Find the Inverse (2x+4)/3

$\frac{2 x + 4}{3}$

Step 1

Interchange the variables.

$x = \frac{2 y + 4}{3}$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{2 y + 4}{3} = x$ .

$\frac{2 y + 4}{3} = x$

Multiply both sides by $3$ .

$\frac{2 y + 4}{3} \cdot 3 = x \cdot 3$

Simplify.

Simplify the left side.

Simplify $\frac{2 y + 4}{3} \cdot 3$ .

Factor $2$ out of $2 y + 4$ .

Factor $2$ out of $2 y$ .

$\frac{2 (y) + 4}{3} \cdot 3 = x \cdot 3$

Factor $2$ out of $4$ .

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Apply the distributive property.

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

Multiply $2$ by $2$ .

$2 y + 4 = x \cdot 3$

Simplify the right side.

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

$2 y + 4 = 3 x$

Solve for $y$ .

Subtract $4$ from both sides of the equation.

$2 y = 3 x - 4$

Divide each term in $2 y = 3 x - 4$ by $2$ and simplify.

Divide each term in $2 y = 3 x - 4$ by $2$ .

$\frac{2 y}{2} = \frac{3 x}{2} + \frac{- 4}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y}{2} = \frac{3 x}{2} + \frac{- 4}{2}$

Divide $y$ by $1$ .

$y = \frac{3 x}{2} + \frac{- 4}{2}$

Simplify the right side.

Divide $- 4$ by $2$ .

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

Step 3

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

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

Step 4

Verify if $f^{- 1} (x) = \frac{3 x}{2} - 2$ is the inverse of $f (x) = \frac{2 x + 4}{3}$ .

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 $f^{- 1} (\frac{2 x + 4}{3})$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Simplify each term.

Factor $2$ out of $2 x + 4$ .

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $4$ .

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

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

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

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

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

Multiply $3$ by $2$ .

$f^{- 1} (\frac{2 x + 4}{3}) = \frac{\frac{6 (x + 2)}{3}}{2} - 2$

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{6 (x + 2)}{3}$ by cancelling the common factors.

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

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

Factor $3$ out of $3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $2 (x + 2)$ by $1$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x + 2$ by $1$ .

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

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

Subtract $2$ from $2$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Simplify the numerator.

Factor $2$ out of $2 (\frac{3 x}{2} - 2) + 4$ .

Factor $2$ out of $4$ .

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

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

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

Add $- 2$ and $2$ .

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

Add $\frac{3 x}{2}$ and $0$ .

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

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

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

Multiply $2$ by $3$ .

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

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{6 x}{2}$ by cancelling the common factors.

Factor $2$ out of $6 x$ .

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

Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $3 x$ by $1$ .

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{3 x}{2} - 2$ is the inverse of $f (x) = \frac{2 x + 4}{3}$ .

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

Find the Inverse (2x+4)/3

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