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

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

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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

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

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.

$2 + y, 1$

Remove parentheses.

$2 + y, 1$

The LCM of one and any expression is the expression.

$2 + y$

Multiply each term in $\frac{3 y + 2}{2 + y} = x$ by $2 + y$ to eliminate the fractions.

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

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

Simplify the left side.

Cancel the common factor of $2 + y$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the right side.

Apply the distributive property.

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

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

$3 y + 2 = 2 x + x y$

Solve the equation.

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

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

Subtract $2$ from both sides of the equation.

$3 y - x y = 2 x - 2$

Factor $y$ out of $3 y - x y$ .

Factor $y$ out of $3 y$ .

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

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

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

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

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

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

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

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

Simplify the left side.

Cancel the common factor of $3 - x$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Combine the numerators over the common denominator.

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

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

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $- 2$ .

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

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

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

Step 4

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

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

Step 5

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

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

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

Simplify the numerator.

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2 + x}{2 + x}$ .

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

Combine $- 1$ and $\frac{2 + x}{2 + x}$ .

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

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

Apply the distributive property.

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

Multiply $- 1$ by $2$ .

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

Subtract $x$ from $3 x$ .

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

Subtract $2$ from $2$ .

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

Add $2 x$ and $0$ .

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

Simplify the denominator.

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

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

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

Apply the distributive property.

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

Multiply $3$ by $2$ .

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

Apply the distributive property.

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

Multiply $3$ by $- 1$ .

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

Multiply $- 1$ by $2$ .

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

Subtract $2$ from $6$ .

$f^{- 1} (\frac{3 x + 2}{2 + x}) = \frac{2 (\frac{2 x}{x + 2})}{\frac{3 x - 3 x + 4}{x + 2}}$

Subtract $3 x$ from $3 x$ .

$f^{- 1} (\frac{3 x + 2}{2 + x}) = \frac{2 (\frac{2 x}{x + 2})}{\frac{0 + 4}{x + 2}}$

Add $0$ and $4$ .

$f^{- 1} (\frac{3 x + 2}{2 + x}) = \frac{2 (\frac{2 x}{x + 2})}{\frac{4}{x + 2}}$

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

$f^{- 1} (\frac{3 x + 2}{2 + x}) = \frac{\frac{2 (2 x)}{x + 2}}{\frac{4}{x + 2}}$

Multiply $2$ by $2$ .

$f^{- 1} (\frac{3 x + 2}{2 + x}) = \frac{\frac{4 x}{x + 2}}{\frac{4}{x + 2}}$

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (\frac{3 x + 2}{2 + x}) = \frac{4 x}{x + 2} \cdot \frac{x + 2}{4}$

Cancel the common factor of $4$ .

Factor $4$ out of $4 x$ .

$f^{- 1} (\frac{3 x + 2}{2 + x}) = \frac{4 (x)}{x + 2} \cdot \frac{x + 2}{4}$

Cancel the common factor.

$f^{- 1} (\frac{3 x + 2}{2 + x}) = \frac{4 x}{x + 2} \cdot \frac{x + 2}{4}$

Rewrite the expression.

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

Cancel the common factor of $x + 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

Set up the composite result function.

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

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

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

Remove parentheses.

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

Multiply the numerator and denominator of the complex fraction by $3 - x$ .

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

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

Combine.

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

Apply the distributive property.

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

Cancel the common factor of $3 - x$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

Factor $3 - x$ out of $(3 - x) \cdot 3 \frac{2 (x - 1)}{3 - x} + (3 - x) \cdot 2$ .

Factor $3 - x$ out of $(3 - x) \cdot 3 \frac{2 (x - 1)}{3 - x}$ .

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

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

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

Factor $3 - x$ out of $(3 - x) (3 \frac{2 (x - 1)}{3 - x}) + (3 - x) (2)$ .

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

Multiply $3 \frac{2 (x - 1)}{3 - x}$ .

Combine $3$ and $\frac{2 (x - 1)}{3 - x}$ .

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

Multiply $2$ by $3$ .

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

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

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

Rewrite $\frac{6 (x - 1) + 2 (3 - x)}{- x + 3}$ in a factored form.

Factor $2$ out of $6 (x - 1) + 2 (3 - x)$ .

Factor $2$ out of $6 (x - 1)$ .

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

Factor $2$ out of $2 (3 (x - 1)) + 2 (3 - x)$ .

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

Apply the distributive property.

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

Multiply $3$ by $- 1$ .

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

Subtract $x$ from $3 x$ .

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

Add $- 3$ and $3$ .

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

Add $2 x$ and $0$ .

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

Multiply $2$ by $2$ .

$f (\frac{2 (x - 1)}{3 - x}) = \frac{(3 - x) (\frac{4 x}{- x + 3})}{(3 - x) \cdot 2 + 2 (x - 1)}$

Simplify the denominator.

Factor $2$ out of $(3 - x) \cdot 2 + 2 (x - 1)$ .

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

$f (\frac{2 (x - 1)}{3 - x}) = \frac{(3 - x) (\frac{4 x}{- x + 3})}{2 \cdot (3 - x) + 2 (x - 1)}$

Factor $2$ out of $2 \cdot (3 - x) + 2 (x - 1)$ .

$f (\frac{2 (x - 1)}{3 - x}) = \frac{(3 - x) (\frac{4 x}{- x + 3})}{2 (3 - x + x - 1)}$

Subtract $1$ from $3$ .

$f (\frac{2 (x - 1)}{3 - x}) = \frac{(3 - x) (\frac{4 x}{- x + 3})}{2 (- x + x + 2)}$

Add $- x$ and $x$ .

$f (\frac{2 (x - 1)}{3 - x}) = \frac{(3 - x) (\frac{4 x}{- x + 3})}{2 (0 + 2)}$

Add $0$ and $2$ .

$f (\frac{2 (x - 1)}{3 - x}) = \frac{(3 - x) (\frac{4 x}{- x + 3})}{2 \cdot 2}$

Simplify terms.

Multiply $2$ by $2$ .

$f (\frac{2 (x - 1)}{3 - x}) = \frac{(3 - x) (\frac{4 x}{- x + 3})}{4}$

Factor $\frac{4 x}{- x + 3}$ out of $\frac{(3 - x) \frac{4 x}{- x + 3}}{4}$ .

$f (\frac{2 (x - 1)}{3 - x}) = \frac{4 x}{- x + 3} \cdot \frac{3 - x}{4}$

Cancel the common factor of $4$ .

Factor $4$ out of $4 x$ .

$f (\frac{2 (x - 1)}{3 - x}) = \frac{4 (x)}{- x + 3} \cdot \frac{3 - x}{4}$

Cancel the common factor.

$f (\frac{2 (x - 1)}{3 - x}) = \frac{4 x}{- x + 3} \cdot \frac{3 - x}{4}$

Rewrite the expression.

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

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

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

Cancel the common factor of $3 - x$ and $- x + 3$ .

Reorder terms.

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

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

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

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