Find the Inverse (9x-2)^3

${(9 x - 2)}^{3}$

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

Rewrite the equation as ${(9 y - 2)}^{3} = x$ .

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

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

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

Add $2$ to both sides of the equation.

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

Divide each term in $9 y = \sqrt[3]{x} + 2$ by $9$ and simplify.

Divide each term in $9 y = \sqrt[3]{x} + 2$ by $9$ .

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

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Step 3

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

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

Step 4

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

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

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

Combine the numerators over the common denominator.

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

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

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

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

Add $- 2$ and $2$ .

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

Add $9 x$ and $0$ .

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

Cancel the common factor of $9$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $9$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $9$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify terms.

Combine the opposite terms in $\sqrt[3]{x} + 2 - 2$ .

Subtract $2$ from $2$ .

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

Add $\sqrt[3]{x}$ and $0$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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

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

Find the Inverse (9x-2)^3

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