Find the Inverse f(x)=-(2x)/(x-1)

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

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

Multiply the equation by $y - 1$ .

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

Simplify the left side.

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

Apply the distributive property.

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

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

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

Simplify the right side.

Cancel the common factor of $y - 1$ .

Move the leading negative in $- \frac{2 y}{y - 1}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

$y x - x = - 2 y$

Solve for $y$ .

Add $2 y$ to both sides of the equation.

$y x - x + 2 y = 0$

Add $x$ to both sides of the equation.

$y x + 2 y = x$

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

Factor $y$ out of $y x$ .

$y (x) + 2 y = x$

Factor $y$ out of $2 y$ .

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

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

$y (x + 2) = x$

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

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

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

Simplify the left side.

Cancel the common factor of $x + 2$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Step 4

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

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

Step 5

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

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

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

Multiply the numerator by the reciprocal of the denominator.

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

Simplify the denominator.

To write $2$ as a fraction with a common denominator, multiply by $\frac{x - 1}{x - 1}$ .

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

Combine the numerators over the common denominator.

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

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

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

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

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

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

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

Add $- x$ and $x$ .

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

Subtract $1$ from $0$ .

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

Multiply $2$ by $- 1$ .

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

Move the negative in front of the fraction.

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

Cancel the common factor of $1$ and $- 1$ .

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

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

Move the negative in front of the fraction.

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

Multiply the numerator by the reciprocal of the denominator.

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

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

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{2 x}{x - 1}$ into the numerator.

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

Move the leading negative in $- \frac{x - 1}{2}$ into the numerator.

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x - 1$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply.

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

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

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

Simplify the denominator.

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{x + 2}{x + 2}$ .

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

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

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

Combine the numerators over the common denominator.

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

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

Apply the distributive property.

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

Multiply $- 1$ by $2$ .

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

Subtract $x$ from $x$ .

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

Subtract $2$ from $0$ .

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

Move the negative in front of the fraction.

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

Multiply the numerator by the reciprocal of the denominator.

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

Rewrite using the commutative property of multiplication.

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

Cancel the common factor of $2$ .

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

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x + 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Find the Inverse f(x)=-(2x)/(x-1)

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