Find the Inverse f(x)=4/(-x-2)+2

$f (x) = \frac{4}{- x - 2} + 2$

Write $f (x) = \frac{4}{- x - 2} + 2$ as an equation.

$y = \frac{4}{- x - 2} + 2$

Interchange the variables.

$x = \frac{4}{- y - 2} + 2$

Solve for $y$ .

Rewrite the equation as $\frac{4}{- y - 2} + 2 = x$ .

$\frac{4}{- y - 2} + 2 = x$

Subtract $2$ from both sides of the equation.

$\frac{4}{- y - 2} = x - 2$

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.

$- y - 2, 1, 1$

Remove parentheses.

$- y - 2, 1, 1$

The LCM of one and any expression is the expression.

$- y - 2$

Multiply each term in $\frac{4}{- y - 2} = x - 2$ by $- y - 2$ to eliminate the fractions.

Multiply each term in $\frac{4}{- y - 2} = x - 2$ by $- y - 2$ .

$\frac{4}{- y - 2} (- y - 2) = x (- y - 2) - 2 (- y - 2)$

Simplify the left side.

Multiply $\frac{4}{- y - 2}$ by $- y - 2$ .

$\frac{4 (- y - 2)}{- y - 2} = x (- y - 2) - 2 (- y - 2)$

Cancel the common factor of $- y - 2$ .

Cancel the common factor.

$\frac{4 (- y - 2)}{- y - 2} = x (- y - 2) - 2 (- y - 2)$

Divide $4$ by $1$ .

$4 = x (- y - 2) - 2 (- y - 2)$

Simplify the right side.

Simplify each term.

Apply the distributive property.

$4 = x (- y) + x \cdot - 2 - 2 (- y - 2)$

Rewrite using the commutative property of multiplication.

$4 = - x y + x \cdot - 2 - 2 (- y - 2)$

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

$4 = - x y - 2 \cdot x - 2 (- y - 2)$

Apply the distributive property.

$4 = - x y - 2 x - 2 (- y) - 2 \cdot - 2$

Multiply $- 1$ by $- 2$ .

$4 = - x y - 2 x + 2 y - 2 \cdot - 2$

Multiply $- 2$ by $- 2$ .

$4 = - x y - 2 x + 2 y + 4$

Solve the equation.

Rewrite the equation as $- x y - 2 x + 2 y + 4 = 4$ .

$- x y - 2 x + 2 y + 4 = 4$

Move all terms not containing $y$ to the right side of the equation.

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

$- x y + 2 y + 4 = 4 + 2 x$

Subtract $4$ from both sides of the equation.

$- x y + 2 y = 4 + 2 x - 4$

Combine the opposite terms in $4 + 2 x - 4$ .

Subtract $4$ from $4$ .

$- x y + 2 y = 2 x + 0$

Add $2 x$ and $0$ .

$- x y + 2 y = 2 x$

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

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

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

Factor $y$ out of $2 y$ .

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

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

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

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

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

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

Simplify the left side.

Cancel the common factor of $- x + 2$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

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

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

Rewrite $2$ as $- 1 (- 2)$ .

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

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

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

Rewrite negatives.

Rewrite $- (x - 2)$ as $- 1 (x - 2)$ .

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

Move the negative in front of the fraction.

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

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

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

Verify if $f^{- 1} (x) = - \frac{2 x}{x - 2}$ is the inverse of $f (x) = \frac{4}{- x - 2} + 2$ .

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 $f^{- 1} (\frac{4}{- x - 2} + 2)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{2 (\frac{4}{- x - 2} + 2)}{(\frac{4}{- x - 2} + 2) - 2}$

Cancel the common factors.

Factor $2$ out of $\frac{4}{- x - 2}$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{2 (\frac{4}{- x - 2} + 2)}{2 (\frac{2}{- x - 2}) + 2 - 2}$

Factor $2$ out of $2$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{2 (\frac{4}{- x - 2} + 2)}{2 (\frac{2}{- x - 2}) + 2 \cdot 1 - 2}$

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

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{2 (\frac{4}{- x - 2} + 2)}{2 (\frac{2}{- x - 2} + 1) - 2}$

Factor $2$ out of $- 2$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{2 (\frac{4}{- x - 2} + 2)}{2 (\frac{2}{- x - 2} + 1) + 2 (- 1)}$

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

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{2 (\frac{4}{- x - 2} + 2)}{2 (\frac{2}{- x - 2} + 1 - 1)}$

Cancel the common factor.

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{2 (\frac{4}{- x - 2} + 2)}{2 (\frac{2}{- x - 2} + 1 - 1)}$

Rewrite the expression.

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{\frac{4}{- x - 2} + 2}{\frac{2}{- x - 2} + 1 - 1}$

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

Multiply $\frac{\frac{4}{- x - 2} + 2}{\frac{2}{- x - 2} + 1 - 1}$ by $\frac{- x - 2}{- x - 2}$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - (\frac{- x - 2}{- x - 2} \cdot \frac{\frac{4}{- x - 2} + 2}{\frac{2}{- x - 2} + 1 - 1})$

Combine.

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{(- x - 2) (\frac{4}{- x - 2} + 2)}{(- x - 2) (\frac{2}{- x - 2} + 1 - 1)}$

Apply the distributive property.

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{(- x - 2) (\frac{4}{- x - 2}) + (- x - 2) \cdot 2}{(- x - 2) (\frac{2}{- x - 2}) + (- x - 2) \cdot 1 + (- x - 2) \cdot - 1}$

Simplify the numerator.

Factor $- x - 2$ out of $(- x - 2) \frac{4}{- x - 2} + (- x - 2) \cdot 2$ .

Factor $- x - 2$ out of $(- x - 2) \frac{4}{- x - 2}$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{(- x - 2) (\frac{4}{- x - 2}) + (- x - 2) \cdot 2}{(- x - 2) (\frac{2}{- x - 2}) + (- x - 2) \cdot 1 + (- x - 2) \cdot - 1}$

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

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{(- x - 2) (\frac{4}{- x - 2}) + (- x - 2) (2)}{(- x - 2) (\frac{2}{- x - 2}) + (- x - 2) \cdot 1 + (- x - 2) \cdot - 1}$

Factor $- x - 2$ out of $(- x - 2) (\frac{4}{- x - 2}) + (- x - 2) (2)$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{(- x - 2) (\frac{4}{- x - 2} + 2)}{(- x - 2) (\frac{2}{- x - 2}) + (- x - 2) \cdot 1 + (- x - 2) \cdot - 1}$

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

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{(- x - 2) (\frac{4}{- x - 2} + \frac{2 (- x - 2)}{- x - 2})}{(- x - 2) (\frac{2}{- x - 2}) + (- x - 2) \cdot 1 + (- x - 2) \cdot - 1}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{(- x - 2) (\frac{4 + 2 (- x - 2)}{- x - 2})}{(- x - 2) (\frac{2}{- x - 2}) + (- x - 2) \cdot 1 + (- x - 2) \cdot - 1}$

Reorder terms.

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{(- x - 2) (\frac{2 (- x - 2) + 4}{- x - 2})}{(- x - 2) (\frac{2}{- x - 2}) + (- x - 2) \cdot 1 + (- x - 2) \cdot - 1}$

Rewrite $\frac{2 (- x - 2) + 4}{- x - 2}$ in a factored form.

Factor $2$ out of $2 (- x - 2) + 4$ .

Factor $2$ out of $4$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{(- x - 2) (\frac{2 (- x - 2) + 2 \cdot 2}{- x - 2})}{(- x - 2) (\frac{2}{- x - 2}) + (- x - 2) \cdot 1 + (- x - 2) \cdot - 1}$

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

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{(- x - 2) (\frac{2 (- x - 2 + 2)}{- x - 2})}{(- x - 2) (\frac{2}{- x - 2}) + (- x - 2) \cdot 1 + (- x - 2) \cdot - 1}$

Add $- 2$ and $2$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{(- x - 2) (\frac{2 (- x + 0)}{- x - 2})}{(- x - 2) (\frac{2}{- x - 2}) + (- x - 2) \cdot 1 + (- x - 2) \cdot - 1}$

Add $- x$ and $0$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{(- x - 2) (\frac{2 \cdot (- 1 x)}{- x - 2})}{(- x - 2) (\frac{2}{- x - 2}) + (- x - 2) \cdot 1 + (- x - 2) \cdot - 1}$

Combine exponents.

Factor out negative.

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{(- x - 2) (\frac{- (2 x)}{- x - 2})}{(- x - 2) (\frac{2}{- x - 2}) + (- x - 2) \cdot 1 + (- x - 2) \cdot - 1}$

Multiply $2$ by $- 1$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{(- x - 2) (\frac{- 2 x}{- x - 2})}{(- x - 2) (\frac{2}{- x - 2}) + (- x - 2) \cdot 1 + (- x - 2) \cdot - 1}$

Move the negative in front of the fraction.

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{(- x - 2) (- \frac{(2) x}{- x - 2})}{(- x - 2) (\frac{2}{- x - 2}) + (- x - 2) \cdot 1 + (- x - 2) \cdot - 1}$

Remove unnecessary parentheses.

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{(- x - 2) \cdot (- 1 \frac{2 x}{- x - 2})}{(- x - 2) (\frac{2}{- x - 2}) + (- x - 2) \cdot 1 + (- x - 2) \cdot - 1}$

Factor out negative.

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{- (- x - 2) \frac{2 x}{- x - 2}}{(- x - 2) (\frac{2}{- x - 2}) + (- x - 2) \cdot 1 + (- x - 2) \cdot - 1}$

Simplify the denominator.

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

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{- (- x - 2) \frac{2 x}{- x - 2}}{\frac{(- x - 2) \cdot 2}{- x - 2} + (- x - 2) \cdot 1 + (- x - 2) \cdot - 1}$

Cancel the common factor of $- x - 2$ .

Cancel the common factor.

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{- (- x - 2) \frac{2 x}{- x - 2}}{\frac{(- x - 2) \cdot 2}{- x - 2} + (- x - 2) \cdot 1 + (- x - 2) \cdot - 1}$

Divide $2$ by $1$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{- (- x - 2) \frac{2 x}{- x - 2}}{2 + (- x - 2) \cdot 1 + (- x - 2) \cdot - 1}$

Multiply $- x - 2$ by $1$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{- (- x - 2) \frac{2 x}{- x - 2}}{2 - x - 2 + (- x - 2) \cdot - 1}$

Apply the distributive property.

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{- (- x - 2) \frac{2 x}{- x - 2}}{2 - x - 2 - x \cdot - 1 - 2 \cdot - 1}$

Multiply $- x \cdot - 1$ .

Multiply $- 1$ by $- 1$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{- (- x - 2) \frac{2 x}{- x - 2}}{2 - x - 2 + 1 x - 2 \cdot - 1}$

Multiply $x$ by $1$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{- (- x - 2) \frac{2 x}{- x - 2}}{2 - x - 2 + x - 2 \cdot - 1}$

Multiply $- 2$ by $- 1$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{- (- x - 2) \frac{2 x}{- x - 2}}{2 - x - 2 + x + 2}$

Subtract $2$ from $2$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{- (- x - 2) \frac{2 x}{- x - 2}}{- x + 0 + x + 2}$

Add $- x$ and $0$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{- (- x - 2) \frac{2 x}{- x - 2}}{- x + x + 2}$

Add $- x$ and $x$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{- (- x - 2) \frac{2 x}{- x - 2}}{0 + 2}$

Add $0$ and $2$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{- (- x - 2) \frac{2 x}{- x - 2}}{2}$

Simplify terms.

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

$f^{- 1} (\frac{4}{- x - 2} + 2) = - (\frac{2 x}{- x - 2} \cdot \frac{- (- x - 2)}{2})$

Cancel the common factor of $2$ .

Factor $2$ out of $2 x$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - (\frac{2 (x)}{- x - 2} \cdot \frac{- (- x - 2)}{2})$

Cancel the common factor.

$f^{- 1} (\frac{4}{- x - 2} + 2) = - (\frac{2 x}{- x - 2} \cdot \frac{- (- x - 2)}{2})$

Rewrite the expression.

$f^{- 1} (\frac{4}{- x - 2} + 2) = - (\frac{x}{- x - 2} \cdot (- (- x - 2)))$

Rewrite using the commutative property of multiplication.

$f^{- 1} (\frac{4}{- x - 2} + 2) = - (- \frac{x}{- x - 2} \cdot (- x - 2))$

Apply the distributive property.

$f^{- 1} (\frac{4}{- x - 2} + 2) = - (- \frac{x}{- x - 2} \cdot (- x) - \frac{x}{- x - 2} \cdot - 2)$

Multiply $- \frac{x}{- x - 2} (- x)$ .

Multiply $- 1$ by $- 1$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - (1 (\frac{x}{- x - 2}) \cdot x - \frac{x}{- x - 2} \cdot - 2)$

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

$f^{- 1} (\frac{4}{- x - 2} + 2) = - (\frac{x}{- x - 2} \cdot x - \frac{x}{- x - 2} \cdot - 2)$

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

$f^{- 1} (\frac{4}{- x - 2} + 2) = - (\frac{x \cdot x}{- x - 2} - \frac{x}{- x - 2} \cdot - 2)$

Raise $x$ to the power of $1$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - (\frac{x \cdot x}{- x - 2} - \frac{x}{- x - 2} \cdot - 2)$

Raise $x$ to the power of $1$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - (\frac{x \cdot x}{- x - 2} - \frac{x}{- x - 2} \cdot - 2)$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f^{- 1} (\frac{4}{- x - 2} + 2) = - (\frac{x^{1 + 1}}{- x - 2} - \frac{x}{- x - 2} \cdot - 2)$

Add $1$ and $1$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - (\frac{x^{2}}{- x - 2} - \frac{x}{- x - 2} \cdot - 2)$

Multiply $- \frac{x}{- x - 2} \cdot - 2$ .

Multiply $- 2$ by $- 1$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - (\frac{x^{2}}{- x - 2} + 2 (\frac{x}{- x - 2}))$

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

$f^{- 1} (\frac{4}{- x - 2} + 2) = - (\frac{x^{2}}{- x - 2} + \frac{2 x}{- x - 2})$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{x^{2} + 2 x}{- x - 2}$

Factor $x$ out of $x^{2} + 2 x$ .

Factor $x$ out of $x^{2}$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{x \cdot x + 2 x}{- x - 2}$

Factor $x$ out of $2 x$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{x \cdot x + x \cdot 2}{- x - 2}$

Factor $x$ out of $x \cdot x + x \cdot 2$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{x (x + 2)}{- x - 2}$

Cancel the common factor of $x + 2$ and $- x - 2$ .

Factor $- 1$ out of $x$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{x (- 1 (- x) + 2)}{- x - 2}$

Rewrite $2$ as $- 1 (- 2)$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{x (- 1 (- x) - 1 \cdot - 2)}{- x - 2}$

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

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{x (- 1 (- x - 2))}{- x - 2}$

Cancel the common factor.

$f^{- 1} (\frac{4}{- x - 2} + 2) = - \frac{x (- 1 (- x - 2))}{- x - 2}$

Divide $x \cdot (- 1)$ by $1$ .

$f^{- 1} (\frac{4}{- x - 2} + 2) = - (x \cdot (- 1))$

Simplify the expression.

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

$f^{- 1} (\frac{4}{- x - 2} + 2) = - (- 1 x)$

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

$f^{- 1} (\frac{4}{- x - 2} + 2) = x$

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

Set up the composite result function.

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

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

$f (- \frac{2 x}{x - 2}) = \frac{4}{- (- \frac{2 x}{x - 2}) - 2} + 2$

Simplify each term.

Simplify the denominator.

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

Move parentheses.

$f (- \frac{2 x}{x - 2}) = \frac{4}{1 (\frac{2 x}{x - 2}) - 2} + 2$

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

$f (- \frac{2 x}{x - 2}) = \frac{4}{\frac{2 x}{x - 2} - 2} + 2$

Rewrite $- 2$ as $- 1 (2)$ .

$f (- \frac{2 x}{x - 2}) = \frac{4}{\frac{2 x}{x - 2} - 1 \cdot 2} + 2$

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

$f (- \frac{2 x}{x - 2}) = \frac{4}{- (- \frac{2 x}{x - 2} + 2)} + 2$

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

$f (- \frac{2 x}{x - 2}) = \frac{4}{- (- \frac{2 x}{x - 2} + \frac{2 (x - 2)}{x - 2})} + 2$

Combine the numerators over the common denominator.

$f (- \frac{2 x}{x - 2}) = \frac{4}{- \frac{- 2 x + 2 (x - 2)}{x - 2}} + 2$

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

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

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

$f (- \frac{2 x}{x - 2}) = \frac{4}{- \frac{2 (- x) + 2 (x - 2)}{x - 2}} + 2$

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

$f (- \frac{2 x}{x - 2}) = \frac{4}{- \frac{2 (- x + x - 2)}{x - 2}} + 2$

Add $- x$ and $x$ .

$f (- \frac{2 x}{x - 2}) = \frac{4}{- \frac{2 (0 - 2)}{x - 2}} + 2$

Subtract $2$ from $0$ .

$f (- \frac{2 x}{x - 2}) = \frac{4}{- \frac{2 \cdot - 2}{x - 2}} + 2$

Multiply $2$ by $- 2$ .

$f (- \frac{2 x}{x - 2}) = \frac{4}{- \frac{- 4}{x - 2}} + 2$

Move the negative in front of the fraction.

$f (- \frac{2 x}{x - 2}) = \frac{4}{1 (\frac{4}{x - 2})} + 2$

Combine exponents.

Factor out negative.

$f (- \frac{2 x}{x - 2}) = \frac{4}{\frac{4}{x - 2}} + 2$

Multiply $- 1$ by $- 1$ .

$f (- \frac{2 x}{x - 2}) = \frac{4}{1 (\frac{4}{x - 2})} + 2$

Multiply $\frac{4}{x - 2}$ by $1$ .

$f (- \frac{2 x}{x - 2}) = \frac{4}{\frac{4}{x - 2}} + 2$

Multiply the numerator by the reciprocal of the denominator.

$f (- \frac{2 x}{x - 2}) = 4 (\frac{x - 2}{4}) + 2$

Cancel the common factor of $4$ .

Cancel the common factor.

$f (- \frac{2 x}{x - 2}) = 4 (\frac{x - 2}{4}) + 2$

Rewrite the expression.

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

Combine the opposite terms in $x - 2 + 2$ .

Add $- 2$ and $2$ .

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

Add $x$ and $0$ .

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

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - \frac{2 x}{x - 2}$ is the inverse of $f (x) = \frac{4}{- x - 2} + 2$ .

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

Find the Inverse f(x)=4/(-x-2)+2

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