Algebra Examples

$4 x^{3} - 9$

Step 1

Interchange the variables.

$x = 4 y^{3} - 9$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $4 y^{3} - 9 = x$ .

$4 y^{3} - 9 = x$

Step 2.2

Add $9$ to both sides of the equation.

$4 y^{3} = x + 9$

Step 2.3

Divide each term in $4 y^{3} = x + 9$ by $4$ and simplify.

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Step 2.3.1

Divide each term in $4 y^{3} = x + 9$ by $4$ .

$\frac{4 y^{3}}{4} = \frac{x}{4} + \frac{9}{4}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $4$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{4 y^{3}}{4} = \frac{x}{4} + \frac{9}{4}$

Step 2.3.2.1.2

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

$y^{3} = \frac{x}{4} + \frac{9}{4}$

Step 2.4

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

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

Step 2.5

Simplify $\sqrt[3]{\frac{x}{4} + \frac{9}{4}}$ .

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Step 2.5.1

Combine the numerators over the common denominator.

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

Step 2.5.2

Rewrite $\sqrt[3]{\frac{x + 9}{4}}$ as $\frac{\sqrt[3]{x + 9}}{\sqrt[3]{4}}$ .

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

Step 2.5.3

Multiply $\frac{\sqrt[3]{x + 9}}{\sqrt[3]{4}}$ by $\frac{{\sqrt[3]{4}}^{2}}{{\sqrt[3]{4}}^{2}}$ .

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

Step 2.5.4

Combine and simplify the denominator.

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Step 2.5.4.1

Multiply $\frac{\sqrt[3]{x + 9}}{\sqrt[3]{4}}$ by $\frac{{\sqrt[3]{4}}^{2}}{{\sqrt[3]{4}}^{2}}$ .

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

Step 2.5.4.2

Raise $\sqrt[3]{4}$ to the power of $1$ .

$y = \frac{\sqrt[3]{x + 9} {\sqrt[3]{4}}^{2}}{{\sqrt[3]{4}}^{1} {\sqrt[3]{4}}^{2}}$

Step 2.5.4.3

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

$y = \frac{\sqrt[3]{x + 9} {\sqrt[3]{4}}^{2}}{{\sqrt[3]{4}}^{1 + 2}}$

Step 2.5.4.4

Add $1$ and $2$ .

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

Step 2.5.4.5

Rewrite ${\sqrt[3]{4}}^{3}$ as $4$ .

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Step 2.5.4.5.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{4}$ as $4^{\frac{1}{3}}$ .

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

Step 2.5.4.5.2

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

$y = \frac{\sqrt[3]{x + 9} {\sqrt[3]{4}}^{2}}{4^{\frac{1}{3} \cdot 3}}$

Step 2.5.4.5.3

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

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

Step 2.5.4.5.4

Cancel the common factor of $3$ .

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Step 2.5.4.5.4.1

Cancel the common factor.

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

Step 2.5.4.5.4.2

Rewrite the expression.

$y = \frac{\sqrt[3]{x + 9} {\sqrt[3]{4}}^{2}}{4^{1}}$

Step 2.5.4.5.5

Evaluate the exponent.

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

Step 2.5.5

Simplify the numerator.

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Step 2.5.5.1

Rewrite ${\sqrt[3]{4}}^{2}$ as $\sqrt[3]{4^{2}}$ .

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

Step 2.5.5.2

Raise $4$ to the power of $2$ .

$y = \frac{\sqrt[3]{x + 9} \sqrt[3]{16}}{4}$

Step 2.5.5.3

Rewrite $16$ as $2^{3} \cdot 2$ .

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Step 2.5.5.3.1

Factor $8$ out of $16$ .

$y = \frac{\sqrt[3]{x + 9} \sqrt[3]{8 (2)}}{4}$

Step 2.5.5.3.2

Rewrite $8$ as $2^{3}$ .

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

Step 2.5.5.4

Pull terms out from under the radical.

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

Step 2.5.5.5

Combine using the product rule for radicals.

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

Step 2.5.6

Reduce the expression by cancelling the common factors.

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Step 2.5.6.1

Cancel the common factor of $2$ and $4$ .

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Step 2.5.6.1.1

Factor $2$ out of $2 \sqrt[3]{(x + 9) \cdot 2}$ .

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

Step 2.5.6.1.2

Cancel the common factors.

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Step 2.5.6.1.2.1

Factor $2$ out of $4$ .

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

Step 2.5.6.1.2.2

Cancel the common factor.

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

Step 2.5.6.1.2.3

Rewrite the expression.

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

Step 2.5.6.2

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

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

Step 3

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

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

Step 4

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

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Step 4.1

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Step 4.2

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

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Step 4.2.1

Set up the composite result function.

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

Step 4.2.2

Evaluate $f^{- 1} (4 x^{3} - 9)$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Step 4.2.3

Simplify the numerator.

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Step 4.2.3.1

Add $- 9$ and $9$ .

$f^{- 1} (4 x^{3} - 9) = \frac{\sqrt[3]{2 (4 x^{3} + 0)}}{2}$

Step 4.2.3.2

Add $4 x^{3}$ and $0$ .

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

Step 4.2.3.3

Multiply $2$ by $4$ .

$f^{- 1} (4 x^{3} - 9) = \frac{\sqrt[3]{8 x^{3}}}{2}$

Step 4.2.3.4

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

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

Step 4.2.3.5

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

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

Step 4.2.4

Cancel the common factor of $2$ .

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Step 4.2.4.1

Cancel the common factor.

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

Step 4.2.4.2

Divide $x$ by $1$ .

$f^{- 1} (4 x^{3} - 9) = x$

Step 4.3

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

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Step 4.3.1

Set up the composite result function.

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

Step 4.3.2

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

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

Step 4.3.3

Simplify each term.

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Step 4.3.3.1

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

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

Step 4.3.3.2

Simplify the numerator.

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Step 4.3.3.2.1

Rewrite ${\sqrt[3]{2 (x + 9)}}^{3}$ as $2 (x + 9)$ .

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Step 4.3.3.2.1.1

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

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

Step 4.3.3.2.1.2

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

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

Step 4.3.3.2.1.3

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

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

Step 4.3.3.2.1.4

Cancel the common factor of $3$ .

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Step 4.3.3.2.1.4.1

Cancel the common factor.

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

Step 4.3.3.2.1.4.2

Rewrite the expression.

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

Step 4.3.3.2.1.5

Simplify.

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

Step 4.3.3.2.2

Apply the distributive property.

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

Step 4.3.3.2.3

Multiply $2$ by $9$ .

$f (\frac{\sqrt[3]{2 (x + 9)}}{2}) = 4 (\frac{2 x + 18}{2^{3}}) - 9$

Step 4.3.3.2.4

Factor $2$ out of $2 x + 18$ .

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Step 4.3.3.2.4.1

Factor $2$ out of $2 x$ .

$f (\frac{\sqrt[3]{2 (x + 9)}}{2}) = 4 (\frac{2 (x) + 18}{2^{3}}) - 9$

Step 4.3.3.2.4.2

Factor $2$ out of $18$ .

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

Step 4.3.3.2.4.3

Factor $2$ out of $2 x + 2 \cdot 9$ .

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

Step 4.3.3.3

Raise $2$ to the power of $3$ .

$f (\frac{\sqrt[3]{2 (x + 9)}}{2}) = 4 (\frac{2 (x + 9)}{8}) - 9$

Step 4.3.3.4

Cancel the common factor of $4$ .

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Step 4.3.3.4.1

Factor $4$ out of $8$ .

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

Step 4.3.3.4.2

Cancel the common factor.

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

Step 4.3.3.4.3

Rewrite the expression.

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

Step 4.3.3.5

Cancel the common factor of $2$ .

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Step 4.3.3.5.1

Cancel the common factor.

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

Step 4.3.3.5.2

Divide $x + 9$ by $1$ .

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

Step 4.3.4

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

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Step 4.3.4.1

Subtract $9$ from $9$ .

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

Step 4.3.4.2

Add $x$ and $0$ .

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

Step 4.4

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

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