Precalculus Examples

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

Step 1

Write $f (x) = \frac{x^{3} - 1}{2}$ as an equation.

$y = \frac{x^{3} - 1}{2}$

Step 2

Interchange the variables.

$x = \frac{y^{3} - 1}{2}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{y^{3} - 1}{2} = x$ .

$\frac{y^{3} - 1}{2} = x$

Step 3.2

Multiply both sides of the equation by $2$ .

$2 \frac{y^{3} - 1}{2} = 2 x$

Step 3.3

Simplify the left side.

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Step 3.3.1

Cancel the common factor of $2$ .

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Step 3.3.1.1

Cancel the common factor.

$2 \frac{y^{3} - 1}{2} = 2 x$

Step 3.3.1.2

Rewrite the expression.

$y^{3} - 1 = 2 x$

Step 3.4

Add $1$ to both sides of the equation.

$y^{3} = 2 x + 1$

Step 3.5

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

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

Step 4

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

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

Step 5

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

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

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

Step 5.2.3

Simplify the numerator.

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Step 5.2.3.1

Rewrite $1$ as $1^{3}$ .

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

Step 5.2.3.2

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 1$ .

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

Step 5.2.3.3

Simplify.

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Step 5.2.3.3.1

Multiply $x$ by $1$ .

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

Step 5.2.3.3.2

One to any power is one.

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

Step 5.2.4

Cancel the common factor of $2$ .

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Step 5.2.4.1

Cancel the common factor.

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

Step 5.2.4.2

Rewrite the expression.

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

Step 5.2.5

Expand $(x - 1) (x^{2} + x + 1)$ by multiplying each term in the first expression by each term in the second expression.

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

Step 5.2.6

Simplify terms.

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Step 5.2.6.1

Combine the opposite terms in $x \cdot x^{2} + x \cdot x + x \cdot 1 - 1 x^{2} - 1 x - 1 \cdot 1$ .

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Step 5.2.6.1.1

Reorder the factors in the terms $x \cdot 1$ and $- 1 x$ .

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

Step 5.2.6.1.2

Subtract $1 x$ from $1 x$ .

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

Step 5.2.6.1.3

Add $x \cdot x^{2} + x \cdot x - 1 x^{2}$ and $0$ .

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

Step 5.2.6.2

Simplify each term.

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Step 5.2.6.2.1

Multiply $x$ by $x^{2}$ by adding the exponents.

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Step 5.2.6.2.1.1

Multiply $x$ by $x^{2}$ .

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Step 5.2.6.2.1.1.1

Raise $x$ to the power of $1$ .

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

Step 5.2.6.2.1.1.2

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

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

Step 5.2.6.2.1.2

Add $1$ and $2$ .

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

Step 5.2.6.2.2

Multiply $x$ by $x$ .

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

Step 5.2.6.2.3

Rewrite $- 1 x^{2}$ as $- x^{2}$ .

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

Step 5.2.6.2.4

Multiply $- 1$ by $1$ .

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

Step 5.2.6.3

Simplify by adding terms.

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Step 5.2.6.3.1

Combine the opposite terms in $x^{3} + x^{2} - x^{2} - 1$ .

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Step 5.2.6.3.1.1

Subtract $x^{2}$ from $x^{2}$ .

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

Step 5.2.6.3.1.2

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

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

Step 5.2.6.3.2

Simplify by adding numbers.

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Step 5.2.6.3.2.1

Add $- 1$ and $1$ .

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

Step 5.2.6.3.2.2

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

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

Step 5.2.7

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

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

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

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

Step 5.3.3

Simplify the numerator.

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Step 5.3.3.1

Rewrite $1$ as $1^{3}$ .

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

Step 5.3.3.2

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = \sqrt[3]{2 x + 1}$ and $b = 1$ .

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

Step 5.3.3.3

Simplify.

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Step 5.3.3.3.1

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

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

Step 5.3.3.3.2

Multiply $\sqrt[3]{2 x + 1}$ by $1$ .

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

Step 5.3.3.3.3

One to any power is one.

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

Step 5.4

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

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