Find the Inverse f(x)=3/(-x-1)+1

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

Step 1

Write $f (x) = \frac{3}{- x - 1} + 1$ as an equation.

$y = \frac{3}{- x - 1} + 1$

Step 2

Interchange the variables.

$x = \frac{3}{- y - 1} + 1$

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{3}{- y - 1} + 1 = x$ .

$\frac{3}{- y - 1} + 1 = x$

Subtract $1$ from both sides of the equation.

$\frac{3}{- y - 1} = x - 1$

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.

$- y - 1, 1, 1$

Remove parentheses.

$- y - 1, 1, 1$

The LCM of one and any expression is the expression.

$- y - 1$

Multiply each term in $\frac{3}{- y - 1} = x - 1$ by $- y - 1$ to eliminate the fractions.

Multiply each term in $\frac{3}{- y - 1} = x - 1$ by $- y - 1$ .

$\frac{3}{- y - 1} (- y - 1) = x (- y - 1) - (- y - 1)$

Simplify the left side.

Multiply $\frac{3}{- y - 1}$ by $- y - 1$ .

$\frac{3 (- y - 1)}{- y - 1} = x (- y - 1) - (- y - 1)$

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

Cancel the common factor.

$\frac{3 (- y - 1)}{- y - 1} = x (- y - 1) - (- y - 1)$

Divide $3$ by $1$ .

$3 = x (- y - 1) - (- y - 1)$

Simplify the right side.

Simplify each term.

Apply the distributive property.

$3 = x (- y) + x \cdot - 1 - (- y - 1)$

Rewrite using the commutative property of multiplication.

$3 = - x y + x \cdot - 1 - (- y - 1)$

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

$3 = - x y - 1 \cdot x - (- y - 1)$

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

$3 = - x y - x - (- y - 1)$

Apply the distributive property.

$3 = - x y - x - - y - - 1$

Multiply $- - y$ .

Multiply $- 1$ by $- 1$ .

$3 = - x y - x + 1 y - - 1$

Multiply $y$ by $1$ .

$3 = - x y - x + y - - 1$

Multiply $- 1$ by $- 1$ .

$3 = - x y - x + y + 1$

Solve the equation.

Rewrite the equation as $- x y - x + y + 1 = 3$ .

$- x y - x + y + 1 = 3$

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

Add $x$ to both sides of the equation.

$- x y + y + 1 = 3 + x$

Subtract $1$ from both sides of the equation.

$- x y + y = 3 + x - 1$

Subtract $1$ from $3$ .

$- x y + y = x + 2$

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

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

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

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

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

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

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

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

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

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

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

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

Simplify the left side.

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

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Combine the numerators over the common denominator.

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

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

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

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

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

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

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

Rewrite negatives.

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

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

Move the negative in front of the fraction.

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

Step 4

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

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

Step 5

Verify if $f^{- 1} (x) = - \frac{x + 2}{x - 1}$ is the inverse of $f (x) = \frac{3}{- x - 1} + 1$ .

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 $f^{- 1} (\frac{3}{- x - 1} + 1)$ by substituting in the value of $f$ into $f^{- 1}$ .

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

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

Multiply $\frac{\frac{3}{- x - 1} + 1 + 2}{\frac{3}{- x - 1} + 1 - 1}$ by $\frac{- x - 1}{- x - 1}$ .

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

Combine.

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

Apply the distributive property.

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

Simplify the numerator.

Reorder $- x - 1$ and $2$ .

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

Add $1 \cdot (- x - 1)$ and $2 \cdot (- x - 1)$ .

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

Factor $- x - 1$ out of $(- x - 1) \frac{3}{- x - 1} + 3 \cdot (- x - 1)$ .

Factor $- x - 1$ out of $3 \cdot (- x - 1)$ .

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

Factor $- x - 1$ out of $(- x - 1) (\frac{3}{- x - 1}) + (- x - 1) \cdot 3$ .

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

To write $3$ as a fraction with a common denominator, multiply by $\frac{- x - 1}{- x - 1}$ .

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

Rewrite $\frac{3 (- x - 1) + 3}{- x - 1}$ in a factored form.

Factor $3$ out of $3 (- x - 1) + 3$ .

Factor $3$ out of $3$ .

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

Factor $3$ out of $3 (- x - 1) + 3 (1)$ .

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

Add $- 1$ and $1$ .

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

Add $- x$ and $0$ .

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

Combine exponents.

Factor out negative.

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

Multiply $3$ by $- 1$ .

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

Move the negative in front of the fraction.

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

Remove unnecessary parentheses.

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

Factor out negative.

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

Simplify the denominator.

Multiply $- x - 1$ by $\frac{3}{- x - 1}$ .

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

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

Cancel the common factor.

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

Divide $3$ by $1$ .

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

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

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

Apply the distributive property.

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

Multiply $- x \cdot - 1$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Multiply $- 1$ by $- 1$ .

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

Subtract $1$ from $3$ .

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

Add $- x$ and $x$ .

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

Add $0$ and $2$ .

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

Add $2$ and $1$ .

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

Simplify terms.

Factor $\frac{3 x}{- x - 1}$ out of $\frac{- (- x - 1) \frac{3 x}{- x - 1}}{3}$ .

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

Cancel the common factor of $3$ .

Factor $3$ out of $3 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Rewrite using the commutative property of multiplication.

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

Apply the distributive property.

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

Multiply $- \frac{x}{- x - 1} (- x)$ .

Multiply $- 1$ by $- 1$ .

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

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

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

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

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

Multiply $- \frac{x}{- x - 1} \cdot - 1$ .

Multiply $- 1$ by $- 1$ .

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

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

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

Combine the numerators over the common denominator.

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

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

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

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

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

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

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

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

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

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

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

Factor $- 1$ out of $x$ .

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

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

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

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

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

Cancel the common factor.

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

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

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

Simplify the expression.

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

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

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

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Simplify the denominator.

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

Move parentheses.

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

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

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

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

Apply the distributive property.

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

Multiply $- 1$ by $2$ .

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

Subtract $x$ from $x$ .

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

Subtract $1$ from $0$ .

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

Subtract $2$ from $- 1$ .

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

Move the negative in front of the fraction.

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

Combine exponents.

Factor out negative.

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

Multiply $- 1$ by $- 1$ .

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

Multiply $\frac{3}{x - 1}$ by $1$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Combine the opposite terms in $x - 1 + 1$ .

Add $- 1$ and $1$ .

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

Add $x$ and $0$ .

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

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - \frac{x + 2}{x - 1}$ is the inverse of $f (x) = \frac{3}{- x - 1} + 1$ .

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

Find the Inverse f(x)=3/(-x-1)+1

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