Find the Inverse cube root of 9x-2

$\sqrt[3]{9 x - 2}$

Step 1

Interchange the variables.

$x = \sqrt[3]{9 y - 2}$

Step 2

Solve for $y$ .

Rewrite the equation as $\sqrt[3]{9 y - 2} = x$ .

$\sqrt[3]{9 y - 2} = x$

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

${\sqrt[3]{9 y - 2}}^{3} = x^{3}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{9 y - 2}$ as ${(9 y - 2)}^{\frac{1}{3}}$ .

${({(9 y - 2)}^{\frac{1}{3}})}^{3} = x^{3}$

Simplify the left side.

Simplify ${({(9 y - 2)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(9 y - 2)}^{\frac{1}{3}})}^{3}$ .

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

${(9 y - 2)}^{\frac{1}{3} \cdot 3} = x^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(9 y - 2)}^{\frac{1}{3} \cdot 3} = x^{3}$

Rewrite the expression.

${(9 y - 2)}^{1} = x^{3}$

Simplify.

$9 y - 2 = x^{3}$

Solve for $y$ .

Add $2$ to both sides of the equation.

$9 y = x^{3} + 2$

Divide each term in $9 y = x^{3} + 2$ by $9$ and simplify.

Divide each term in $9 y = x^{3} + 2$ by $9$ .

$\frac{9 y}{9} = \frac{x^{3}}{9} + \frac{2}{9}$

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{9 y}{9} = \frac{x^{3}}{9} + \frac{2}{9}$

Divide $y$ by $1$ .

$y = \frac{x^{3}}{9} + \frac{2}{9}$

Step 3

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

$f^{- 1} (x) = \frac{x^{3}}{9} + \frac{2}{9}$

Step 4

Verify if $f^{- 1} (x) = \frac{x^{3}}{9} + \frac{2}{9}$ is the inverse of $f (x) = \sqrt[3]{9 x - 2}$ .

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 $f^{- 1} (\sqrt[3]{9 x - 2})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\sqrt[3]{9 x - 2}) = \frac{{(\sqrt[3]{9 x - 2})}^{3}}{9} + \frac{2}{9}$

Combine the numerators over the common denominator.

$f^{- 1} (\sqrt[3]{9 x - 2}) = \frac{{(\sqrt[3]{9 x - 2})}^{3} + 2}{9}$

Rewrite ${\sqrt[3]{9 x - 2}}^{3}$ as $9 x - 2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{9 x - 2}$ as ${(9 x - 2)}^{\frac{1}{3}}$ .

$f^{- 1} (\sqrt[3]{9 x - 2}) = \frac{{({(9 x - 2)}^{\frac{1}{3}})}^{3} + 2}{9}$

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

$f^{- 1} (\sqrt[3]{9 x - 2}) = \frac{{(9 x - 2)}^{\frac{1}{3} \cdot 3} + 2}{9}$

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

$f^{- 1} (\sqrt[3]{9 x - 2}) = \frac{{(9 x - 2)}^{\frac{3}{3}} + 2}{9}$

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} (\sqrt[3]{9 x - 2}) = \frac{{(9 x - 2)}^{\frac{3}{3}} + 2}{9}$

Rewrite the expression.

$f^{- 1} (\sqrt[3]{9 x - 2}) = \frac{(9 x - 2) + 2}{9}$

Simplify.

$f^{- 1} (\sqrt[3]{9 x - 2}) = \frac{9 x - 2 + 2}{9}$

Simplify terms.

Combine the opposite terms in $9 x - 2 + 2$ .

Add $- 2$ and $2$ .

$f^{- 1} (\sqrt[3]{9 x - 2}) = \frac{9 x + 0}{9}$

Add $9 x$ and $0$ .

$f^{- 1} (\sqrt[3]{9 x - 2}) = \frac{9 x}{9}$

Cancel the common factor of $9$ .

Cancel the common factor.

$f^{- 1} (\sqrt[3]{9 x - 2}) = \frac{9 x}{9}$

Divide $x$ by $1$ .

$f^{- 1} (\sqrt[3]{9 x - 2}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{x^{3}}{9} + \frac{2}{9})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x^{3}}{9} + \frac{2}{9}) = \sqrt[3]{9 (\frac{x^{3}}{9} + \frac{2}{9}) - 2}$

Apply the distributive property.

$f (\frac{x^{3}}{9} + \frac{2}{9}) = \sqrt[3]{9 (\frac{x^{3}}{9}) + 9 (\frac{2}{9}) - 2}$

Cancel the common factor of $9$ .

Cancel the common factor.

$f (\frac{x^{3}}{9} + \frac{2}{9}) = \sqrt[3]{9 (\frac{x^{3}}{9}) + 9 (\frac{2}{9}) - 2}$

Rewrite the expression.

$f (\frac{x^{3}}{9} + \frac{2}{9}) = \sqrt[3]{x^{3} + 9 (\frac{2}{9}) - 2}$

Cancel the common factor of $9$ .

Cancel the common factor.

$f (\frac{x^{3}}{9} + \frac{2}{9}) = \sqrt[3]{x^{3} + 9 (\frac{2}{9}) - 2}$

Rewrite the expression.

$f (\frac{x^{3}}{9} + \frac{2}{9}) = \sqrt[3]{x^{3} + 2 - 2}$

Simplify by subtracting numbers.

Subtract $2$ from $2$ .

$f (\frac{x^{3}}{9} + \frac{2}{9}) = \sqrt[3]{x^{3} + 0}$

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

$f (\frac{x^{3}}{9} + \frac{2}{9}) = \sqrt[3]{x^{3}}$

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

$f (\frac{x^{3}}{9} + \frac{2}{9}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x^{3}}{9} + \frac{2}{9}$ is the inverse of $f (x) = \sqrt[3]{9 x - 2}$ .

$f^{- 1} (x) = \frac{x^{3}}{9} + \frac{2}{9}$

Find the Inverse cube root of 9x-2

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