Find the Inverse cube root of 3x+2

$\sqrt[3]{3 x + 2}$

Step 1

Interchange the variables.

$x = \sqrt[3]{3 y + 2}$

Step 2

Solve for $y$ .

Rewrite the equation as $\sqrt[3]{3 y + 2} = x$ .

$\sqrt[3]{3 y + 2} = x$

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

${\sqrt[3]{3 y + 2}}^{3} = x^{3}$

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Solve for $y$ .

Subtract $2$ from both sides of the equation.

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

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

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

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 3

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

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

Step 4

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

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

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

Combine the numerators over the common denominator.

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

Rewrite ${\sqrt[3]{3 x + 2}}^{3}$ as $3 x + 2$ .

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

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

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

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

Combine $\frac{1}{3}$ and $3$ .

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify terms.

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

Subtract $2$ from $2$ .

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

Add $3 x$ and $0$ .

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Apply the distributive property.

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $3$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify by adding numbers.

Add $- 2$ and $2$ .

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

Add $x^{3}$ and $0$ .

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

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

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

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

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

Find the Inverse cube root of 3x+2

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