Precalculus Examples

$f (x) = 2 x^{3} - 4$

Step 1

Write $f (x) = 2 x^{3} - 4$ as an equation.

$y = 2 x^{3} - 4$

Step 2

Interchange the variables.

$x = 2 y^{3} - 4$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $2 y^{3} - 4 = x$ .

$2 y^{3} - 4 = x$

Step 3.2

Add $4$ to both sides of the equation.

$2 y^{3} = x + 4$

Step 3.3

Divide each term in $2 y^{3} = x + 4$ by $2$ and simplify.

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Step 3.3.1

Divide each term in $2 y^{3} = x + 4$ by $2$ .

$\frac{2 y^{3}}{2} = \frac{x}{2} + \frac{4}{2}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $2$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{2 y^{3}}{2} = \frac{x}{2} + \frac{4}{2}$

Step 3.3.2.1.2

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

$y^{3} = \frac{x}{2} + \frac{4}{2}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Divide $4$ by $2$ .

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

Step 3.4

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

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

Step 3.5

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

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Step 3.5.1

To write $2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 3.5.2

Combine $2$ and $\frac{2}{2}$ .

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

Step 3.5.3

Combine the numerators over the common denominator.

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

Step 3.5.4

Multiply $2$ by $2$ .

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

Step 3.5.5

Rewrite $\sqrt[3]{\frac{x + 4}{2}}$ as $\frac{\sqrt[3]{x + 4}}{\sqrt[3]{2}}$ .

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

Step 3.5.6

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

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

Step 3.5.7

Combine and simplify the denominator.

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Step 3.5.7.1

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

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

Step 3.5.7.2

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

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

Step 3.5.7.3

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

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

Step 3.5.7.4

Add $1$ and $2$ .

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

Step 3.5.7.5

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

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Step 3.5.7.5.1

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

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

Step 3.5.7.5.2

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

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

Step 3.5.7.5.3

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

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

Step 3.5.7.5.4

Cancel the common factor of $3$ .

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Step 3.5.7.5.4.1

Cancel the common factor.

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

Step 3.5.7.5.4.2

Rewrite the expression.

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

Step 3.5.7.5.5

Evaluate the exponent.

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

Step 3.5.8

Simplify the numerator.

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Step 3.5.8.1

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

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

Step 3.5.8.2

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

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

Step 3.5.9

Simplify with factoring out.

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Step 3.5.9.1

Combine using the product rule for radicals.

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

Step 3.5.9.2

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

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

Step 4

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

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

Step 5

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

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Step 5.1

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

Step 5.2

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

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Step 5.2.1

Set up the composite result function.

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

Step 5.2.2

Evaluate $f^{- 1} (2 x^{3} - 4)$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Step 5.2.3

Simplify the numerator.

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Step 5.2.3.1

Add $- 4$ and $4$ .

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

Step 5.2.3.2

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

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

Step 5.2.3.3

Multiply $4$ by $2$ .

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

Step 5.2.3.4

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

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

Step 5.2.3.5

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

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

Step 5.2.4

Cancel the common factor of $2$ .

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Step 5.2.4.1

Cancel the common factor.

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

Step 5.2.4.2

Divide $x$ by $1$ .

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

Step 5.3

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

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Step 5.3.1

Set up the composite result function.

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

Step 5.3.2

Evaluate $f (\frac{\sqrt[3]{4 (x + 4)}}{2})$ by substituting in the value of $f^{- 1}$ into $f$ .

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

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Apply the product rule to $\frac{\sqrt[3]{4 (x + 4)}}{2}$ .

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

Step 5.3.3.2

Simplify the numerator.

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Step 5.3.3.2.1

Rewrite ${\sqrt[3]{4 (x + 4)}}^{3}$ as $4 (x + 4)$ .

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Step 5.3.3.2.1.1

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

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

Step 5.3.3.2.1.2

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

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

Step 5.3.3.2.1.3

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

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

Step 5.3.3.2.1.4

Cancel the common factor of $3$ .

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Step 5.3.3.2.1.4.1

Cancel the common factor.

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

Step 5.3.3.2.1.4.2

Rewrite the expression.

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

Step 5.3.3.2.1.5

Simplify.

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

Step 5.3.3.2.2

Apply the distributive property.

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

Step 5.3.3.2.3

Multiply $4$ by $4$ .

$f (\frac{\sqrt[3]{4 (x + 4)}}{2}) = 2 (\frac{4 x + 16}{2^{3}}) - 4$

Step 5.3.3.2.4

Factor $4$ out of $4 x + 16$ .

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Step 5.3.3.2.4.1

Factor $4$ out of $4 x$ .

$f (\frac{\sqrt[3]{4 (x + 4)}}{2}) = 2 (\frac{4 (x) + 16}{2^{3}}) - 4$

Step 5.3.3.2.4.2

Factor $4$ out of $16$ .

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

Step 5.3.3.2.4.3

Factor $4$ out of $4 x + 4 \cdot 4$ .

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

Step 5.3.3.3

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

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

Step 5.3.3.4

Cancel the common factor of $2$ .

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Step 5.3.3.4.1

Factor $2$ out of $8$ .

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

Step 5.3.3.4.2

Cancel the common factor.

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

Step 5.3.3.4.3

Rewrite the expression.

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

Step 5.3.3.5

Cancel the common factor of $4$ .

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Step 5.3.3.5.1

Cancel the common factor.

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

Step 5.3.3.5.2

Divide $x + 4$ by $1$ .

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

Step 5.3.4

Combine the opposite terms in $x + 4 - 4$ .

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Step 5.3.4.1

Subtract $4$ from $4$ .

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

Step 5.3.4.2

Add $x$ and $0$ .

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

Step 5.4

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

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