Find the Inverse f(x)=(x+1)/(x-1)

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

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

Multiply the equation by $y - 1$ .

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

Simplify the left side.

Simplify $(y - 1) (x)$ .

Apply the distributive property.

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

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

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

Simplify the right side.

Cancel the common factor of $y - 1$ .

Cancel the common factor.

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

Rewrite the expression.

$y x - x = y + 1$

Solve for $y$ .

Subtract $y$ from both sides of the equation.

$y x - x - y = 1$

Add $x$ to both sides of the equation.

$y x - y = 1 + x$

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

Factor $y$ out of $y x$ .

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

Factor $y$ out of $- y$ .

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

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

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

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

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

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

Simplify the left side.

Cancel the common factor of $x - 1$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Combine the numerators over the common denominator.

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

Step 4

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

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

Step 5

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

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

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

Remove parentheses.

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

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

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

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

Combine.

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

Apply the distributive property.

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

Simplify by cancelling.

Cancel the common factor of $x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

Multiply $x - 1$ by $1$ .

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

Add $x$ and $x$ .

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

Add $- 1$ and $1$ .

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

Add $2 x$ and $0$ .

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

Simplify the denominator.

Apply the distributive property.

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

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

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

Multiply $- 1$ by $- 1$ .

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

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

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

Subtract $x$ from $x$ .

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

Add $0$ and $1$ .

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

Add $1$ and $1$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

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

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

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

Combine.

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

Apply the distributive property.

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

Simplify by cancelling.

Cancel the common factor of $x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

Multiply $x - 1$ by $1$ .

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

Subtract $1$ from $1$ .

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

Add $x + x$ and $0$ .

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

Add $x$ and $x$ .

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

Simplify the denominator.

Apply the distributive property.

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

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

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

Multiply $- 1$ by $- 1$ .

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

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

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

Add $1$ and $1$ .

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

Subtract $x$ from $x$ .

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

Add $0$ and $2$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

Find the Inverse f(x)=(x+1)/(x-1)

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