3 + x) \cdot 1 + 2 - x}$

Step 5.2.7.1.2

Factor $3 + x$ out of $(3 + x) \cdot - 2$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) (3 (\frac{2 - x}{3 + x})) + (3 + x) (- 2)}{(3 + x) \cdot 1 + 2 - x}$

Step 5.2.7.1.3

Factor $3 + x$ out of $(3 + x) (3 \frac{2 - x}{3 + x}) + (3 + x) (- 2)$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) (3 (\frac{2 - x}{3 + x}) - 2)}{(3 + x) \cdot 1 + 2 - x}$

Step 5.2.7.2

Combine $3$ and $\frac{2 - x}{3 + x}$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) (\frac{3 (2 - x)}{3 + x} - 2)}{(3 + x) \cdot 1 + 2 - x}$

Step 5.2.7.3

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{3 + x}{3 + x}$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) (\frac{3 (2 - x)}{3 + x} - 2 \cdot \frac{3 + x}{3 + x})}{(3 + x) \cdot 1 + 2 - x}$

Step 5.2.7.4

Combine $- 2$ and $\frac{3 + x}{3 + x}$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) (\frac{3 (2 - x)}{3 + x} + \frac{- 2 (3 + x)}{3 + x})}{(3 + x) \cdot 1 + 2 - x}$

Step 5.2.7.5

Combine the numerators over the common denominator.

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) (\frac{3 (2 - x) - 2 (3 + x)}{3 + x})}{(3 + x) \cdot 1 + 2 - x}$

Step 5.2.7.6

Reorder terms.

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) (\frac{3 (2 - x) - 2 (3 + x)}{x + 3})}{(3 + x) \cdot 1 + 2 - x}$

Step 5.2.7.7

Rewrite $\frac{3 (2 - x) - 2 (3 + x)}{x + 3}$ in a factored form.

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Step 5.2.7.7.1

Apply the distributive property.

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) (\frac{3 \cdot 2 + 3 (- x) - 2 (3 + x)}{x + 3})}{(3 + x) \cdot 1 + 2 - x}$

Step 5.2.7.7.2

Multiply $3$ by $2$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) (\frac{6 + 3 (- x) - 2 (3 + x)}{x + 3})}{(3 + x) \cdot 1 + 2 - x}$

Step 5.2.7.7.3

Multiply $- 1$ by $3$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) (\frac{6 - 3 x - 2 (3 + x)}{x + 3})}{(3 + x) \cdot 1 + 2 - x}$

Step 5.2.7.7.4

Apply the distributive property.

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) (\frac{6 - 3 x - 2 \cdot 3 - 2 x}{x + 3})}{(3 + x) \cdot 1 + 2 - x}$

Step 5.2.7.7.5

Multiply $- 2$ by $3$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) (\frac{6 - 3 x - 6 - 2 x}{x + 3})}{(3 + x) \cdot 1 + 2 - x}$

Step 5.2.7.7.6

Subtract $6$ from $6$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) (\frac{- 3 x + 0 - 2 x}{x + 3})}{(3 + x) \cdot 1 + 2 - x}$

Step 5.2.7.7.7

Add $- 3 x$ and $0$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) (\frac{- 3 x - 2 x}{x + 3})}{(3 + x) \cdot 1 + 2 - x}$

Step 5.2.7.7.8

Subtract $2 x$ from $- 3 x$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) (\frac{- 5 x}{x + 3})}{(3 + x) \cdot 1 + 2 - x}$

Step 5.2.7.8

Move the negative in front of the fraction.

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) (- \frac{(5) x}{x + 3})}{(3 + x) \cdot 1 + 2 - x}$

Step 5.2.7.9

Remove unnecessary parentheses.

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{(3 + x) \cdot (- 1 \frac{5 x}{x + 3})}{(3 + x) \cdot 1 + 2 - x}$

Step 5.2.7.10

Factor out negative.

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{- (3 + x) \frac{5 x}{x + 3}}{(3 + x) \cdot 1 + 2 - x}$

Step 5.2.8

Simplify the denominator.

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Step 5.2.8.1

Multiply $3 + x$ by $1$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{- (3 + x) \frac{5 x}{x + 3}}{3 + x + 2 - x}$

Step 5.2.8.2

Add $3$ and $2$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{- (3 + x) \frac{5 x}{x + 3}}{x + 5 - x}$

Step 5.2.8.3

Subtract $x$ from $x$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{- (3 + x) \frac{5 x}{x + 3}}{0 + 5}$

Step 5.2.8.4

Add $0$ and $5$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{- (3 + x) \frac{5 x}{x + 3}}{5}$

Step 5.2.9

Simplify terms.

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Step 5.2.9.1

Factor $\frac{5 x}{x + 3}$ out of $\frac{- (3 + x) \frac{5 x}{x + 3}}{5}$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - (\frac{5 x}{x + 3} \cdot \frac{- (3 + x)}{5})$

Step 5.2.9.2

Cancel the common factor of $5$ .

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Step 5.2.9.2.1

Factor $5$ out of $5 x$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - (\frac{5 (x)}{x + 3} \cdot \frac{- (3 + x)}{5})$

Step 5.2.9.2.2

Cancel the common factor.

$f^{- 1} (\frac{2 - x}{3 + x}) = - (\frac{5 x}{x + 3} \cdot \frac{- (3 + x)}{5})$

Step 5.2.9.2.3

Rewrite the expression.

$f^{- 1} (\frac{2 - x}{3 + x}) = - (\frac{x}{x + 3} \cdot (- (3 + x)))$

Step 5.2.9.3

Rewrite using the commutative property of multiplication.

$f^{- 1} (\frac{2 - x}{3 + x}) = - (- \frac{x}{x + 3} \cdot (3 + x))$

Step 5.2.9.4

Apply the distributive property.

$f^{- 1} (\frac{2 - x}{3 + x}) = - (- \frac{x}{x + 3} \cdot 3 - \frac{x}{x + 3} \cdot x)$

Step 5.2.10

Multiply $- \frac{x}{x + 3} \cdot 3$ .

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Step 5.2.10.1

Multiply $3$ by $- 1$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - (- 3 \frac{x}{x + 3} - \frac{x}{x + 3} \cdot x)$

Step 5.2.10.2

Combine $- 3$ and $\frac{x}{x + 3}$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - (\frac{- 3 x}{x + 3} - \frac{x}{x + 3} \cdot x)$

Step 5.2.11

Multiply $- \frac{x}{x + 3} x$ .

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Step 5.2.11.1

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

$f^{- 1} (\frac{2 - x}{3 + x}) = - (\frac{- 3 x}{x + 3} - \frac{x \cdot x}{x + 3})$

Step 5.2.11.2

Raise $x$ to the power of $1$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - (\frac{- 3 x}{x + 3} - \frac{x \cdot x}{x + 3})$

Step 5.2.11.3

Raise $x$ to the power of $1$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - (\frac{- 3 x}{x + 3} - \frac{x \cdot x}{x + 3})$

Step 5.2.11.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f^{- 1} (\frac{2 - x}{3 + x}) = - (\frac{- 3 x}{x + 3} - \frac{x^{1 + 1}}{x + 3})$

Step 5.2.11.5

Add $1$ and $1$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - (\frac{- 3 x}{x + 3} - \frac{x^{2}}{x + 3})$

Step 5.2.12

Combine the numerators over the common denominator.

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{- 3 x - x^{2}}{x + 3}$

Step 5.2.13

Reorder $- 3 x$ and $- x^{2}$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{- x^{2} - 3 x}{x + 3}$

Step 5.2.14

Factor $x$ out of $- x^{2} - 3 x$ .

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Step 5.2.14.1

Factor $x$ out of $- x^{2}$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{x (- x) - 3 x}{x + 3}$

Step 5.2.14.2

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

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{x (- x) + x \cdot - 3}{x + 3}$

Step 5.2.14.3

Factor $x$ out of $x (- x) + x \cdot - 3$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{x (- x - 3)}{x + 3}$

Step 5.2.15

Cancel the common factor of $- x - 3$ and $x + 3$ .

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Step 5.2.15.1

Factor $- 1$ out of $- x$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{x (- (x) - 3)}{x + 3}$

Step 5.2.15.2

Rewrite $- 3$ as $- 1 (3)$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{x (- (x) - 1 \cdot 3)}{x + 3}$

Step 5.2.15.3

Factor $- 1$ out of $- (x) - 1 (3)$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{x (- (x + 3))}{x + 3}$

Step 5.2.15.4

Rewrite $- (x + 3)$ as $- 1 (x + 3)$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{x (- 1 (x + 3))}{x + 3}$

Step 5.2.15.5

Cancel the common factor.

$f^{- 1} (\frac{2 - x}{3 + x}) = - \frac{x (- 1 (x + 3))}{x + 3}$

Step 5.2.15.6

Divide $x \cdot (- 1)$ by $1$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - (x \cdot (- 1))$

Step 5.2.16

Simplify the expression.

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Step 5.2.16.1

Move $- 1$ to the left of $x$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = - (- 1 x)$

Step 5.2.16.2

Rewrite $- 1 x$ as $- x$ .

$f^{- 1} (\frac{2 - x}{3 + x}) = x$

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 $f (- \frac{3 x - 2}{1 + x})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (- \frac{3 x - 2}{1 + x}) = \frac{2 - (- \frac{3 x - 2}{1 + x})}{3 - \frac{3 x - 2}{1 + x}}$

Step 5.3.3

Remove parentheses.

$f (- \frac{3 x - 2}{1 + x}) = \frac{2 - (- \frac{3 x - 2}{1 + x})}{3 - \frac{3 x - 2}{1 + x}}$

Step 5.3.4

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

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Step 5.3.4.1

Multiply $\frac{2 - - \frac{3 x - 2}{1 + x}}{3 - \frac{3 x - 2}{1 + x}}$ by $\frac{1 + x}{1 + x}$ .

$f (- \frac{3 x - 2}{1 + x}) = \frac{1 + x}{1 + x} \cdot \frac{2 + \frac{3 x - 2}{1 + x}}{3 - \frac{3 x - 2}{1 + x}}$

Step 5.3.4.2

Combine.

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (2 + \frac{3 x - 2}{1 + x})}{(1 + x) (3 - \frac{3 x - 2}{1 + x})}$

Step 5.3.5

Apply the distributive property.

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) \cdot 2 + (1 + x) (\frac{3 x - 2}{1 + x})}{(1 + x) \cdot 3 + (1 + x) (- \frac{3 x - 2}{1 + x})}$

Step 5.3.6

Cancel the common factor of $1 + x$ .

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Step 5.3.6.1

Move the leading negative in $- \frac{3 x - 2}{1 + x}$ into the numerator.

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) \cdot 2 + (1 + x) (\frac{3 x - 2}{1 + x})}{(1 + x) \cdot 3 + (1 + x) (\frac{- (3 x - 2)}{1 + x})}$

Step 5.3.6.2

Cancel the common factor.

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) \cdot 2 + (1 + x) (\frac{3 x - 2}{1 + x})}{(1 + x) \cdot 3 + (1 + x) (\frac{- (3 x - 2)}{1 + x})}$

Step 5.3.6.3

Rewrite the expression.

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) \cdot 2 + (1 + x) (\frac{3 x - 2}{1 + x})}{(1 + x) \cdot 3 - (3 x - 2)}$

Step 5.3.7

Simplify the numerator.

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Step 5.3.7.1

Factor $1 + x$ out of $(1 + x) \cdot 2 + (1 + x) (- - \frac{3 x - 2}{1 + x})$ .

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Step 5.3.7.1.1

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

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (2) + (1 + x) (\frac{3 x - 2}{1 + x})}{(1 + x) \cdot 3 - (3 x - 2)}$

Step 5.3.7.1.2

Factor $1 + x$ out of $(1 + x) (2) + (1 + x) (- - \frac{3 x - 2}{1 + x})$ .

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (2 + \frac{3 x - 2}{1 + x})}{(1 + x) \cdot 3 - (3 x - 2)}$

Step 5.3.7.2

Multiply $- - \frac{3 x - 2}{1 + x}$ .

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Step 5.3.7.2.1

Multiply $- 1$ by $- 1$ .

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (2 + 1 (\frac{3 x - 2}{1 + x}))}{(1 + x) \cdot 3 - (3 x - 2)}$

Step 5.3.7.2.2

Multiply $\frac{3 x - 2}{1 + x}$ by $1$ .

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (2 + \frac{3 x - 2}{1 + x})}{(1 + x) \cdot 3 - (3 x - 2)}$

Step 5.3.7.3

To write $2$ as a fraction with a common denominator, multiply by $\frac{1 + x}{1 + x}$ .

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (\frac{2 (1 + x)}{1 + x} + \frac{3 x - 2}{1 + x})}{(1 + x) \cdot 3 - (3 x - 2)}$

Step 5.3.7.4

Combine the numerators over the common denominator.

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (\frac{2 (1 + x) + 3 x - 2}{1 + x})}{(1 + x) \cdot 3 - (3 x - 2)}$

Step 5.3.7.5

Reorder terms.

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (\frac{3 x + 2 (1 + x) - 2}{x + 1})}{(1 + x) \cdot 3 - (3 x - 2)}$

Step 5.3.7.6

Rewrite $\frac{3 x + 2 (1 + x) - 2}{x + 1}$ in a factored form.

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Step 5.3.7.6.1

Apply the distributive property.

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (\frac{3 x + 2 \cdot 1 + 2 x - 2}{x + 1})}{(1 + x) \cdot 3 - (3 x - 2)}$

Step 5.3.7.6.2

Multiply $2$ by $1$ .

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (\frac{3 x + 2 + 2 x - 2}{x + 1})}{(1 + x) \cdot 3 - (3 x - 2)}$

Step 5.3.7.6.3

Add $3 x$ and $2 x$ .

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (\frac{5 x + 2 - 2}{x + 1})}{(1 + x) \cdot 3 - (3 x - 2)}$

Step 5.3.7.6.4

Subtract $2$ from $2$ .

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (\frac{5 x + 0}{x + 1})}{(1 + x) \cdot 3 - (3 x - 2)}$

Step 5.3.7.6.5

Add $5 x$ and $0$ .

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (\frac{5 x}{x + 1})}{(1 + x) \cdot 3 - (3 x - 2)}$

Step 5.3.8

Simplify the denominator.

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Step 5.3.8.1

Apply the distributive property.

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (\frac{5 x}{x + 1})}{1 \cdot 3 + x \cdot 3 - (3 x - 2)}$

Step 5.3.8.2

Multiply $3$ by $1$ .

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (\frac{5 x}{x + 1})}{3 + x \cdot 3 - (3 x - 2)}$

Step 5.3.8.3

Move $3$ to the left of $x$ .

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (\frac{5 x}{x + 1})}{3 + 3 \cdot x - (3 x - 2)}$

Step 5.3.8.4

Apply the distributive property.

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (\frac{5 x}{x + 1})}{3 + 3 x - (3 x) + 2}$

Step 5.3.8.5

Multiply $3$ by $- 1$ .

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (\frac{5 x}{x + 1})}{3 + 3 x - 3 x + 2}$

Step 5.3.8.6

Multiply $- 1$ by $- 2$ .

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (\frac{5 x}{x + 1})}{3 + 3 x - 3 x + 2}$

Step 5.3.8.7

Add $3$ and $2$ .

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (\frac{5 x}{x + 1})}{3 x - 3 x + 5}$

Step 5.3.8.8

Subtract $3 x$ from $3 x$ .

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (\frac{5 x}{x + 1})}{0 + 5}$

Step 5.3.8.9

Add $0$ and $5$ .

$f (- \frac{3 x - 2}{1 + x}) = \frac{(1 + x) (\frac{5 x}{x + 1})}{5}$

Step 5.3.9

Simplify terms.

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Step 5.3.9.1

Factor $\frac{5 x}{x + 1}$ out of $\frac{(1 + x) \frac{5 x}{x + 1}}{5}$ .

$f (- \frac{3 x - 2}{1 + x}) = \frac{5 x}{x + 1} \cdot \frac{1 + x}{5}$

Step 5.3.9.2

Cancel the common factor of $5$ .

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Step 5.3.9.2.1

Factor $5$ out of $5 x$ .

$f (- \frac{3 x - 2}{1 + x}) = \frac{5 (x)}{x + 1} \cdot \frac{1 + x}{5}$

Step 5.3.9.2.2

Cancel the common factor.

$f (- \frac{3 x - 2}{1 + x}) = \frac{5 x}{x + 1} \cdot \frac{1 + x}{5}$

Step 5.3.9.2.3

Rewrite the expression.

$f (- \frac{3 x - 2}{1 + x}) = \frac{x}{x + 1} \cdot (1 + x)$

Step 5.3.9.3

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

$f (- \frac{3 x - 2}{1 + x}) = \frac{x (1 + x)}{x + 1}$

Step 5.3.9.4

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

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Step 5.3.9.4.1

Reorder terms.

$f (- \frac{3 x - 2}{1 + x}) = \frac{x (x + 1)}{x + 1}$

Step 5.3.9.4.2

Cancel the common factor.

$f (- \frac{3 x - 2}{1 + x}) = \frac{x (x + 1)}{x + 1}$

Step 5.3.9.4.3

Divide $x$ by $1$ .

$f (- \frac{3 x - 2}{1 + x}) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - \frac{3 x - 2}{1 + x}$ is the inverse of $f (x) = \frac{2 - x}{3 + x}$ .

$f^{- 1} (x) = - \frac{3 x - 2}{1 + x}$