Precalculus Examples

$f (x) = 5 \sqrt[5]{x + 7}$ ; find $f^{- 1} (x)$

Step 1

Write $f (x) = 5 \sqrt[5]{x + 7}$ as an equation.

$y = 5 \sqrt[5]{x + 7}$

Step 2

Interchange the variables.

$x = 5 \sqrt[5]{y + 7}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $5 \sqrt[5]{y + 7} = x$ .

$5 \sqrt[5]{y + 7} = x$

Step 3.2

Divide each term in $5 \sqrt[5]{y + 7} = x$ by $5$ and simplify.

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Step 3.2.1

Divide each term in $5 \sqrt[5]{y + 7} = x$ by $5$ .

$\frac{5 \sqrt[5]{y + 7}}{5} = \frac{x}{5}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $5$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{5 \sqrt[5]{y + 7}}{5} = \frac{x}{5}$

Step 3.2.2.1.2

Divide $\sqrt[5]{y + 7}$ by $1$ .

$\sqrt[5]{y + 7} = \frac{x}{5}$

Step 3.3

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

${\sqrt[5]{y + 7}}^{5} = {(\frac{x}{5})}^{5}$

Step 3.4

Simplify each side of the equation.

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Step 3.4.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[5]{y + 7}$ as ${(y + 7)}^{\frac{1}{5}}$ .

${({(y + 7)}^{\frac{1}{5}})}^{5} = {(\frac{x}{5})}^{5}$

Step 3.4.2

Simplify the left side.

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Step 3.4.2.1

Simplify ${({(y + 7)}^{\frac{1}{5}})}^{5}$ .

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Step 3.4.2.1.1

Multiply the exponents in ${({(y + 7)}^{\frac{1}{5}})}^{5}$ .

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Step 3.4.2.1.1.1

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

${(y + 7)}^{\frac{1}{5} \cdot 5} = {(\frac{x}{5})}^{5}$

Step 3.4.2.1.1.2

Cancel the common factor of $5$ .

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Step 3.4.2.1.1.2.1

Cancel the common factor.

${(y + 7)}^{\frac{1}{5} \cdot 5} = {(\frac{x}{5})}^{5}$

Step 3.4.2.1.1.2.2

Rewrite the expression.

${(y + 7)}^{1} = {(\frac{x}{5})}^{5}$

Step 3.4.2.1.2

Simplify.

$y + 7 = {(\frac{x}{5})}^{5}$

Step 3.4.3

Simplify the right side.

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Step 3.4.3.1

Simplify ${(\frac{x}{5})}^{5}$ .

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Step 3.4.3.1.1

Apply the product rule to $\frac{x}{5}$ .

$y + 7 = \frac{x^{5}}{5^{5}}$

Step 3.4.3.1.2

Raise $5$ to the power of $5$ .

$y + 7 = \frac{x^{5}}{3125}$

Step 3.5

Subtract $7$ from both sides of the equation.

$y = \frac{x^{5}}{3125} - 7$

Step 4

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

$f^{- 1} (x) = \frac{x^{5}}{3125} - 7$

Step 5

Verify if $f^{- 1} (x) = \frac{x^{5}}{3125} - 7$ is the inverse of $f (x) = 5 \sqrt[5]{x + 7}$ .

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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 $f^{- 1} (5 \sqrt[5]{x + 7})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (5 \sqrt[5]{x + 7}) = \frac{{(5 \sqrt[5]{x + 7})}^{5}}{3125} - 7$

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Simplify the numerator.

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Step 5.2.3.1.1

Apply the product rule to $5 \sqrt[5]{x + 7}$ .

$f^{- 1} (5 \sqrt[5]{x + 7}) = \frac{5^{5} {\sqrt[5]{x + 7}}^{5}}{3125} - 7$

Step 5.2.3.1.2

Raise $5$ to the power of $5$ .

$f^{- 1} (5 \sqrt[5]{x + 7}) = \frac{3125 {\sqrt[5]{x + 7}}^{5}}{3125} - 7$

Step 5.2.3.1.3

Rewrite ${\sqrt[5]{x + 7}}^{5}$ as $x + 7$ .

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Step 5.2.3.1.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[5]{x + 7}$ as ${(x + 7)}^{\frac{1}{5}}$ .

$f^{- 1} (5 \sqrt[5]{x + 7}) = \frac{3125 {({(x + 7)}^{\frac{1}{5}})}^{5}}{3125} - 7$

Step 5.2.3.1.3.2

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

$f^{- 1} (5 \sqrt[5]{x + 7}) = \frac{3125 {(x + 7)}^{\frac{1}{5} \cdot 5}}{3125} - 7$

Step 5.2.3.1.3.3

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

$f^{- 1} (5 \sqrt[5]{x + 7}) = \frac{3125 {(x + 7)}^{\frac{5}{5}}}{3125} - 7$

Step 5.2.3.1.3.4

Cancel the common factor of $5$ .

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Step 5.2.3.1.3.4.1

Cancel the common factor.

$f^{- 1} (5 \sqrt[5]{x + 7}) = \frac{3125 {(x + 7)}^{\frac{5}{5}}}{3125} - 7$

Step 5.2.3.1.3.4.2

Rewrite the expression.

$f^{- 1} (5 \sqrt[5]{x + 7}) = \frac{3125 (x + 7)}{3125} - 7$

Step 5.2.3.1.3.5

Simplify.

$f^{- 1} (5 \sqrt[5]{x + 7}) = \frac{3125 (x + 7)}{3125} - 7$

Step 5.2.3.2

Cancel the common factor of $3125$ .

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Step 5.2.3.2.1

Cancel the common factor.

$f^{- 1} (5 \sqrt[5]{x + 7}) = \frac{3125 (x + 7)}{3125} - 7$

Step 5.2.3.2.2

Divide $x + 7$ by $1$ .

$f^{- 1} (5 \sqrt[5]{x + 7}) = x + 7 - 7$

Step 5.2.4

Combine the opposite terms in $x + 7 - 7$ .

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Step 5.2.4.1

Subtract $7$ from $7$ .

$f^{- 1} (5 \sqrt[5]{x + 7}) = x + 0$

Step 5.2.4.2

Add $x$ and $0$ .

$f^{- 1} (5 \sqrt[5]{x + 7}) = x$

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 $f (\frac{x^{5}}{3125} - 7)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x^{5}}{3125} - 7) = 5 \sqrt[5]{(\frac{x^{5}}{3125} - 7) + 7}$

Step 5.3.3

Add $- 7$ and $7$ .

$f (\frac{x^{5}}{3125} - 7) = 5 \sqrt[5]{\frac{x^{5}}{3125} + 0}$

Step 5.3.4

Add $\frac{x^{5}}{3125}$ and $0$ .

$f (\frac{x^{5}}{3125} - 7) = 5 \sqrt[5]{\frac{x^{5}}{3125}}$

Step 5.3.5

Rewrite $3125$ as $5^{5}$ .

$f (\frac{x^{5}}{3125} - 7) = 5 \sqrt[5]{\frac{x^{5}}{5^{5}}}$

Step 5.3.6

Rewrite $\frac{x^{5}}{5^{5}}$ as ${(\frac{x}{5})}^{5}$ .

$f (\frac{x^{5}}{3125} - 7) = 5 \sqrt[5]{{(\frac{x}{5})}^{5}}$

Step 5.3.7

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

$f (\frac{x^{5}}{3125} - 7) = 5 (\frac{x}{5})$

Step 5.3.8

Cancel the common factor of $5$ .

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Step 5.3.8.1

Cancel the common factor.

$f (\frac{x^{5}}{3125} - 7) = 5 (\frac{x}{5})$

Step 5.3.8.2

Rewrite the expression.

$f (\frac{x^{5}}{3125} - 7) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x^{5}}{3125} - 7$ is the inverse of $f (x) = 5 \sqrt[5]{x + 7}$ .

$f^{- 1} (x) = \frac{x^{5}}{3125} - 7$