Precalculus Examples

$f (x) = \sqrt[5]{x + 2}$ ; find $f^{- 1} (x)$

Step 1

Write $f (x) = \sqrt[5]{x + 2}$ as an equation.

$y = \sqrt[5]{x + 2}$

Step 2

Interchange the variables.

$x = \sqrt[5]{y + 2}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\sqrt[5]{y + 2} = x$ .

$\sqrt[5]{y + 2} = x$

Step 3.2

To remove the radical on the left side of the equation, raise both sides of the equation to the power of $5$ .

${\sqrt[5]{y + 2}}^{5} = x^{5}$

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 $\sqrt[5]{y + 2}$ as ${(y + 2)}^{\frac{1}{5}}$ .

${({(y + 2)}^{\frac{1}{5}})}^{5} = x^{5}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Simplify ${({(y + 2)}^{\frac{1}{5}})}^{5}$ .

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Step 3.3.2.1.1

Multiply the exponents in ${({(y + 2)}^{\frac{1}{5}})}^{5}$ .

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Step 3.3.2.1.1.1

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

${(y + 2)}^{\frac{1}{5} \cdot 5} = x^{5}$

Step 3.3.2.1.1.2

Cancel the common factor of $5$ .

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Step 3.3.2.1.1.2.1

Cancel the common factor.

${(y + 2)}^{\frac{1}{5} \cdot 5} = x^{5}$

Step 3.3.2.1.1.2.2

Rewrite the expression.

${(y + 2)}^{1} = x^{5}$

Step 3.3.2.1.2

Simplify.

$y + 2 = x^{5}$

Step 3.4

Subtract $2$ from both sides of the equation.

$y = x^{5} - 2$

Step 4

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

$f^{- 1} (x) = x^{5} - 2$

Step 5

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

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

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

Step 5.2.3

Rewrite ${\sqrt[5]{x + 2}}^{5}$ as $x + 2$ .

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Step 5.2.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[5]{x + 2}$ as ${(x + 2)}^{\frac{1}{5}}$ .

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

Step 5.2.3.2

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

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

Step 5.2.3.3

Combine $\frac{1}{5}$ and $5$ .

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

Step 5.2.3.4

Cancel the common factor of $5$ .

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Step 5.2.3.4.1

Cancel the common factor.

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

Step 5.2.3.4.2

Rewrite the expression.

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

Step 5.2.3.5

Simplify.

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

Step 5.2.4

Combine the opposite terms in $x + 2 - 2$ .

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Step 5.2.4.1

Subtract $2$ from $2$ .

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

Step 5.2.4.2

Add $x$ and $0$ .

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

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 $f (x^{5} - 2)$ by substituting in the value of $f^{- 1}$ into $f$ .

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

Step 5.3.3

Add $- 2$ and $2$ .

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

Step 5.3.4

Add $x^{5}$ and $0$ .

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

Step 5.3.5

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

$f (x^{5} - 2) = x$

Step 5.4

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

$f^{- 1} (x) = x^{5} - 2$