Find the Inverse (x-5)/(x+1)

$\frac{x - 5}{x + 1}$

Step 1

Interchange the variables.

$x = \frac{y - 5}{y + 1}$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{y - 5}{y + 1} = x$ .

$\frac{y - 5}{y + 1} = x$

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.

$y + 1, 1$

Remove parentheses.

$y + 1, 1$

The LCM of one and any expression is the expression.

$y + 1$

Multiply each term in $\frac{y - 5}{y + 1} = x$ by $y + 1$ to eliminate the fractions.

Multiply each term in $\frac{y - 5}{y + 1} = x$ by $y + 1$ .

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

Simplify the left side.

Cancel the common factor of $y + 1$ .

Cancel the common factor.

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

Rewrite the expression.

$y - 5 = x (y + 1)$

Simplify the right side.

Apply the distributive property.

$y - 5 = x y + x \cdot 1$

Multiply $x$ by $1$ .

$y - 5 = x y + x$

Solve the equation.

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

$y - 5 - x y = x$

Add $5$ to both sides of the equation.

$y - x y = x + 5$

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

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

$y \cdot 1 - x y = x + 5$

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

$y \cdot 1 + y (- x) = x + 5$

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

$y (1 - x) = x + 5$

Divide each term in $y (1 - x) = x + 5$ by $1 - x$ and simplify.

Divide each term in $y (1 - x) = x + 5$ by $1 - x$ .

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

Simplify the left side.

Cancel the common factor of $1 - x$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

$y = \frac{x}{1 - x} + \frac{5}{1 - x}$

Simplify the right side.

Combine the numerators over the common denominator.

$y = \frac{x + 5}{1 - x}$

Step 3

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

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

Step 4

Verify if $f^{- 1} (x) = \frac{x + 5}{1 - x}$ is the inverse of $f (x) = \frac{x - 5}{x + 1}$ .

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 $f^{- 1} (\frac{x - 5}{x + 1})$ by substituting in the value of $f$ into $f^{- 1}$ .

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

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

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

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

Combine.

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

Apply the distributive property.

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

Simplify by cancelling.

Cancel the common factor of $x + 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x + 1$ .

Move the leading negative in $- \frac{x - 5}{x + 1}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

Apply the distributive property.

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

Move $5$ to the left of $x$ .

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

Multiply $5$ by $1$ .

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

Add $x$ and $5 x$ .

$f^{- 1} (\frac{x - 5}{x + 1}) = \frac{6 x - 5 + 5}{(x + 1) \cdot 1 - (x - 5)}$

Add $- 5$ and $5$ .

$f^{- 1} (\frac{x - 5}{x + 1}) = \frac{6 x + 0}{(x + 1) \cdot 1 - (x - 5)}$

Add $6 x$ and $0$ .

$f^{- 1} (\frac{x - 5}{x + 1}) = \frac{6 x}{(x + 1) \cdot 1 - (x - 5)}$

Simplify the denominator.

Multiply $x + 1$ by $1$ .

$f^{- 1} (\frac{x - 5}{x + 1}) = \frac{6 x}{x + 1 - (x - 5)}$

Apply the distributive property.

$f^{- 1} (\frac{x - 5}{x + 1}) = \frac{6 x}{x + 1 - x + 5}$

Multiply $- 1$ by $- 5$ .

$f^{- 1} (\frac{x - 5}{x + 1}) = \frac{6 x}{x + 1 - x + 5}$

Subtract $x$ from $x$ .

$f^{- 1} (\frac{x - 5}{x + 1}) = \frac{6 x}{0 + 1 + 5}$

Add $0$ and $1$ .

$f^{- 1} (\frac{x - 5}{x + 1}) = \frac{6 x}{1 + 5}$

Add $1$ and $5$ .

$f^{- 1} (\frac{x - 5}{x + 1}) = \frac{6 x}{6}$

Cancel the common factor of $6$ .

Cancel the common factor.

$f^{- 1} (\frac{x - 5}{x + 1}) = \frac{6 x}{6}$

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

Evaluate $f (\frac{x + 5}{1 - x})$ by substituting in the value of $f^{- 1}$ into $f$ .

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

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

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

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

Combine.

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

Apply the distributive property.

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

Simplify by cancelling.

Cancel the common factor of $1 - x$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $1 - x$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $- 5$ by $1$ .

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

Multiply $- 5$ by $- 1$ .

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

Add $x$ and $5 x$ .

$f (\frac{x + 5}{1 - x}) = \frac{6 x + 5 - 5}{x + 5 + (1 - x) \cdot 1}$

Subtract $5$ from $5$ .

$f (\frac{x + 5}{1 - x}) = \frac{6 x + 0}{x + 5 + (1 - x) \cdot 1}$

Add $6 x$ and $0$ .

$f (\frac{x + 5}{1 - x}) = \frac{6 x}{x + 5 + (1 - x) \cdot 1}$

Simplify the denominator.

Multiply $1 - x$ by $1$ .

$f (\frac{x + 5}{1 - x}) = \frac{6 x}{x + 5 + 1 - x}$

Subtract $x$ from $x$ .

$f (\frac{x + 5}{1 - x}) = \frac{6 x}{0 + 5 + 1}$

Add $0$ and $5$ .

$f (\frac{x + 5}{1 - x}) = \frac{6 x}{5 + 1}$

Add $5$ and $1$ .

$f (\frac{x + 5}{1 - x}) = \frac{6 x}{6}$

Cancel the common factor of $6$ .

Cancel the common factor.

$f (\frac{x + 5}{1 - x}) = \frac{6 x}{6}$

Divide $x$ by $1$ .

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

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x + 5}{1 - x}$ is the inverse of $f (x) = \frac{x - 5}{x + 1}$ .

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

Find the Inverse (x-5)/(x+1)

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