Find the Inverse f(x)=(2x-3)/(x+1)

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

Step 1

Write $f (x) = \frac{2 x - 3}{x + 1}$ as an equation.

$y = \frac{2 x - 3}{x + 1}$

Step 2

Interchange the variables.

$x = \frac{2 y - 3}{y + 1}$

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{2 y - 3}{y + 1} = x$ .

$\frac{2 y - 3}{y + 1} = x$

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 $\frac{2 y - 3}{y + 1} = x$ by $y + 1$ to eliminate the fractions.

Multiply each term in $\frac{2 y - 3}{y + 1} = x$ by $y + 1$ .

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

Simplify the left side.

Cancel the common factor of $y + 1$ .

Cancel the common factor.

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

Rewrite the expression.

$2 y - 3 = x (y + 1)$

Simplify the right side.

Apply the distributive property.

$2 y - 3 = x y + x \cdot 1$

Multiply $x$ by $1$ .

$2 y - 3 = x y + x$

Solve the equation.

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

$2 y - 3 - x y = x$

Add $3$ to both sides of the equation.

$2 y - x y = x + 3$

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

Factor $y$ out of $2 y$ .

$y \cdot 2 - x y = x + 3$

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

$y \cdot 2 + y (- x) = x + 3$

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

$y (2 - x) = x + 3$

Divide each term in $y (2 - x) = x + 3$ by $2 - x$ and simplify.

Divide each term in $y (2 - x) = x + 3$ by $2 - x$ .

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

Simplify the left side.

Cancel the common factor of $2 - x$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

$y = \frac{x}{2 - x} + \frac{3}{2 - x}$

Simplify the right side.

Combine the numerators over the common denominator.

$y = \frac{x + 3}{2 - x}$

Step 4

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

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

Step 5

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

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 $f^{- 1} (\frac{2 x - 3}{x + 1})$ by substituting in the value of $f$ into $f^{- 1}$ .

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

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

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

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

Combine.

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

Apply the distributive property.

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

Simplify by cancelling.

Cancel the common factor of $x + 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x + 1$ .

Move the leading negative in $- \frac{2 x - 3}{x + 1}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

Apply the distributive property.

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

Move $3$ to the left of $x$ .

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

Multiply $3$ by $1$ .

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

Add $2 x$ and $3 x$ .

$f^{- 1} (\frac{2 x - 3}{x + 1}) = \frac{5 x - 3 + 3}{(x + 1) \cdot 2 - (2 x - 3)}$

Add $- 3$ and $3$ .

$f^{- 1} (\frac{2 x - 3}{x + 1}) = \frac{5 x + 0}{(x + 1) \cdot 2 - (2 x - 3)}$

Add $5 x$ and $0$ .

$f^{- 1} (\frac{2 x - 3}{x + 1}) = \frac{5 x}{(x + 1) \cdot 2 - (2 x - 3)}$

Simplify the denominator.

Apply the distributive property.

$f^{- 1} (\frac{2 x - 3}{x + 1}) = \frac{5 x}{x \cdot 2 + 1 \cdot 2 - (2 x - 3)}$

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

$f^{- 1} (\frac{2 x - 3}{x + 1}) = \frac{5 x}{2 \cdot x + 1 \cdot 2 - (2 x - 3)}$

Multiply $2$ by $1$ .

$f^{- 1} (\frac{2 x - 3}{x + 1}) = \frac{5 x}{2 \cdot x + 2 - (2 x - 3)}$

Apply the distributive property.

$f^{- 1} (\frac{2 x - 3}{x + 1}) = \frac{5 x}{2 x + 2 - (2 x) + 3}$

Multiply $2$ by $- 1$ .

$f^{- 1} (\frac{2 x - 3}{x + 1}) = \frac{5 x}{2 x + 2 - 2 x + 3}$

Multiply $- 1$ by $- 3$ .

$f^{- 1} (\frac{2 x - 3}{x + 1}) = \frac{5 x}{2 x + 2 - 2 x + 3}$

Subtract $2 x$ from $2 x$ .

$f^{- 1} (\frac{2 x - 3}{x + 1}) = \frac{5 x}{0 + 2 + 3}$

Add $0$ and $2$ .

$f^{- 1} (\frac{2 x - 3}{x + 1}) = \frac{5 x}{2 + 3}$

Add $2$ and $3$ .

$f^{- 1} (\frac{2 x - 3}{x + 1}) = \frac{5 x}{5}$

Cancel the common factor of $5$ .

Cancel the common factor.

$f^{- 1} (\frac{2 x - 3}{x + 1}) = \frac{5 x}{5}$

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

Evaluate $f (\frac{x + 3}{2 - x})$ by substituting in the value of $f^{- 1}$ into $f$ .

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

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

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

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

Combine.

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

Apply the distributive property.

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

Cancel the common factor of $2 - x$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

Factor $2 - x$ out of $(2 - x) \cdot 2 \frac{x + 3}{2 - x} + (2 - x) \cdot - 3$ .

Factor $2 - x$ out of $(2 - x) \cdot 2 \frac{x + 3}{2 - x}$ .

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

Factor $2 - x$ out of $(2 - x) \cdot - 3$ .

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

Factor $2 - x$ out of $(2 - x) (2 \frac{x + 3}{2 - x}) + (2 - x) (- 3)$ .

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

Combine $2$ and $\frac{x + 3}{2 - x}$ .

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

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{2 - x}{2 - x}$ .

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

Combine $- 3$ and $\frac{2 - x}{2 - x}$ .

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

Rewrite $\frac{2 (x + 3) - 3 (2 - x)}{- x + 2}$ in a factored form.

Apply the distributive property.

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

Multiply $2$ by $3$ .

$f (\frac{x + 3}{2 - x}) = \frac{(2 - x) (\frac{2 x + 6 - 3 (2 - x)}{- x + 2})}{x + 3 + (2 - x) \cdot 1}$

Apply the distributive property.

$f (\frac{x + 3}{2 - x}) = \frac{(2 - x) (\frac{2 x + 6 - 3 \cdot 2 - 3 (- x)}{- x + 2})}{x + 3 + (2 - x) \cdot 1}$

Multiply $- 3$ by $2$ .

$f (\frac{x + 3}{2 - x}) = \frac{(2 - x) (\frac{2 x + 6 - 6 - 3 (- x)}{- x + 2})}{x + 3 + (2 - x) \cdot 1}$

Multiply $- 1$ by $- 3$ .

$f (\frac{x + 3}{2 - x}) = \frac{(2 - x) (\frac{2 x + 6 - 6 + 3 x}{- x + 2})}{x + 3 + (2 - x) \cdot 1}$

Add $2 x$ and $3 x$ .

$f (\frac{x + 3}{2 - x}) = \frac{(2 - x) (\frac{5 x + 6 - 6}{- x + 2})}{x + 3 + (2 - x) \cdot 1}$

Subtract $6$ from $6$ .

$f (\frac{x + 3}{2 - x}) = \frac{(2 - x) (\frac{5 x + 0}{- x + 2})}{x + 3 + (2 - x) \cdot 1}$

Add $5 x$ and $0$ .

$f (\frac{x + 3}{2 - x}) = \frac{(2 - x) (\frac{5 x}{- x + 2})}{x + 3 + (2 - x) \cdot 1}$

Simplify the denominator.

Multiply $2 - x$ by $1$ .

$f (\frac{x + 3}{2 - x}) = \frac{(2 - x) (\frac{5 x}{- x + 2})}{x + 3 + 2 - x}$

Subtract $x$ from $x$ .

$f (\frac{x + 3}{2 - x}) = \frac{(2 - x) (\frac{5 x}{- x + 2})}{0 + 3 + 2}$

Add $0$ and $3$ .

$f (\frac{x + 3}{2 - x}) = \frac{(2 - x) (\frac{5 x}{- x + 2})}{3 + 2}$

Add $3$ and $2$ .

$f (\frac{x + 3}{2 - x}) = \frac{(2 - x) (\frac{5 x}{- x + 2})}{5}$

Simplify terms.

Factor $\frac{5 x}{- x + 2}$ out of $\frac{(2 - x) \frac{5 x}{- x + 2}}{5}$ .

$f (\frac{x + 3}{2 - x}) = \frac{5 x}{- x + 2} \cdot \frac{2 - x}{5}$

Cancel the common factor of $5$ .

Factor $5$ out of $5 x$ .

$f (\frac{x + 3}{2 - x}) = \frac{5 (x)}{- x + 2} \cdot \frac{2 - x}{5}$

Cancel the common factor.

$f (\frac{x + 3}{2 - x}) = \frac{5 x}{- x + 2} \cdot \frac{2 - x}{5}$

Rewrite the expression.

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

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

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

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

Reorder terms.

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

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

Find the Inverse f(x)=(2x-3)/(x+1)

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