Precalculus Examples

$f (x) = 8 x^{3} + 5$

Step 1

Write $f (x) = 8 x^{3} + 5$ as an equation.

$y = 8 x^{3} + 5$

Step 2

Interchange the variables.

$x = 8 y^{3} + 5$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $8 y^{3} + 5 = x$ .

$8 y^{3} + 5 = x$

Step 3.2

Subtract $5$ from both sides of the equation.

$8 y^{3} = x - 5$

Step 3.3

Divide each term in $8 y^{3} = x - 5$ by $8$ and simplify.

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Step 3.3.1

Divide each term in $8 y^{3} = x - 5$ by $8$ .

$\frac{8 y^{3}}{8} = \frac{x}{8} + \frac{- 5}{8}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $8$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{8 y^{3}}{8} = \frac{x}{8} + \frac{- 5}{8}$

Step 3.3.2.1.2

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

$y^{3} = \frac{x}{8} + \frac{- 5}{8}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Move the negative in front of the fraction.

$y^{3} = \frac{x}{8} - \frac{5}{8}$

Step 3.4

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

$y = \sqrt[3]{\frac{x}{8} - \frac{5}{8}}$

Step 3.5

Simplify $\sqrt[3]{\frac{x}{8} - \frac{5}{8}}$ .

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Step 3.5.1

Combine the numerators over the common denominator.

$y = \sqrt[3]{\frac{x - 5}{8}}$

Step 3.5.2

Rewrite $\frac{x - 5}{8}$ as ${(\frac{1}{2})}^{3} (x - 5)$ .

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Step 3.5.2.1

Factor the perfect power $1^{3}$ out of $x - 5$ .

$y = \sqrt[3]{\frac{1^{3} (x - 5)}{8}}$

Step 3.5.2.2

Factor the perfect power $2^{3}$ out of $8$ .

$y = \sqrt[3]{\frac{1^{3} (x - 5)}{2^{3} \cdot 1}}$

Step 3.5.2.3

Rearrange the fraction $\frac{1^{3} (x - 5)}{2^{3} \cdot 1}$ .

$y = \sqrt[3]{{(\frac{1}{2})}^{3} (x - 5)}$

Step 3.5.3

Pull terms out from under the radical.

$y = \frac{1}{2} \sqrt[3]{x - 5}$

Step 3.5.4

Combine $\frac{1}{2}$ and $\sqrt[3]{x - 5}$ .

$y = \frac{\sqrt[3]{x - 5}}{2}$

Step 4

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

$f^{- 1} (x) = \frac{\sqrt[3]{x - 5}}{2}$

Step 5

Verify if $f^{- 1} (x) = \frac{\sqrt[3]{x - 5}}{2}$ is the inverse of $f (x) = 8 x^{3} + 5$ .

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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 $f^{- 1} (8 x^{3} + 5)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (8 x^{3} + 5) = \frac{\sqrt[3]{(8 x^{3} + 5) - 5}}{2}$

Step 5.2.3

Simplify the numerator.

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Step 5.2.3.1

Subtract $5$ from $5$ .

$f^{- 1} (8 x^{3} + 5) = \frac{\sqrt[3]{8 x^{3} + 0}}{2}$

Step 5.2.3.2

Add $8 x^{3}$ and $0$ .

$f^{- 1} (8 x^{3} + 5) = \frac{\sqrt[3]{8 x^{3}}}{2}$

Step 5.2.3.3

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

$f^{- 1} (8 x^{3} + 5) = \frac{\sqrt[3]{{(2 x)}^{3}}}{2}$

Step 5.2.3.4

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

$f^{- 1} (8 x^{3} + 5) = \frac{2 x}{2}$

Step 5.2.4

Cancel the common factor of $2$ .

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Step 5.2.4.1

Cancel the common factor.

$f^{- 1} (8 x^{3} + 5) = \frac{2 x}{2}$

Step 5.2.4.2

Divide $x$ by $1$ .

$f^{- 1} (8 x^{3} + 5) = x$

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 $f (\frac{\sqrt[3]{x - 5}}{2})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\sqrt[3]{x - 5}}{2}) = 8 {(\frac{\sqrt[3]{x - 5}}{2})}^{3} + 5$

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Apply the product rule to $\frac{\sqrt[3]{x - 5}}{2}$ .

$f (\frac{\sqrt[3]{x - 5}}{2}) = 8 (\frac{{\sqrt[3]{x - 5}}^{3}}{2^{3}}) + 5$

Step 5.3.3.2

Rewrite ${\sqrt[3]{x - 5}}^{3}$ as $x - 5$ .

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Step 5.3.3.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x - 5}$ as ${(x - 5)}^{\frac{1}{3}}$ .

$f (\frac{\sqrt[3]{x - 5}}{2}) = 8 (\frac{{({(x - 5)}^{\frac{1}{3}})}^{3}}{2^{3}}) + 5$

Step 5.3.3.2.2

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

$f (\frac{\sqrt[3]{x - 5}}{2}) = 8 (\frac{{(x - 5)}^{\frac{1}{3} \cdot 3}}{2^{3}}) + 5$

Step 5.3.3.2.3

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

$f (\frac{\sqrt[3]{x - 5}}{2}) = 8 (\frac{{(x - 5)}^{\frac{3}{3}}}{2^{3}}) + 5$

Step 5.3.3.2.4

Cancel the common factor of $3$ .

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Step 5.3.3.2.4.1

Cancel the common factor.

$f (\frac{\sqrt[3]{x - 5}}{2}) = 8 (\frac{{(x - 5)}^{\frac{3}{3}}}{2^{3}}) + 5$

Step 5.3.3.2.4.2

Rewrite the expression.

$f (\frac{\sqrt[3]{x - 5}}{2}) = 8 (\frac{x - 5}{2^{3}}) + 5$

Step 5.3.3.2.5

Simplify.

$f (\frac{\sqrt[3]{x - 5}}{2}) = 8 (\frac{x - 5}{2^{3}}) + 5$

Step 5.3.3.3

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

$f (\frac{\sqrt[3]{x - 5}}{2}) = 8 (\frac{x - 5}{8}) + 5$

Step 5.3.3.4

Cancel the common factor of $8$ .

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Step 5.3.3.4.1

Cancel the common factor.

$f (\frac{\sqrt[3]{x - 5}}{2}) = 8 (\frac{x - 5}{8}) + 5$

Step 5.3.3.4.2

Rewrite the expression.

$f (\frac{\sqrt[3]{x - 5}}{2}) = x - 5 + 5$

Step 5.3.4

Combine the opposite terms in $x - 5 + 5$ .

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Step 5.3.4.1

Add $- 5$ and $5$ .

$f (\frac{\sqrt[3]{x - 5}}{2}) = x + 0$

Step 5.3.4.2

Add $x$ and $0$ .

$f (\frac{\sqrt[3]{x - 5}}{2}) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\sqrt[3]{x - 5}}{2}$ is the inverse of $f (x) = 8 x^{3} + 5$ .

$f^{- 1} (x) = \frac{\sqrt[3]{x - 5}}{2}$