Find the Inverse 2x^3-9

$2 x^{3} - 9$

Interchange the variables.

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

Solve for $y$ .

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

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

Add $9$ to both sides of the equation.

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

Divide each term by $2$ and simplify.

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

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

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

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

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

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

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

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Combine the numerators over the common denominator.

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

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

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

Multiply $\frac{\sqrt[3]{x + 9}}{\sqrt[3]{2}}$ by $\frac{{\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{2}}$ .

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

Combine and simplify the denominator.

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Multiply $\frac{\sqrt[3]{x + 9}}{\sqrt[3]{2}}$ and $\frac{{\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{2}}$ .

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

Raise $\sqrt[3]{2}$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $2$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $3$ .

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Cancel the common factor.

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

Divide $1$ by $1$ .

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

Evaluate the exponent.

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

Simplify the numerator.

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Rewrite ${\sqrt[3]{2}}^{2}$ as ${(2^{2})}^{\frac{1}{3}}$ .

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

Raise $2$ to the power of $2$ .

$y = \frac{\sqrt[3]{x + 9} \sqrt[3]{4}}{2}$

Simplify with factoring out.

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Combine using the product rule for radicals.

$y = \frac{\sqrt[3]{(x + 9) \cdot 4}}{2}$

Reorder factors in $\frac{\sqrt[3]{(x + 9) \cdot 4}}{2}$ .

$y = \frac{\sqrt[3]{4 (x + 9)}}{2}$

Solve for $y$ and replace with $f^{- 1} (x)$ .

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

$f^{- 1} (x) = \frac{\sqrt[3]{4 (x + 9)}}{2}$

Set up the composite result function.

$g (f (x))$

Evaluate $g (f (x))$ by substituting in the value of $f$ into $g$ .

$g (2 x^{3} - 9) = \frac{\sqrt[3]{4 ((2 x^{3} - 9) + 9)}}{2}$

Simplify the numerator.

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Add $- 9$ and $9$ .

$g (2 x^{3} - 9) = \frac{\sqrt[3]{4 (2 x^{3} + 0)}}{2}$

Add $2 x^{3}$ and $0$ .

$g (2 x^{3} - 9) = \frac{\sqrt[3]{4 \cdot (2 x^{3})}}{2}$

Multiply $4$ by $2$ .

$g (2 x^{3} - 9) = \frac{\sqrt[3]{8 x^{3}}}{2}$

Rewrite $8 x^{3}$ as ${(2 x)}^{3}$ .

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

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

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

Divide $x$ by $1$ .

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

Since $g (f (x)) = x$ , $f^{- 1} (x) = \frac{\sqrt[3]{4 (x + 9)}}{2}$ is the inverse of $f (x) = 2 x^{3} - 9$ .

$f^{- 1} (x) = \frac{\sqrt[3]{4 (x + 9)}}{2}$

Find the Inverse 2x^3-9

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