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

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

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

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

Interchange the variables.

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

Solve for $y$ .

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

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

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 $\frac{y}{y + 1} = x$ by $y + 1$ to eliminate the fractions.

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

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

Simplify the left side.

Cancel the common factor of $y + 1$ .

Cancel the common factor.

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

Rewrite the expression.

$y = x (y + 1)$

Simplify the right side.

Apply the distributive property.

$y = x y + x \cdot 1$

Multiply $x$ by $1$ .

$y = x y + x$

Solve the equation.

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

$y - x y = x$

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

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

$y \cdot 1 - x y = x$

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

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

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

$y (1 - x) = x$

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

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

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

Simplify the left side.

Cancel the common factor of $1 - x$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

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

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

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

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

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

Multiply the numerator by the reciprocal of the denominator.

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

Simplify the denominator.

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

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

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

Subtract $x$ from $x$ .

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

Add $0$ and $1$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Cancel the common factor of $x + 1$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Set up the composite result function.

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

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

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

Multiply the numerator by the reciprocal of the denominator.

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

Simplify the denominator.

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

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

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

Subtract $x$ from $x$ .

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

Add $0$ and $1$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $- x + 1$ by $1$ .

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

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

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

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

Reorder terms.

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

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

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

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