Find the Inverse f(x)=(4x-2)/3

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

Write $f (x) = \frac{4 x - 2}{3}$ as an equation.

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

Interchange the variables.

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

Solve for $y$ .

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

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

Multiply both sides by $3$ .

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

Simplify.

Simplify the left side.

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

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

Factor $2$ out of $4 y$ .

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

Factor $2$ out of $- 2$ .

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

$2 (2 y - 1) = x \cdot 3$

Apply the distributive property.

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

Multiply.

Multiply $2$ by $2$ .

$4 y + 2 \cdot - 1 = x \cdot 3$

Multiply $2$ by $- 1$ .

$4 y - 2 = x \cdot 3$

Simplify the right side.

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

$4 y - 2 = 3 x$

Solve for $y$ .

Add $2$ to both sides of the equation.

$4 y = 3 x + 2$

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

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

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

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Cancel the common factor of $2$ and $4$ .

Factor $2$ out of $2$ .

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

Cancel the common factors.

Factor $2$ out of $4$ .

$y = \frac{3 x}{4} + \frac{2 \cdot 1}{2 \cdot 2}$

Cancel the common factor.

$y = \frac{3 x}{4} + \frac{2 \cdot 1}{2 \cdot 2}$

Rewrite the expression.

$y = \frac{3 x}{4} + \frac{1}{2}$

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

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

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

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

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

Simplify each term.

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

Factor $2$ out of $4 x$ .

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

Factor $2$ out of $- 2$ .

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

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

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

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

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

Multiply $3$ by $2$ .

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

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{6 (2 x - 1)}{3}$ by cancelling the common factors.

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

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

Factor $3$ out of $3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Cancel the common factors.

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify terms.

Combine the numerators over the common denominator.

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

Combine the opposite terms in $2 x - 1 + 1$ .

Add $- 1$ and $1$ .

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

Add $2 x$ and $0$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify the numerator.

Factor $2$ out of $4 (\frac{3 x}{4} + \frac{1}{2}) - 2$ .

Factor $2$ out of $4 (\frac{3 x}{4} + \frac{1}{2})$ .

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

Factor $2$ out of $- 2$ .

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

Factor $2$ out of $2 (2 (\frac{3 x}{4} + \frac{1}{2})) + 2 (- 1)$ .

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

Apply the distributive property.

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

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Subtract $1$ from $1$ .

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

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

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

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

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

Multiply $2$ by $3$ .

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

Reduce the expression by cancelling the common factors.

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

Factor $2$ out of $6 x$ .

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

Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $3 x$ by $1$ .

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

Find the Inverse f(x)=(4x-2)/3

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