Find the Inverse (x-4)/(x+4)

$\frac{x - 4}{x + 4}$

Step 1

Interchange the variables.

$x = \frac{y - 4}{y + 4}$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{y - 4}{y + 4} = x$ .

$\frac{y - 4}{y + 4} = x$

Find the LCD of the terms in the equation.

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

$y + 4, 1$

Remove parentheses.

$y + 4, 1$

The LCM of one and any expression is the expression.

$y + 4$

Multiply each term in $\frac{y - 4}{y + 4} = x$ by $y + 4$ to eliminate the fractions.

Multiply each term in $\frac{y - 4}{y + 4} = x$ by $y + 4$ .

$\frac{y - 4}{y + 4} (y + 4) = x (y + 4)$

Simplify the left side.

Cancel the common factor of $y + 4$ .

Cancel the common factor.

$\frac{y - 4}{y + 4} (y + 4) = x (y + 4)$

Rewrite the expression.

$y - 4 = x (y + 4)$

Simplify the right side.

Apply the distributive property.

$y - 4 = x y + x \cdot 4$

Move $4$ to the left of $x$ .

$y - 4 = x y + 4 x$

Solve the equation.

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

$y - 4 - x y = 4 x$

Add $4$ to both sides of the equation.

$y - x y = 4 x + 4$

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

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

$y \cdot 1 - x y = 4 x + 4$

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

$y \cdot 1 + y (- x) = 4 x + 4$

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

$y (1 - x) = 4 x + 4$

Divide each term in $y (1 - x) = 4 x + 4$ by $1 - x$ and simplify.

Divide each term in $y (1 - x) = 4 x + 4$ by $1 - x$ .

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

Simplify the left side.

Cancel the common factor of $1 - x$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

$y = \frac{4 x}{1 - x} + \frac{4}{1 - x}$

Simplify the right side.

Combine the numerators over the common denominator.

$y = \frac{4 x + 4}{1 - x}$

Factor $4$ out of $4 x + 4$ .

Factor $4$ out of $4 x$ .

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

Factor $4$ out of $4$ .

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

Factor $4$ out of $4 (x) + 4 (1)$ .

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

Step 3

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

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

Step 4

Verify if $f^{- 1} (x) = \frac{4 (x + 1)}{1 - x}$ is the inverse of $f (x) = \frac{x - 4}{x + 4}$ .

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Evaluate $f^{- 1} (f (x))$ .

Set up the composite result function.

$f^{- 1} (f (x))$

Evaluate $f^{- 1} (\frac{x - 4}{x + 4})$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Simplify the numerator.

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

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

Rewrite $\frac{x + x - 4 + 4}{x + 4}$ in a factored form.

Add $x$ and $x$ .

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

Add $- 4$ and $4$ .

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

Add $2 x$ and $0$ .

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

Simplify the denominator.

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

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

Combine the numerators over the common denominator.

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

Rewrite $\frac{x + 4 - (x - 4)}{x + 4}$ in a factored form.

Apply the distributive property.

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

Multiply $- 1$ by $- 4$ .

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

Subtract $x$ from $x$ .

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

Add $0$ and $4$ .

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

Add $4$ and $4$ .

$f^{- 1} (\frac{x - 4}{x + 4}) = \frac{4 (\frac{2 x}{x + 4})}{\frac{8}{x + 4}}$

Combine $4$ and $\frac{2 x}{x + 4}$ .

$f^{- 1} (\frac{x - 4}{x + 4}) = \frac{\frac{4 (2 x)}{x + 4}}{\frac{8}{x + 4}}$

Multiply $4$ by $2$ .

$f^{- 1} (\frac{x - 4}{x + 4}) = \frac{\frac{8 x}{x + 4}}{\frac{8}{x + 4}}$

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (\frac{x - 4}{x + 4}) = \frac{8 x}{x + 4} \cdot \frac{x + 4}{8}$

Cancel the common factor of $8$ .

Factor $8$ out of $8 x$ .

$f^{- 1} (\frac{x - 4}{x + 4}) = \frac{8 (x)}{x + 4} \cdot \frac{x + 4}{8}$

Cancel the common factor.

$f^{- 1} (\frac{x - 4}{x + 4}) = \frac{8 x}{x + 4} \cdot \frac{x + 4}{8}$

Rewrite the expression.

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

Cancel the common factor of $x + 4$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate $f (f^{- 1} (x))$ .

Set up the composite result function.

$f (f^{- 1} (x))$

Evaluate $f (\frac{4 (x + 1)}{1 - x})$ by substituting in the value of $f^{- 1}$ into $f$ .

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

Cancel the common factor of $(\frac{4 (x + 1)}{1 - x}) - 4$ and $(\frac{4 (x + 1)}{1 - x}) + 4$ .

Factor $4$ out of $\frac{4 (x + 1)}{1 - x}$ .

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

Factor $4$ out of $- 4$ .

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

Factor $4$ out of $4 \frac{x + 1}{1 - x} + 4 (- 1)$ .

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

Cancel the common factors.

Factor $4$ out of $\frac{4 (x + 1)}{1 - x}$ .

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

Factor $4$ out of $4$ .

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

Factor $4$ out of $4 \frac{x + 1}{1 - x} + 4 (1)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

Combine.

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

Apply the distributive property.

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

Simplify by cancelling.

Cancel the common factor of $1 - x$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $1 - x$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $- 1$ by $1$ .

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

Multiply $- x \cdot - 1$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Add $x$ and $x$ .

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

Subtract $1$ from $1$ .

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

Add $2 x$ and $0$ .

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

Simplify the denominator.

Multiply $1 - x$ by $1$ .

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

Subtract $x$ from $x$ .

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

Add $0$ and $1$ .

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

Add $1$ and $1$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{4 (x + 1)}{1 - x}$ is the inverse of $f (x) = \frac{x - 4}{x + 4}$ .

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

Find the Inverse (x-4)/(x+4)

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