Find the Inverse (3x+2)/4

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

Multiply both sides by $4$ .

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

Simplify.

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

$3 y + 2 = x \cdot 4$

Simplify the right side.

Move $4$ to the left of $x$ .

$3 y + 2 = 4 x$

Solve for $y$ .

Subtract $2$ from both sides of the equation.

$3 y = 4 x - 2$

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

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

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 3

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

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

Step 4

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

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

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

Combine the numerators over the common denominator.

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

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

Combine the opposite terms in $3 x + 2 - 2$ .

Subtract $2$ from $2$ .

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

Add $3 x$ and $0$ .

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify the numerator.

Apply the distributive property.

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $3$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Add $- 2$ and $2$ .

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

Add $4 x$ and $0$ .

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

Find the Inverse (3x+2)/4

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