Trigonometry Examples

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

Step 1

Write $f (x) = \sqrt[3]{4 x - 1}$ as an equation.

$y = \sqrt[3]{4 x - 1}$

Step 2

Interchange the variables.

$x = \sqrt[3]{4 y - 1}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\sqrt[3]{4 y - 1} = x$ .

$\sqrt[3]{4 y - 1} = x$

Step 3.2

To remove the radical on the left side of the equation, cube both sides of the equation.

${\sqrt[3]{4 y - 1}}^{3} = x^{3}$

Step 3.3

Simplify each side of the equation.

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Step 3.3.1

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

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

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Simplify ${({(4 y - 1)}^{\frac{1}{3}})}^{3}$ .

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Step 3.3.2.1.1

Multiply the exponents in ${({(4 y - 1)}^{\frac{1}{3}})}^{3}$ .

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Step 3.3.2.1.1.1

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

${(4 y - 1)}^{\frac{1}{3} \cdot 3} = x^{3}$

Step 3.3.2.1.1.2

Cancel the common factor of $3$ .

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Step 3.3.2.1.1.2.1

Cancel the common factor.

${(4 y - 1)}^{\frac{1}{3} \cdot 3} = x^{3}$

Step 3.3.2.1.1.2.2

Rewrite the expression.

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

Step 3.3.2.1.2

Simplify.

$4 y - 1 = x^{3}$

Step 3.4

Solve for $y$ .

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Step 3.4.1

Add $1$ to both sides of the equation.

$4 y = x^{3} + 1$

Step 3.4.2

Divide each term in $4 y = x^{3} + 1$ by $4$ and simplify.

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Step 3.4.2.1

Divide each term in $4 y = x^{3} + 1$ by $4$ .

$\frac{4 y}{4} = \frac{x^{3}}{4} + \frac{1}{4}$

Step 3.4.2.2

Simplify the left side.

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Step 3.4.2.2.1

Cancel the common factor of $4$ .

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Step 3.4.2.2.1.1

Cancel the common factor.

$\frac{4 y}{4} = \frac{x^{3}}{4} + \frac{1}{4}$

Step 3.4.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{x^{3}}{4} + \frac{1}{4}$

Step 4

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

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

Step 5

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

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

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

Step 5.2.3

Combine the numerators over the common denominator.

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

Step 5.2.4

Rewrite ${\sqrt[3]{4 x - 1}}^{3}$ as $4 x - 1$ .

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Step 5.2.4.1

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

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

Step 5.2.4.2

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

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

Step 5.2.4.3

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

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

Step 5.2.4.4

Cancel the common factor of $3$ .

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Step 5.2.4.4.1

Cancel the common factor.

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

Step 5.2.4.4.2

Rewrite the expression.

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

Step 5.2.4.5

Simplify.

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

Step 5.2.5

Simplify terms.

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Step 5.2.5.1

Combine the opposite terms in $4 x - 1 + 1$ .

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Step 5.2.5.1.1

Add $- 1$ and $1$ .

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

Step 5.2.5.1.2

Add $4 x$ and $0$ .

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

Step 5.2.5.2

Cancel the common factor of $4$ .

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Step 5.2.5.2.1

Cancel the common factor.

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

Step 5.2.5.2.2

Divide $x$ by $1$ .

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

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

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

Step 5.3.3

Apply the distributive property.

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

Step 5.3.4

Cancel the common factor of $4$ .

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Step 5.3.4.1

Cancel the common factor.

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

Step 5.3.4.2

Rewrite the expression.

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

Step 5.3.5

Cancel the common factor of $4$ .

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Step 5.3.5.1

Cancel the common factor.

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

Step 5.3.5.2

Rewrite the expression.

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

Step 5.3.6

Simplify by subtracting numbers.

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Step 5.3.6.1

Subtract $1$ from $1$ .

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

Step 5.3.6.2

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

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

Step 5.3.7

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

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

Step 5.4

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

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