Precalculus Examples

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

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

Move all terms not containing $y$ to the right side of the equation.

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Step 3.2.1

Add $9$ to both sides of the equation.

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

Step 3.2.2

Add $x$ to both sides of the equation.

$y^{\frac{1}{3}} = 9 + x$

Step 3.3

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

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

Step 3.4

Simplify the exponent.

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Step 3.4.1

Simplify the left side.

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Step 3.4.1.1

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

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Step 3.4.1.1.1

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

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Step 3.4.1.1.1.1

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

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

Step 3.4.1.1.1.2

Cancel the common factor of $3$ .

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Step 3.4.1.1.1.2.1

Cancel the common factor.

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

Step 3.4.1.1.1.2.2

Rewrite the expression.

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

Step 3.4.1.1.2

Simplify.

$y = {(9 + x)}^{3}$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify ${(9 + x)}^{3}$ .

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Step 3.4.2.1.1

Use the Binomial Theorem.

$y = 9^{3} + 3 \cdot 9^{2} x + 3 \cdot 9 x^{2} + x^{3}$

Step 3.4.2.1.2

Simplify each term.

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Step 3.4.2.1.2.1

Raise $9$ to the power of $3$ .

$y = 729 + 3 \cdot 9^{2} x + 3 \cdot 9 x^{2} + x^{3}$

Step 3.4.2.1.2.2

Raise $9$ to the power of $2$ .

$y = 729 + 3 \cdot 81 x + 3 \cdot 9 x^{2} + x^{3}$

Step 3.4.2.1.2.3

Multiply $3$ by $81$ .

$y = 729 + 243 x + 3 \cdot 9 x^{2} + x^{3}$

Step 3.4.2.1.2.4

Multiply $3$ by $9$ .

$y = 729 + 243 x + 27 x^{2} + x^{3}$

Step 3.5

Simplify $729 + 243 x + 27 x^{2} + x^{3}$ .

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Step 3.5.1

Move $729$ .

$y = 243 x + 27 x^{2} + x^{3} + 729$

Step 3.5.2

Move $243 x$ .

$y = 27 x^{2} + x^{3} + 243 x + 729$

Step 3.5.3

Reorder $27 x^{2}$ and $x^{3}$ .

$y = x^{3} + 27 x^{2} + 243 x + 729$

Step 4

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

$f^{- 1} (x) = x^{3} + 27 x^{2} + 243 x + 729$

Step 5

Verify if $f^{- 1} (x) = x^{3} + 27 x^{2} + 243 x + 729$ is the inverse of $f (x) = x^{\frac{1}{3}} - 9$ .

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

$f^{- 1} (x^{\frac{1}{3}} - 9) = {(x^{\frac{1}{3}} - 9)}^{3} + 27 {(x^{\frac{1}{3}} - 9)}^{2} + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Use the Binomial Theorem.

$f^{- 1} (x^{\frac{1}{3}} - 9) = {(x^{\frac{1}{3}})}^{3} + 3 {(x^{\frac{1}{3}})}^{2} \cdot - 9 + 3 x^{\frac{1}{3}} {(- 9)}^{2} + {(- 9)}^{3} + 27 {(x^{\frac{1}{3}} - 9)}^{2} + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.2

Simplify each term.

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Step 5.2.3.2.1

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

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Step 5.2.3.2.1.1

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

$f^{- 1} (x^{\frac{1}{3}} - 9) = x^{\frac{1}{3} \cdot 3} + 3 {(x^{\frac{1}{3}})}^{2} \cdot - 9 + 3 x^{\frac{1}{3}} {(- 9)}^{2} + {(- 9)}^{3} + 27 {(x^{\frac{1}{3}} - 9)}^{2} + 243 (x^{\frac{1}{3}} - 9) + 729$

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.

Step 5.2.3.2.1.2.2

Rewrite the expression.

$f^{- 1} (x^{\frac{1}{3}} - 9) = x + 3 {(x^{\frac{1}{3}})}^{2} \cdot - 9 + 3 x^{\frac{1}{3}} {(- 9)}^{2} + {(- 9)}^{3} + 27 {(x^{\frac{1}{3}} - 9)}^{2} + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.2.2

Simplify.

$f^{- 1} (x^{\frac{1}{3}} - 9) = x + 3 {(x^{\frac{1}{3}})}^{2} \cdot - 9 + 3 x^{\frac{1}{3}} {(- 9)}^{2} + {(- 9)}^{3} + 27 {(x^{\frac{1}{3}} - 9)}^{2} + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.2.3

Multiply the exponents in ${(x^{\frac{1}{3}})}^{2}$ .

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Step 5.2.3.2.3.1

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

$f^{- 1} (x^{\frac{1}{3}} - 9) = x + 3 x^{\frac{1}{3} \cdot 2} \cdot - 9 + 3 x^{\frac{1}{3}} {(- 9)}^{2} + {(- 9)}^{3} + 27 {(x^{\frac{1}{3}} - 9)}^{2} + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.2.3.2

Combine $\frac{1}{3}$ and $2$ .

$f^{- 1} (x^{\frac{1}{3}} - 9) = x + 3 x^{\frac{2}{3}} \cdot - 9 + 3 x^{\frac{1}{3}} {(- 9)}^{2} + {(- 9)}^{3} + 27 {(x^{\frac{1}{3}} - 9)}^{2} + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.2.4

Multiply $- 9$ by $3$ .

$f^{- 1} (x^{\frac{1}{3}} - 9) = x - 27 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} {(- 9)}^{2} + {(- 9)}^{3} + 27 {(x^{\frac{1}{3}} - 9)}^{2} + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.2.5

Raise $- 9$ to the power of $2$ .

$f^{- 1} (x^{\frac{1}{3}} - 9) = x - 27 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} \cdot 81 + {(- 9)}^{3} + 27 {(x^{\frac{1}{3}} - 9)}^{2} + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.2.6

Multiply $81$ by $3$ .

$f^{- 1} (x^{\frac{1}{3}} - 9) = x - 27 x^{\frac{2}{3}} + 243 x^{\frac{1}{3}} + {(- 9)}^{3} + 27 {(x^{\frac{1}{3}} - 9)}^{2} + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.2.7

Raise $- 9$ to the power of $3$ .

$f^{- 1} (x^{\frac{1}{3}} - 9) = x - 27 x^{\frac{2}{3}} + 243 x^{\frac{1}{3}} - 729 + 27 {(x^{\frac{1}{3}} - 9)}^{2} + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.3

Rewrite ${(x^{\frac{1}{3}} - 9)}^{2}$ as $(x^{\frac{1}{3}} - 9) (x^{\frac{1}{3}} - 9)$ .

$f^{- 1} (x^{\frac{1}{3}} - 9) = x - 27 x^{\frac{2}{3}} + 243 x^{\frac{1}{3}} - 729 + 27 ((x^{\frac{1}{3}} - 9) (x^{\frac{1}{3}} - 9)) + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.4

Expand $(x^{\frac{1}{3}} - 9) (x^{\frac{1}{3}} - 9)$ using the FOIL Method.

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Step 5.2.3.4.1

Apply the distributive property.

$f^{- 1} (x^{\frac{1}{3}} - 9) = x - 27 x^{\frac{2}{3}} + 243 x^{\frac{1}{3}} - 729 + 27 (x^{\frac{1}{3}} (x^{\frac{1}{3}} - 9) - 9 (x^{\frac{1}{3}} - 9)) + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.4.2

Apply the distributive property.

$f^{- 1} (x^{\frac{1}{3}} - 9) = x - 27 x^{\frac{2}{3}} + 243 x^{\frac{1}{3}} - 729 + 27 (x^{\frac{1}{3}} x^{\frac{1}{3}} + x^{\frac{1}{3}} \cdot - 9 - 9 (x^{\frac{1}{3}} - 9)) + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.4.3

Apply the distributive property.

$f^{- 1} (x^{\frac{1}{3}} - 9) = x - 27 x^{\frac{2}{3}} + 243 x^{\frac{1}{3}} - 729 + 27 (x^{\frac{1}{3}} x^{\frac{1}{3}} + x^{\frac{1}{3}} \cdot - 9 - 9 x^{\frac{1}{3}} - 9 \cdot - 9) + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.5

Simplify and combine like terms.

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Step 5.2.3.5.1

Simplify each term.

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Step 5.2.3.5.1.1

Multiply $x^{\frac{1}{3}}$ by $x^{\frac{1}{3}}$ by adding the exponents.

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Step 5.2.3.5.1.1.1

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

$f^{- 1} (x^{\frac{1}{3}} - 9) = x - 27 x^{\frac{2}{3}} + 243 x^{\frac{1}{3}} - 729 + 27 (x^{\frac{1}{3} + \frac{1}{3}} + x^{\frac{1}{3}} \cdot - 9 - 9 x^{\frac{1}{3}} - 9 \cdot - 9) + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.5.1.1.2

Combine the numerators over the common denominator.

$f^{- 1} (x^{\frac{1}{3}} - 9) = x - 27 x^{\frac{2}{3}} + 243 x^{\frac{1}{3}} - 729 + 27 (x^{\frac{1 + 1}{3}} + x^{\frac{1}{3}} \cdot - 9 - 9 x^{\frac{1}{3}} - 9 \cdot - 9) + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.5.1.1.3

Add $1$ and $1$ .

$f^{- 1} (x^{\frac{1}{3}} - 9) = x - 27 x^{\frac{2}{3}} + 243 x^{\frac{1}{3}} - 729 + 27 (x^{\frac{2}{3}} + x^{\frac{1}{3}} \cdot - 9 - 9 x^{\frac{1}{3}} - 9 \cdot - 9) + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.5.1.2

Move $- 9$ to the left of $x^{\frac{1}{3}}$ .

$f^{- 1} (x^{\frac{1}{3}} - 9) = x - 27 x^{\frac{2}{3}} + 243 x^{\frac{1}{3}} - 729 + 27 (x^{\frac{2}{3}} - 9 \cdot x^{\frac{1}{3}} - 9 x^{\frac{1}{3}} - 9 \cdot - 9) + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.5.1.3

Multiply $- 9$ by $- 9$ .

$f^{- 1} (x^{\frac{1}{3}} - 9) = x - 27 x^{\frac{2}{3}} + 243 x^{\frac{1}{3}} - 729 + 27 (x^{\frac{2}{3}} - 9 x^{\frac{1}{3}} - 9 x^{\frac{1}{3}} + 81) + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.5.2

Subtract $9 x^{\frac{1}{3}}$ from $- 9 x^{\frac{1}{3}}$ .

$f^{- 1} (x^{\frac{1}{3}} - 9) = x - 27 x^{\frac{2}{3}} + 243 x^{\frac{1}{3}} - 729 + 27 (x^{\frac{2}{3}} - 18 x^{\frac{1}{3}} + 81) + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.6

Apply the distributive property.

$f^{- 1} (x^{\frac{1}{3}} - 9) = x - 27 x^{\frac{2}{3}} + 243 x^{\frac{1}{3}} - 729 + 27 x^{\frac{2}{3}} + 27 (- 18 x^{\frac{1}{3}}) + 27 \cdot 81 + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.7

Simplify.

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Step 5.2.3.7.1

Multiply $- 18$ by $27$ .

$f^{- 1} (x^{\frac{1}{3}} - 9) = x - 27 x^{\frac{2}{3}} + 243 x^{\frac{1}{3}} - 729 + 27 x^{\frac{2}{3}} - 486 x^{\frac{1}{3}} + 27 \cdot 81 + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.7.2

Multiply $27$ by $81$ .

$f^{- 1} (x^{\frac{1}{3}} - 9) = x - 27 x^{\frac{2}{3}} + 243 x^{\frac{1}{3}} - 729 + 27 x^{\frac{2}{3}} - 486 x^{\frac{1}{3}} + 2187 + 243 (x^{\frac{1}{3}} - 9) + 729$

Step 5.2.3.8

Apply the distributive property.

$f^{- 1} (x^{\frac{1}{3}} - 9) = x - 27 x^{\frac{2}{3}} + 243 x^{\frac{1}{3}} - 729 + 27 x^{\frac{2}{3}} - 486 x^{\frac{1}{3}} + 2187 + 243 x^{\frac{1}{3}} + 243 \cdot - 9 + 729$

Step 5.2.3.9

Multiply $243$ by $- 9$ .

$f^{- 1} (x^{\frac{1}{3}} - 9) = x - 27 x^{\frac{2}{3}} + 243 x^{\frac{1}{3}} - 729 + 27 x^{\frac{2}{3}} - 486 x^{\frac{1}{3}} + 2187 + 243 x^{\frac{1}{3}} - 2187 + 729$

Step 5.2.4

Simplify by adding terms.

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Step 5.2.4.1

Combine the opposite terms in $x - 27 x^{\frac{2}{3}} + 243 x^{\frac{1}{3}} - 729 + 27 x^{\frac{2}{3}} - 486 x^{\frac{1}{3}} + 2187 + 243 x^{\frac{1}{3}} - 2187 + 729$ .

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Step 5.2.4.1.1

Add $- 27 x^{\frac{2}{3}}$ and $27 x^{\frac{2}{3}}$ .

$f^{- 1} (x^{\frac{1}{3}} - 9) = x + 243 x^{\frac{1}{3}} - 729 + 0 - 486 x^{\frac{1}{3}} + 2187 + 243 x^{\frac{1}{3}} - 2187 + 729$

Step 5.2.4.1.2

Add $x + 243 x^{\frac{1}{3}} - 729$ and $0$ .

$f^{- 1} (x^{\frac{1}{3}} - 9) = x + 243 x^{\frac{1}{3}} - 729 - 486 x^{\frac{1}{3}} + 2187 + 243 x^{\frac{1}{3}} - 2187 + 729$

Step 5.2.4.1.3

Subtract $2187$ from $2187$ .

$f^{- 1} (x^{\frac{1}{3}} - 9) = x + 243 x^{\frac{1}{3}} - 729 - 486 x^{\frac{1}{3}} + 243 x^{\frac{1}{3}} + 0 + 729$

Step 5.2.4.1.4

Add $x + 243 x^{\frac{1}{3}} - 729 - 486 x^{\frac{1}{3}} + 243 x^{\frac{1}{3}}$ and $0$ .

$f^{- 1} (x^{\frac{1}{3}} - 9) = x + 243 x^{\frac{1}{3}} - 729 - 486 x^{\frac{1}{3}} + 243 x^{\frac{1}{3}} + 729$

Step 5.2.4.1.5

Add $- 729$ and $729$ .

$f^{- 1} (x^{\frac{1}{3}} - 9) = x + 243 x^{\frac{1}{3}} - 486 x^{\frac{1}{3}} + 243 x^{\frac{1}{3}} + 0$

Step 5.2.4.1.6

Add $x + 243 x^{\frac{1}{3}} - 486 x^{\frac{1}{3}} + 243 x^{\frac{1}{3}}$ and $0$ .

$f^{- 1} (x^{\frac{1}{3}} - 9) = x + 243 x^{\frac{1}{3}} - 486 x^{\frac{1}{3}} + 243 x^{\frac{1}{3}}$

Step 5.2.4.2

Subtract $486 x^{\frac{1}{3}}$ from $243 x^{\frac{1}{3}}$ .

$f^{- 1} (x^{\frac{1}{3}} - 9) = x - 243 x^{\frac{1}{3}} + 243 x^{\frac{1}{3}}$

Step 5.2.4.3

Combine the opposite terms in $x - 243 x^{\frac{1}{3}} + 243 x^{\frac{1}{3}}$ .

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Step 5.2.4.3.1

Add $- 243 x^{\frac{1}{3}}$ and $243 x^{\frac{1}{3}}$ .

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

Step 5.2.4.3.2

Add $x$ and $0$ .

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

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 $f (x^{3} + 27 x^{2} + 243 x + 729)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (x^{3} + 27 x^{2} + 243 x + 729) = {(x^{3} + 27 x^{2} + 243 x + 729)}^{\frac{1}{3}} - 9$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = x^{3} + 27 x^{2} + 243 x + 729$ is the inverse of $f (x) = x^{\frac{1}{3}} - 9$ .

$f^{- 1} (x) = x^{3} + 27 x^{2} + 243 x + 729$