Find the Inverse ( cube root of x+2)/9

$\frac{\sqrt[3]{x + 2}}{9}$

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

Multiply both sides of the equation by $9$ .

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

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

Simplify ${(9 x)}^{3}$ .

Apply the product rule to $9 x$ .

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

Raise $9$ to the power of $3$ .

$y + 2 = 729 x^{3}$

Subtract $2$ from both sides of the equation.

$y = 729 x^{3} - 2$

Step 3

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

$f^{- 1} (x) = 729 x^{3} - 2$

Step 4

Verify if $f^{- 1} (x) = 729 x^{3} - 2$ is the inverse of $f (x) = \frac{\sqrt[3]{x + 2}}{9}$ .

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

$f^{- 1} (\frac{\sqrt[3]{x + 2}}{9}) = 729 {(\frac{\sqrt[3]{x + 2}}{9})}^{3} - 2$

Simplify each term.

Apply the product rule to $\frac{\sqrt[3]{x + 2}}{9}$ .

$f^{- 1} (\frac{\sqrt[3]{x + 2}}{9}) = 729 (\frac{{\sqrt[3]{x + 2}}^{3}}{9^{3}}) - 2$

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

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

$f^{- 1} (\frac{\sqrt[3]{x + 2}}{9}) = 729 (\frac{{({(x + 2)}^{\frac{1}{3}})}^{3}}{9^{3}}) - 2$

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

$f^{- 1} (\frac{\sqrt[3]{x + 2}}{9}) = 729 (\frac{{(x + 2)}^{\frac{1}{3} \cdot 3}}{9^{3}}) - 2$

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

$f^{- 1} (\frac{\sqrt[3]{x + 2}}{9}) = 729 (\frac{{(x + 2)}^{\frac{3}{3}}}{9^{3}}) - 2$

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} (\frac{\sqrt[3]{x + 2}}{9}) = 729 (\frac{{(x + 2)}^{\frac{3}{3}}}{9^{3}}) - 2$

Rewrite the expression.

$f^{- 1} (\frac{\sqrt[3]{x + 2}}{9}) = 729 (\frac{x + 2}{9^{3}}) - 2$

Simplify.

$f^{- 1} (\frac{\sqrt[3]{x + 2}}{9}) = 729 (\frac{x + 2}{9^{3}}) - 2$

Raise $9$ to the power of $3$ .

$f^{- 1} (\frac{\sqrt[3]{x + 2}}{9}) = 729 (\frac{x + 2}{729}) - 2$

Cancel the common factor of $729$ .

Cancel the common factor.

$f^{- 1} (\frac{\sqrt[3]{x + 2}}{9}) = 729 (\frac{x + 2}{729}) - 2$

Rewrite the expression.

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

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

Subtract $2$ from $2$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

Evaluate $f (729 x^{3} - 2)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (729 x^{3} - 2) = \frac{\sqrt[3]{(729 x^{3} - 2) + 2}}{9}$

Simplify the numerator.

Add $- 2$ and $2$ .

$f (729 x^{3} - 2) = \frac{\sqrt[3]{729 x^{3} + 0}}{9}$

Add $729 x^{3}$ and $0$ .

$f (729 x^{3} - 2) = \frac{\sqrt[3]{729 x^{3}}}{9}$

Rewrite $729 x^{3}$ as ${(9 x)}^{3}$ .

$f (729 x^{3} - 2) = \frac{\sqrt[3]{{(9 x)}^{3}}}{9}$

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

$f (729 x^{3} - 2) = \frac{9 x}{9}$

Cancel the common factor of $9$ .

Cancel the common factor.

$f (729 x^{3} - 2) = \frac{9 x}{9}$

Divide $x$ by $1$ .

$f (729 x^{3} - 2) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 729 x^{3} - 2$ is the inverse of $f (x) = \frac{\sqrt[3]{x + 2}}{9}$ .

$f^{- 1} (x) = 729 x^{3} - 2$

Find the Inverse ( cube root of x+2)/9

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