Find the Inverse y=(2x+1)/(x-1)

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

Multiply the equation by $y - 1$ .

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

Simplify the left side.

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

Apply the distributive property.

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

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

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

Simplify the right side.

Cancel the common factor of $y - 1$ .

Cancel the common factor.

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

Rewrite the expression.

$y x - x = 2 y + 1$

Solve for $y$ .

Subtract $2 y$ from both sides of the equation.

$y x - x - 2 y = 1$

Add $x$ to both sides of the equation.

$y x - 2 y = 1 + x$

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

Factor $y$ out of $y x$ .

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

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

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

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

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

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

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

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

Simplify the left side.

Cancel the common factor of $x - 2$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Combine the numerators over the common denominator.

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

Step 3

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

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

Step 4

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

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

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

Remove parentheses.

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

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

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

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

Combine.

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

Apply the distributive property.

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

Simplify by cancelling.

Cancel the common factor of $x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

Multiply $x - 1$ by $1$ .

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

Add $x$ and $2 x$ .

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

Add $- 1$ and $1$ .

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

Add $3 x$ and $0$ .

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

Simplify the denominator.

Apply the distributive property.

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

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

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

Multiply $- 1$ by $- 2$ .

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

Subtract $2 x$ from $2 x$ .

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

Add $0$ and $1$ .

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

Add $1$ and $2$ .

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Multiply the numerator and denominator of the complex fraction by $x - 2$ .

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

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

Combine.

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

Apply the distributive property.

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

Cancel the common factor of $x - 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

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

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

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

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

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

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

Apply the distributive property.

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

Multiply $2$ by $1$ .

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

Add $x$ and $2 x$ .

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

Subtract $2$ from $2$ .

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

Add $3 x$ and $0$ .

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

Simplify the denominator.

Apply the distributive property.

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

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

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

Multiply $- 2$ by $- 1$ .

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

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

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

Add $1$ and $2$ .

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

Subtract $x$ from $x$ .

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

Add $0$ and $3$ .

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

Reduce the expression by cancelling the common factors.

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

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

Cancel the common factor of $3$ .

Factor $3$ out of $3 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x - 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Find the Inverse y=(2x+1)/(x-1)

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