Find the Inverse f(x) = cube root of 9(x-9)

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

Write $f (x) = \sqrt[3]{9 (x - 9)}$ as an equation.

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

Interchange the variables.

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

Solve for $y$ .

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

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

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Apply the distributive property.

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

Multiply.

Multiply $9$ by $- 9$ .

${(9 y - 81)}^{1} = x^{3}$

Simplify.

$9 y - 81 = x^{3}$

Solve for $y$ .

Add $81$ to both sides of the equation.

$9 y = x^{3} + 81$

Divide each term in $9 y = x^{3} + 81$ by $9$ and simplify.

Divide each term in $9 y = x^{3} + 81$ by $9$ .

$\frac{9 y}{9} = \frac{x^{3}}{9} + \frac{81}{9}$

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{9 y}{9} = \frac{x^{3}}{9} + \frac{81}{9}$

Divide $y$ by $1$ .

$y = \frac{x^{3}}{9} + \frac{81}{9}$

Simplify the right side.

Divide $81$ by $9$ .

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

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

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

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

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

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

Simplify each term.

Simplify the numerator.

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

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Apply the distributive property.

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

Multiply $9$ by $- 9$ .

$f^{- 1} (\sqrt[3]{9 (x - 9)}) = \frac{9 x - 81}{9} + 9$

Factor $9$ out of $9 x - 81$ .

Factor $9$ out of $9 x$ .

$f^{- 1} (\sqrt[3]{9 (x - 9)}) = \frac{9 (x) - 81}{9} + 9$

Factor $9$ out of $- 81$ .

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

Factor $9$ out of $9 x + 9 \cdot - 9$ .

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

Cancel the common factor of $9$ .

Cancel the common factor.

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

Divide $x - 9$ by $1$ .

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

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

Add $- 9$ and $9$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Simplify by subtracting numbers.

Subtract $9$ from $9$ .

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

Add $\frac{x^{3}}{9}$ and $0$ .

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

Combine $9$ and $\frac{x^{3}}{9}$ .

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

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{9 x^{3}}{9}$ by cancelling the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

Divide $x^{3}$ by $1$ .

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

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

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

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

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

Find the Inverse f(x) = cube root of 9(x-9)

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