Precalculus Examples

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

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

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

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Step 3.3.1

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

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

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

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

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

Step 3.3.2.2

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

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Step 3.3.2.2.1

Cancel the common factor.

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

Step 3.3.2.2.2

Divide $- 2 y - 1$ by $1$ .

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

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Apply the distributive property.

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

Step 3.3.3.2

Reorder.

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Step 3.3.3.2.1

Rewrite using the commutative property of multiplication.

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

Step 3.3.3.2.2

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

$- 2 y - 1 = - x y + 2 x$

Step 3.4

Solve the equation.

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Step 3.4.1

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

$- 2 y - 1 + x y = 2 x$

Step 3.4.2

Add $1$ to both sides of the equation.

$- 2 y + x y = 2 x + 1$

Step 3.4.3

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

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Step 3.4.3.1

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

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

Step 3.4.3.2

Factor $y$ out of $x y$ .

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

Step 3.4.3.3

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

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

Step 3.4.4

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

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Step 3.4.4.1

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

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

Step 3.4.4.2

Simplify the left side.

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Step 3.4.4.2.1

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

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Step 3.4.4.2.1.1

Cancel the common factor.

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

Step 3.4.4.2.1.2

Divide $y$ by $1$ .

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

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.

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

Step 3.4.4.3.2

Rewrite $- 2$ as $- 1 (2)$ .

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

Step 3.4.4.3.3

Factor $- 1$ out of $x$ .

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

Step 3.4.4.3.4

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

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

Step 3.4.4.3.5

Move the negative in front of the fraction.

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

Step 4

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

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

Step 5

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

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

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

Step 5.2.3

Simplify the numerator.

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Step 5.2.3.1

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

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

Step 5.2.3.2

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

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

Step 5.2.3.3

Combine the numerators over the common denominator.

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

Step 5.2.3.4

Reorder terms.

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

Step 5.2.3.5

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

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Step 5.2.3.5.1

Apply the distributive property.

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

Step 5.2.3.5.2

Multiply $- 2$ by $2$ .

$f^{- 1} (\frac{- 2 x - 1}{- x + 2}) = - \frac{\frac{- x - 4 x + 2 \cdot - 1 + 2}{- x + 2}}{2 - \frac{- 2 x - 1}{- x + 2}}$

Step 5.2.3.5.3

Multiply $2$ by $- 1$ .

$f^{- 1} (\frac{- 2 x - 1}{- x + 2}) = - \frac{\frac{- x - 4 x - 2 + 2}{- x + 2}}{2 - \frac{- 2 x - 1}{- x + 2}}$

Step 5.2.3.5.4

Subtract $4 x$ from $- x$ .

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

Step 5.2.3.5.5

Add $- 2$ and $2$ .

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

Step 5.2.3.5.6

Add $- 5 x$ and $0$ .

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

Step 5.2.3.6

Move the negative in front of the fraction.

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

Step 5.2.4

Simplify the denominator.

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Step 5.2.4.1

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

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

Step 5.2.4.2

Combine the numerators over the common denominator.

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

Step 5.2.4.3

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

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Step 5.2.4.3.1

Apply the distributive property.

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

Step 5.2.4.3.2

Multiply $- 1$ by $2$ .

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

Step 5.2.4.3.3

Multiply $2$ by $2$ .

$f^{- 1} (\frac{- 2 x - 1}{- x + 2}) = - \frac{- \frac{5 x}{- x + 2}}{\frac{- 2 x + 4 - (- 2 x - 1)}{- x + 2}}$

Step 5.2.4.3.4

Apply the distributive property.

$f^{- 1} (\frac{- 2 x - 1}{- x + 2}) = - \frac{- \frac{5 x}{- x + 2}}{\frac{- 2 x + 4 - (- 2 x) + 1}{- x + 2}}$

Step 5.2.4.3.5

Multiply $- 2$ by $- 1$ .

$f^{- 1} (\frac{- 2 x - 1}{- x + 2}) = - \frac{- \frac{5 x}{- x + 2}}{\frac{- 2 x + 4 + 2 x + 1}{- x + 2}}$

Step 5.2.4.3.6

Multiply $- 1$ by $- 1$ .

$f^{- 1} (\frac{- 2 x - 1}{- x + 2}) = - \frac{- \frac{5 x}{- x + 2}}{\frac{- 2 x + 4 + 2 x + 1}{- x + 2}}$

Step 5.2.4.3.7

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

$f^{- 1} (\frac{- 2 x - 1}{- x + 2}) = - \frac{- \frac{5 x}{- x + 2}}{\frac{0 + 4 + 1}{- x + 2}}$

Step 5.2.4.3.8

Add $0$ and $4$ .

$f^{- 1} (\frac{- 2 x - 1}{- x + 2}) = - \frac{- \frac{5 x}{- x + 2}}{\frac{4 + 1}{- x + 2}}$

Step 5.2.4.3.9

Add $4$ and $1$ .

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

Step 5.2.5

Multiply the numerator by the reciprocal of the denominator.

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

Step 5.2.6

Cancel the common factor of $5$ .

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Step 5.2.6.1

Move the leading negative in $- \frac{5 x}{- x + 2}$ into the numerator.

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

Step 5.2.6.2

Factor $5$ out of $- 5 x$ .

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

Step 5.2.6.3

Cancel the common factor.

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

Step 5.2.6.4

Rewrite the expression.

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

Step 5.2.7

Move the negative in front of the fraction.

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

Step 5.2.8

Apply the distributive property.

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

Step 5.2.9

Multiply $- \frac{x}{- x + 2} (- x)$ .

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Step 5.2.9.1

Multiply $- 1$ by $- 1$ .

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

Step 5.2.9.2

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

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

Step 5.2.9.3

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

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

Step 5.2.9.4

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

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

Step 5.2.9.5

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

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

Step 5.2.9.6

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

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

Step 5.2.9.7

Add $1$ and $1$ .

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

Step 5.2.10

Multiply $- \frac{x}{- x + 2} \cdot 2$ .

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Step 5.2.10.1

Multiply $2$ by $- 1$ .

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

Step 5.2.10.2

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

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

Step 5.2.11

Combine the numerators over the common denominator.

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

Step 5.2.12

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

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Step 5.2.12.1

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

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

Step 5.2.12.2

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

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

Step 5.2.12.3

Factor $x$ out of $x \cdot x + x \cdot - 2$ .

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

Step 5.2.13

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

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Step 5.2.13.1

Factor $- 1$ out of $x$ .

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

Step 5.2.13.2

Rewrite $- 2$ as $- 1 (2)$ .

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

Step 5.2.13.3

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

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

Step 5.2.13.4

Cancel the common factor.

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

Step 5.2.13.5

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

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

Step 5.2.14

Simplify the expression.

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Step 5.2.14.1

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

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

Step 5.2.14.2

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

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

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

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

Step 5.3.3

Simplify the numerator.

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Step 5.3.3.1

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

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Step 5.3.3.1.1

Reorder $2$ and $- x$ .

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

Step 5.3.3.1.2

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

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

Step 5.3.3.1.3

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

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

Step 5.3.3.1.4

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

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

Step 5.3.3.2

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

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

Step 5.3.3.3

Move the negative in front of the fraction.

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

Step 5.3.3.4

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

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

Step 5.3.3.5

Combine the numerators over the common denominator.

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

Step 5.3.3.6

Reorder terms.

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

Step 5.3.3.7

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

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Step 5.3.3.7.1

Apply the distributive property.

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

Step 5.3.3.7.2

Multiply $2$ by $- 2$ .

$f (- \frac{2 x + 1}{2 - x}) = \frac{- \frac{- x + 2 - 4 x - 2 \cdot 1}{- x + 2}}{\frac{2 x + 1}{2 - x} + 2}$

Step 5.3.3.7.3

Multiply $- 2$ by $1$ .

$f (- \frac{2 x + 1}{2 - x}) = \frac{- \frac{- x + 2 - 4 x - 2}{- x + 2}}{\frac{2 x + 1}{2 - x} + 2}$

Step 5.3.3.7.4

Subtract $4 x$ from $- x$ .

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

Step 5.3.3.7.5

Subtract $2$ from $2$ .

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

Step 5.3.3.7.6

Add $- 5 x$ and $0$ .

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

Step 5.3.3.8

Move the negative in front of the fraction.

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

Step 5.3.3.9

Combine exponents.

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Step 5.3.3.9.1

Factor out negative.

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

Step 5.3.3.9.2

Multiply $- 1$ by $- 1$ .

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

Step 5.3.3.9.3

Multiply $\frac{5 x}{- x + 2}$ by $1$ .

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

Step 5.3.4

Simplify the denominator.

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Step 5.3.4.1

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

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

Step 5.3.4.2

Combine the numerators over the common denominator.

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

Step 5.3.4.3

Reorder terms.

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

Step 5.3.4.4

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

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Step 5.3.4.4.1

Apply the distributive property.

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

Step 5.3.4.4.2

Multiply $2$ by $2$ .

$f (- \frac{2 x + 1}{2 - x}) = \frac{\frac{5 x}{- x + 2}}{\frac{2 x + 4 + 2 (- x) + 1}{- x + 2}}$

Step 5.3.4.4.3

Multiply $- 1$ by $2$ .

$f (- \frac{2 x + 1}{2 - x}) = \frac{\frac{5 x}{- x + 2}}{\frac{2 x + 4 - 2 x + 1}{- x + 2}}$

Step 5.3.4.4.4

Subtract $2 x$ from $2 x$ .

$f (- \frac{2 x + 1}{2 - x}) = \frac{\frac{5 x}{- x + 2}}{\frac{0 + 4 + 1}{- x + 2}}$

Step 5.3.4.4.5

Add $0$ and $4$ .

$f (- \frac{2 x + 1}{2 - x}) = \frac{\frac{5 x}{- x + 2}}{\frac{4 + 1}{- x + 2}}$

Step 5.3.4.4.6

Add $4$ and $1$ .

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

Step 5.3.5

Multiply the numerator by the reciprocal of the denominator.

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

Step 5.3.6

Cancel the common factor of $5$ .

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Step 5.3.6.1

Factor $5$ out of $5 x$ .

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

Step 5.3.6.2

Cancel the common factor.

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

Step 5.3.6.3

Rewrite the expression.

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

Step 5.3.7

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

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

Step 5.3.8

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

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Step 5.3.8.1

Cancel the common factor.

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

Step 5.3.8.2

Divide $x$ by $1$ .

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

Step 5.4

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

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