Precalculus Examples

$f (x) = \frac{{(x - 9)}^{\frac{1}{3}}}{5}$ ; find $f^{- 1} (x)$

Step 1

Write $f (x) = \frac{{(x - 9)}^{\frac{1}{3}}}{5}$ as an equation.

$y = \frac{{(x - 9)}^{\frac{1}{3}}}{5}$

Step 2

Interchange the variables.

$x = \frac{{(y - 9)}^{\frac{1}{3}}}{5}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{{(y - 9)}^{\frac{1}{3}}}{5} = x$ .

$\frac{{(y - 9)}^{\frac{1}{3}}}{5} = x$

Step 3.2

Multiply both sides of the equation by $5$ .

$5 \frac{{(y - 9)}^{\frac{1}{3}}}{5} = 5 x$

Step 3.3

Simplify the left side.

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Step 3.3.1

Cancel the common factor of $5$ .

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Step 3.3.1.1

Cancel the common factor.

$5 \frac{{(y - 9)}^{\frac{1}{3}}}{5} = 5 x$

Step 3.3.1.2

Rewrite the expression.

${(y - 9)}^{\frac{1}{3}} = 5 x$

Step 3.4

Raise each side of the equation to the power of $3$ to eliminate the fractional exponent on the left side.

${({(y - 9)}^{\frac{1}{3}})}^{3} = {(5 x)}^{3}$

Step 3.5

Simplify the exponent.

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Step 3.5.1

Simplify the left side.

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Step 3.5.1.1

Simplify ${({(y - 9)}^{\frac{1}{3}})}^{3}$ .

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Step 3.5.1.1.1

Multiply the exponents in ${({(y - 9)}^{\frac{1}{3}})}^{3}$ .

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Step 3.5.1.1.1.1

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

${(y - 9)}^{\frac{1}{3} \cdot 3} = {(5 x)}^{3}$

Step 3.5.1.1.1.2

Cancel the common factor of $3$ .

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Step 3.5.1.1.1.2.1

Cancel the common factor.

${(y - 9)}^{\frac{1}{3} \cdot 3} = {(5 x)}^{3}$

Step 3.5.1.1.1.2.2

Rewrite the expression.

${(y - 9)}^{1} = {(5 x)}^{3}$

Step 3.5.1.1.2

Simplify.

$y - 9 = {(5 x)}^{3}$

Step 3.5.2

Simplify the right side.

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Step 3.5.2.1

Simplify ${(5 x)}^{3}$ .

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Step 3.5.2.1.1

Apply the product rule to $5 x$ .

$y - 9 = 5^{3} x^{3}$

Step 3.5.2.1.2

Raise $5$ to the power of $3$ .

$y - 9 = 125 x^{3}$

Step 3.6

Add $9$ to both sides of the equation.

$y = 125 x^{3} + 9$

Step 4

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

$f^{- 1} (x) = 125 x^{3} + 9$

Step 5

Verify if $f^{- 1} (x) = 125 x^{3} + 9$ is the inverse of $f (x) = \frac{{(x - 9)}^{\frac{1}{3}}}{5}$ .

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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 $f^{- 1} (\frac{{(x - 9)}^{\frac{1}{3}}}{5})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{{(x - 9)}^{\frac{1}{3}}}{5}) = 125 {(\frac{{(x - 9)}^{\frac{1}{3}}}{5})}^{3} + 9$

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Apply the product rule to $\frac{{(x - 9)}^{\frac{1}{3}}}{5}$ .

$f^{- 1} (\frac{{(x - 9)}^{\frac{1}{3}}}{5}) = 125 (\frac{{({(x - 9)}^{\frac{1}{3}})}^{3}}{5^{3}}) + 9$

Step 5.2.3.2

Simplify the numerator.

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Step 5.2.3.2.1

Multiply the exponents in ${({(x - 9)}^{\frac{1}{3}})}^{3}$ .

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Step 5.2.3.2.1.1

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

$f^{- 1} (\frac{{(x - 9)}^{\frac{1}{3}}}{5}) = 125 (\frac{{(x - 9)}^{\frac{1}{3} \cdot 3}}{5^{3}}) + 9$

Step 5.2.3.2.1.2

Cancel the common factor of $3$ .

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Step 5.2.3.2.1.2.1

Cancel the common factor.

$f^{- 1} (\frac{{(x - 9)}^{\frac{1}{3}}}{5}) = 125 (\frac{{(x - 9)}^{\frac{1}{3} \cdot 3}}{5^{3}}) + 9$

Step 5.2.3.2.1.2.2

Rewrite the expression.

$f^{- 1} (\frac{{(x - 9)}^{\frac{1}{3}}}{5}) = 125 (\frac{x - 9}{5^{3}}) + 9$

Step 5.2.3.2.2

Simplify.

$f^{- 1} (\frac{{(x - 9)}^{\frac{1}{3}}}{5}) = 125 (\frac{x - 9}{5^{3}}) + 9$

Step 5.2.3.3

Raise $5$ to the power of $3$ .

$f^{- 1} (\frac{{(x - 9)}^{\frac{1}{3}}}{5}) = 125 (\frac{x - 9}{125}) + 9$

Step 5.2.3.4

Cancel the common factor of $125$ .

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Step 5.2.3.4.1

Cancel the common factor.

$f^{- 1} (\frac{{(x - 9)}^{\frac{1}{3}}}{5}) = 125 (\frac{x - 9}{125}) + 9$

Step 5.2.3.4.2

Rewrite the expression.

$f^{- 1} (\frac{{(x - 9)}^{\frac{1}{3}}}{5}) = x - 9 + 9$

Step 5.2.4

Combine the opposite terms in $x - 9 + 9$ .

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Step 5.2.4.1

Add $- 9$ and $9$ .

$f^{- 1} (\frac{{(x - 9)}^{\frac{1}{3}}}{5}) = x + 0$

Step 5.2.4.2

Add $x$ and $0$ .

$f^{- 1} (\frac{{(x - 9)}^{\frac{1}{3}}}{5}) = x$

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 $f (125 x^{3} + 9)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (125 x^{3} + 9) = \frac{{((125 x^{3} + 9) - 9)}^{\frac{1}{3}}}{5}$

Step 5.3.3

Simplify the numerator.

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Step 5.3.3.1

Combine the opposite terms in $125 x^{3} + 9 - 9$ .

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Step 5.3.3.1.1

Subtract $9$ from $9$ .

$f (125 x^{3} + 9) = \frac{{(125 x^{3} + 0)}^{\frac{1}{3}}}{5}$

Step 5.3.3.1.2

Add $125 x^{3}$ and $0$ .

$f (125 x^{3} + 9) = \frac{{(125 x^{3})}^{\frac{1}{3}}}{5}$

Step 5.3.3.2

Apply the product rule to $125 x^{3}$ .

$f (125 x^{3} + 9) = \frac{125^{\frac{1}{3}} {(x^{3})}^{\frac{1}{3}}}{5}$

Step 5.3.3.3

Rewrite $125$ as $5^{3}$ .

$f (125 x^{3} + 9) = \frac{{(5^{3})}^{\frac{1}{3}} {(x^{3})}^{\frac{1}{3}}}{5}$

Step 5.3.3.4

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

$f (125 x^{3} + 9) = \frac{5^{3 (\frac{1}{3})} {(x^{3})}^{\frac{1}{3}}}{5}$

Step 5.3.3.5

Cancel the common factor of $3$ .

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Step 5.3.3.5.1

Cancel the common factor.

$f (125 x^{3} + 9) = \frac{5^{3 (\frac{1}{3})} {(x^{3})}^{\frac{1}{3}}}{5}$

Step 5.3.3.5.2

Rewrite the expression.

$f (125 x^{3} + 9) = \frac{5 {(x^{3})}^{\frac{1}{3}}}{5}$

Step 5.3.3.6

Evaluate the exponent.

$f (125 x^{3} + 9) = \frac{5 {(x^{3})}^{\frac{1}{3}}}{5}$

Step 5.3.3.7

Multiply the exponents in ${(x^{3})}^{\frac{1}{3}}$ .

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Step 5.3.3.7.1

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

$f (125 x^{3} + 9) = \frac{5 x^{3 (\frac{1}{3})}}{5}$

Step 5.3.3.7.2

Cancel the common factor of $3$ .

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Step 5.3.3.7.2.1

Cancel the common factor.

$f (125 x^{3} + 9) = \frac{5 x^{3 (\frac{1}{3})}}{5}$

Step 5.3.3.7.2.2

Rewrite the expression.

$f (125 x^{3} + 9) = \frac{5 x}{5}$

Step 5.3.3.8

Simplify.

$f (125 x^{3} + 9) = \frac{5 x}{5}$

Step 5.3.4

Cancel the common factor of $5$ .

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Step 5.3.4.1

Cancel the common factor.

$f (125 x^{3} + 9) = \frac{5 x}{5}$

Step 5.3.4.2

Divide $x$ by $1$ .

$f (125 x^{3} + 9) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 125 x^{3} + 9$ is the inverse of $f (x) = \frac{{(x - 9)}^{\frac{1}{3}}}{5}$ .

$f^{- 1} (x) = 125 x^{3} + 9$