Find the Inverse y=( square root of 3x-2)/2

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

Multiply both sides by $2$ .

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

Simplify.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$\sqrt{3 y - 2} = x \cdot 2$

Simplify the right side.

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

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

Solve for $y$ .

To remove the radical on the left side of the equation, square both sides of the equation.

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

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 y - 2}$ as ${(3 y - 2)}^{\frac{1}{2}}$ .

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

Simplify the left side.

Simplify ${({(3 y - 2)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(3 y - 2)}^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$3 y - 2 = {(2 x)}^{2}$

Simplify the right side.

Simplify ${(2 x)}^{2}$ .

Apply the product rule to $2 x$ .

$3 y - 2 = 2^{2} x^{2}$

Raise $2$ to the power of $2$ .

$3 y - 2 = 4 x^{2}$

Solve for $y$ .

Add $2$ to both sides of the equation.

$3 y = 4 x^{2} + 2$

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

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

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Step 3

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

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

Step 4

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

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

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

Combine the numerators over the common denominator.

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

Simplify each term.

Apply the product rule to $\frac{\sqrt{3 x - 2}}{2}$ .

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

Rewrite ${\sqrt{3 x - 2}}^{2}$ as $3 x - 2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 x - 2}$ as ${(3 x - 2)}^{\frac{1}{2}}$ .

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

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Combine $\frac{1}{2}$ and $2$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Raise $2$ to the power of $2$ .

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify terms.

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

Add $- 2$ and $2$ .

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

Add $3 x$ and $0$ .

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify the numerator.

Combine the numerators over the common denominator.

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

Factor $2$ out of $4 x^{2} + 2$ .

Factor $2$ out of $4 x^{2}$ .

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

Factor $2$ out of $2$ .

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

Factor $2$ out of $2 (2 x^{2}) + 2 (1)$ .

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Apply the distributive property.

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

Multiply $2$ by $2$ .

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

Multiply $2$ by $1$ .

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

Subtract $2$ from $2$ .

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

Add $4 x^{2}$ and $0$ .

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

Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

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

Pull terms out from under the radical, assuming positive real numbers.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

Find the Inverse y=( square root of 3x-2)/2

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