Find the Inverse f(x)=x/(7-x)

$f (x) = \frac{x}{7 - x}$

Write $f (x) = \frac{x}{7 - x}$ as an equation.

$y = \frac{x}{7 - x}$

Interchange the variables.

$x = \frac{y}{7 - y}$

Solve for $y$ .

Rewrite the equation as $\frac{y}{7 - y} = x$ .

$\frac{y}{7 - y} = x$

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.

$7 - y, 1$

Remove parentheses.

$7 - y, 1$

The LCM of one and any expression is the expression.

$7 - y$

Multiply each term in $\frac{y}{7 - y} = x$ by $7 - y$ to eliminate the fractions.

Multiply each term in $\frac{y}{7 - y} = x$ by $7 - y$ .

$\frac{y}{7 - y} (7 - y) = x (7 - y)$

Simplify the left side.

Cancel the common factor of $7 - y$ .

Cancel the common factor.

$\frac{y}{7 - y} (7 - y) = x (7 - y)$

Rewrite the expression.

$y = x (7 - y)$

Simplify the right side.

Apply the distributive property.

$y = x \cdot 7 + x (- y)$

Reorder.

Move $7$ to the left of $x$ .

$y = 7 \cdot x + x (- y)$

Rewrite using the commutative property of multiplication.

$y = 7 x - x y$

Solve the equation.

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

$y + x y = 7 x$

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

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

$y \cdot 1 + x y = 7 x$

Factor $y$ out of $x y$ .

$y \cdot 1 + y x = 7 x$

Factor $y$ out of $y \cdot 1 + y x$ .

$y (1 + x) = 7 x$

Divide each term in $y (1 + x) = 7 x$ by $1 + x$ and simplify.

Divide each term in $y (1 + x) = 7 x$ by $1 + x$ .

$\frac{y (1 + x)}{1 + x} = \frac{7 x}{1 + x}$

Simplify the left side.

Cancel the common factor of $1 + x$ .

Cancel the common factor.

$\frac{y (1 + x)}{1 + x} = \frac{7 x}{1 + x}$

Divide $y$ by $1$ .

$y = \frac{7 x}{1 + x}$

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

$f^{- 1} (x) = \frac{7 x}{1 + x}$

Verify if $f^{- 1} (x) = \frac{7 x}{1 + x}$ is the inverse of $f (x) = \frac{x}{7 - x}$ .

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 $f^{- 1} (\frac{x}{7 - x})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{x}{7 - x}) = \frac{7 (\frac{x}{7 - x})}{1 + \frac{x}{7 - x}}$

Combine fractions.

Remove parentheses.

$f^{- 1} (\frac{x}{7 - x}) = \frac{7 (\frac{x}{7 - x})}{1 + \frac{x}{7 - x}}$

Combine $7$ and $\frac{x}{7 - x}$ .

$f^{- 1} (\frac{x}{7 - x}) = \frac{\frac{7 x}{7 - x}}{1 + \frac{x}{7 - x}}$

Simplify the denominator.

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

$f^{- 1} (\frac{x}{7 - x}) = \frac{\frac{7 x}{7 - x}}{\frac{7 - x}{7 - x} + \frac{x}{7 - x}}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{x}{7 - x}) = \frac{\frac{7 x}{7 - x}}{\frac{7 - x + x}{7 - x}}$

Rewrite $\frac{7 - x + x}{7 - x}$ in a factored form.

Add $- x$ and $x$ .

$f^{- 1} (\frac{x}{7 - x}) = \frac{\frac{7 x}{7 - x}}{\frac{7 + 0}{7 - x}}$

Add $7$ and $0$ .

$f^{- 1} (\frac{x}{7 - x}) = \frac{\frac{7 x}{7 - x}}{\frac{7}{7 - x}}$

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (\frac{x}{7 - x}) = \frac{7 x}{7 - x} \cdot \frac{7 - x}{7}$

Cancel the common factor of $7$ .

Factor $7$ out of $7 x$ .

$f^{- 1} (\frac{x}{7 - x}) = \frac{7 (x)}{7 - x} \cdot \frac{7 - x}{7}$

Cancel the common factor.

$f^{- 1} (\frac{x}{7 - x}) = \frac{7 x}{7 - x} \cdot \frac{7 - x}{7}$

Rewrite the expression.

$f^{- 1} (\frac{x}{7 - x}) = \frac{x}{7 - x} \cdot (7 - x)$

Cancel the common factor of $7 - x$ .

Cancel the common factor.

$f^{- 1} (\frac{x}{7 - x}) = \frac{x}{7 - x} \cdot (7 - x)$

Rewrite the expression.

$f^{- 1} (\frac{x}{7 - x}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{7 x}{1 + x})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{7 x}{1 + x}) = \frac{\frac{7 x}{1 + x}}{7 - (\frac{7 x}{1 + x})}$

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{7 x}{1 + x}) = \frac{7 x}{1 + x} \cdot \frac{1}{7 - \frac{7 x}{1 + x}}$

Simplify the denominator.

To write $7$ as a fraction with a common denominator, multiply by $\frac{1 + x}{1 + x}$ .

$f (\frac{7 x}{1 + x}) = \frac{7 x}{1 + x} \cdot \frac{1}{\frac{7 (1 + x)}{1 + x} - \frac{7 x}{1 + x}}$

Combine the numerators over the common denominator.

$f (\frac{7 x}{1 + x}) = \frac{7 x}{1 + x} \cdot \frac{1}{\frac{7 (1 + x) - 7 x}{1 + x}}$

Rewrite $\frac{7 (1 + x) - 7 x}{1 + x}$ in a factored form.

Factor $7$ out of $7 (1 + x) - 7 x$ .

Factor $7$ out of $- 7 x$ .

$f (\frac{7 x}{1 + x}) = \frac{7 x}{1 + x} \cdot \frac{1}{\frac{7 (1 + x) + 7 (- x)}{1 + x}}$

Factor $7$ out of $7 (1 + x) + 7 (- x)$ .

$f (\frac{7 x}{1 + x}) = \frac{7 x}{1 + x} \cdot \frac{1}{\frac{7 (1 + x - x)}{1 + x}}$

Subtract $x$ from $x$ .

$f (\frac{7 x}{1 + x}) = \frac{7 x}{1 + x} \cdot \frac{1}{\frac{7 (1 + 0)}{1 + x}}$

Add $1$ and $0$ .

$f (\frac{7 x}{1 + x}) = \frac{7 x}{1 + x} \cdot \frac{1}{\frac{7 \cdot 1}{1 + x}}$

Multiply $7$ by $1$ .

$f (\frac{7 x}{1 + x}) = \frac{7 x}{1 + x} \cdot \frac{1}{\frac{7}{1 + x}}$

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{7 x}{1 + x}) = \frac{7 x}{1 + x} \cdot (1 (\frac{1 + x}{7}))$

Multiply $\frac{1 + x}{7}$ by $1$ .

$f (\frac{7 x}{1 + x}) = \frac{7 x}{1 + x} \cdot \frac{1 + x}{7}$

Cancel the common factor of $7$ .

Factor $7$ out of $7 x$ .

$f (\frac{7 x}{1 + x}) = \frac{7 (x)}{1 + x} \cdot \frac{1 + x}{7}$

Cancel the common factor.

$f (\frac{7 x}{1 + x}) = \frac{7 x}{1 + x} \cdot \frac{1 + x}{7}$

Rewrite the expression.

$f (\frac{7 x}{1 + x}) = \frac{x}{1 + x} \cdot (1 + x)$

Cancel the common factor of $1 + x$ .

Cancel the common factor.

$f (\frac{7 x}{1 + x}) = \frac{x}{1 + x} \cdot (1 + x)$

Rewrite the expression.

$f (\frac{7 x}{1 + x}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{7 x}{1 + x}$ is the inverse of $f (x) = \frac{x}{7 - x}$ .

$f^{- 1} (x) = \frac{7 x}{1 + x}$

Find the Inverse f(x)=x/(7-x)

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