Precalculus Examples

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

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

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

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

Step 3.2

Multiply both sides of the equation by $9$ .

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

Step 3.3

Simplify the left side.

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Step 3.3.1

Cancel the common factor of $9$ .

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Step 3.3.1.1

Cancel the common factor.

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

Step 3.3.1.2

Rewrite the expression.

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

Step 3.4

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

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Step 3.4.1

Add $1$ to both sides of the equation.

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

Step 3.4.2

Add $9 x$ to both sides of the equation.

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

Step 3.5

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

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

Step 3.6

Simplify the exponent.

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Step 3.6.1

Simplify the left side.

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Step 3.6.1.1

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

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Step 3.6.1.1.1

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

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Step 3.6.1.1.1.1

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

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

Step 3.6.1.1.1.2

Cancel the common factor of $3$ .

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Step 3.6.1.1.1.2.1

Cancel the common factor.

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

Step 3.6.1.1.1.2.2

Rewrite the expression.

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

Step 3.6.1.1.2

Simplify.

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

Step 3.6.2

Simplify the right side.

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Step 3.6.2.1

Simplify ${(1 + 9 x)}^{3}$ .

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Step 3.6.2.1.1

Use the Binomial Theorem.

$y = 1^{3} + 3 \cdot 1^{2} (9 x) + 3 \cdot 1 {(9 x)}^{2} + {(9 x)}^{3}$

Step 3.6.2.1.2

Simplify each term.

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Step 3.6.2.1.2.1

One to any power is one.

$y = 1 + 3 \cdot 1^{2} (9 x) + 3 \cdot 1 {(9 x)}^{2} + {(9 x)}^{3}$

Step 3.6.2.1.2.2

One to any power is one.

$y = 1 + 3 \cdot 1 (9 x) + 3 \cdot 1 {(9 x)}^{2} + {(9 x)}^{3}$

Step 3.6.2.1.2.3

Multiply $3$ by $1$ .

$y = 1 + 3 (9 x) + 3 \cdot 1 {(9 x)}^{2} + {(9 x)}^{3}$

Step 3.6.2.1.2.4

Multiply $9$ by $3$ .

$y = 1 + 27 x + 3 \cdot 1 {(9 x)}^{2} + {(9 x)}^{3}$

Step 3.6.2.1.2.5

Multiply $3$ by $1$ .

$y = 1 + 27 x + 3 {(9 x)}^{2} + {(9 x)}^{3}$

Step 3.6.2.1.2.6

Apply the product rule to $9 x$ .

$y = 1 + 27 x + 3 (9^{2} x^{2}) + {(9 x)}^{3}$

Step 3.6.2.1.2.7

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

$y = 1 + 27 x + 3 (81 x^{2}) + {(9 x)}^{3}$

Step 3.6.2.1.2.8

Multiply $81$ by $3$ .

$y = 1 + 27 x + 243 x^{2} + {(9 x)}^{3}$

Step 3.6.2.1.2.9

Apply the product rule to $9 x$ .

$y = 1 + 27 x + 243 x^{2} + 9^{3} x^{3}$

Step 3.6.2.1.2.10

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

$y = 1 + 27 x + 243 x^{2} + 729 x^{3}$

Step 3.7

Simplify $1 + 27 x + 243 x^{2} + 729 x^{3}$ .

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Step 3.7.1

Move $1$ .

$y = 27 x + 243 x^{2} + 729 x^{3} + 1$

Step 3.7.2

Move $27 x$ .

$y = 243 x^{2} + 729 x^{3} + 27 x + 1$

Step 3.7.3

Reorder $243 x^{2}$ and $729 x^{3}$ .

$y = 729 x^{3} + 243 x^{2} + 27 x + 1$

Step 4

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

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

Step 5

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

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

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

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

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

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

Step 5.2.3.2

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

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

Step 5.2.3.3

Cancel the common factor of $729$ .

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Step 5.2.3.3.1

Cancel the common factor.

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

Step 5.2.3.3.2

Rewrite the expression.

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

Step 5.2.3.4

Use the Binomial Theorem.

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

Step 5.2.3.5

Simplify each term.

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Step 5.2.3.5.1

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

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Step 5.2.3.5.1.1

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

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

Step 5.2.3.5.1.2

Cancel the common factor of $3$ .

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Step 5.2.3.5.1.2.1

Cancel the common factor.

Step 5.2.3.5.1.2.2

Rewrite the expression.

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

Step 5.2.3.5.2

Simplify.

Step 5.2.3.5.3

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

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Step 5.2.3.5.3.1

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

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

Step 5.2.3.5.3.2

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

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

Step 5.2.3.5.4

Multiply $- 1$ by $3$ .

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

Step 5.2.3.5.5

Raise $- 1$ to the power of $2$ .

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

Step 5.2.3.5.6

Multiply $3$ by $1$ .

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

Step 5.2.3.5.7

Raise $- 1$ to the power of $3$ .

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

Step 5.2.3.6

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

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

Step 5.2.3.7

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

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

Step 5.2.3.8

Cancel the common factor of $81$ .

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Step 5.2.3.8.1

Factor $81$ out of $243$ .

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

Step 5.2.3.8.2

Cancel the common factor.

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

Step 5.2.3.8.3

Rewrite the expression.

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

Step 5.2.3.9

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

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

Step 5.2.3.10

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

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Step 5.2.3.10.1

Apply the distributive property.

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

Step 5.2.3.10.2

Apply the distributive property.

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

Step 5.2.3.10.3

Apply the distributive property.

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

Step 5.2.3.11

Simplify and combine like terms.

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Step 5.2.3.11.1

Simplify each term.

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Step 5.2.3.11.1.1

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

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Step 5.2.3.11.1.1.1

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

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

Step 5.2.3.11.1.1.2

Combine the numerators over the common denominator.

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

Step 5.2.3.11.1.1.3

Add $1$ and $1$ .

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

Step 5.2.3.11.1.2

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

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

Step 5.2.3.11.1.3

Rewrite $- 1 x^{\frac{1}{3}}$ as $- x^{\frac{1}{3}}$ .

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

Step 5.2.3.11.1.4

Rewrite $- 1 x^{\frac{1}{3}}$ as $- x^{\frac{1}{3}}$ .

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

Step 5.2.3.11.1.5

Multiply $- 1$ by $- 1$ .

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

Step 5.2.3.11.2

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

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

Step 5.2.3.12

Apply the distributive property.

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

Step 5.2.3.13

Simplify.

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Step 5.2.3.13.1

Multiply $- 2$ by $3$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x - 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 1 + 3 x^{\frac{2}{3}} - 6 x^{\frac{1}{3}} + 3 \cdot 1 + 27 (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

Step 5.2.3.13.2

Multiply $3$ by $1$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x - 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 1 + 3 x^{\frac{2}{3}} - 6 x^{\frac{1}{3}} + 3 + 27 (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

Step 5.2.3.14

Cancel the common factor of $9$ .

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Step 5.2.3.14.1

Factor $9$ out of $27$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x - 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 1 + 3 x^{\frac{2}{3}} - 6 x^{\frac{1}{3}} + 3 + 9 (3) (\frac{x^{\frac{1}{3}} - 1}{9}) + 1$

Step 5.2.3.14.2

Cancel the common factor.

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x - 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 1 + 3 x^{\frac{2}{3}} - 6 x^{\frac{1}{3}} + 3 + 9 \cdot (3 (\frac{x^{\frac{1}{3}} - 1}{9})) + 1$

Step 5.2.3.14.3

Rewrite the expression.

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x - 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 1 + 3 x^{\frac{2}{3}} - 6 x^{\frac{1}{3}} + 3 + 3 (x^{\frac{1}{3}} - 1) + 1$

Step 5.2.3.15

Apply the distributive property.

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x - 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 1 + 3 x^{\frac{2}{3}} - 6 x^{\frac{1}{3}} + 3 + 3 x^{\frac{1}{3}} + 3 \cdot - 1 + 1$

Step 5.2.3.16

Multiply $3$ by $- 1$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x - 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 1 + 3 x^{\frac{2}{3}} - 6 x^{\frac{1}{3}} + 3 + 3 x^{\frac{1}{3}} - 3 + 1$

Step 5.2.4

Simplify by adding terms.

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Step 5.2.4.1

Combine the opposite terms in $x - 3 x^{\frac{2}{3}} + 3 x^{\frac{1}{3}} - 1 + 3 x^{\frac{2}{3}} - 6 x^{\frac{1}{3}} + 3 + 3 x^{\frac{1}{3}} - 3 + 1$ .

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Step 5.2.4.1.1

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

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x + 3 x^{\frac{1}{3}} - 1 + 0 - 6 x^{\frac{1}{3}} + 3 + 3 x^{\frac{1}{3}} - 3 + 1$

Step 5.2.4.1.2

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

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x + 3 x^{\frac{1}{3}} - 1 - 6 x^{\frac{1}{3}} + 3 + 3 x^{\frac{1}{3}} - 3 + 1$

Step 5.2.4.1.3

Subtract $3$ from $3$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x + 3 x^{\frac{1}{3}} - 1 - 6 x^{\frac{1}{3}} + 3 x^{\frac{1}{3}} + 0 + 1$

Step 5.2.4.1.4

Add $x + 3 x^{\frac{1}{3}} - 1 - 6 x^{\frac{1}{3}} + 3 x^{\frac{1}{3}}$ and $0$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x + 3 x^{\frac{1}{3}} - 1 - 6 x^{\frac{1}{3}} + 3 x^{\frac{1}{3}} + 1$

Step 5.2.4.1.5

Add $- 1$ and $1$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x + 3 x^{\frac{1}{3}} - 6 x^{\frac{1}{3}} + 3 x^{\frac{1}{3}} + 0$

Step 5.2.4.1.6

Add $x + 3 x^{\frac{1}{3}} - 6 x^{\frac{1}{3}} + 3 x^{\frac{1}{3}}$ and $0$ .

$f^{- 1} (\frac{x^{\frac{1}{3}} - 1}{9}) = x + 3 x^{\frac{1}{3}} - 6 x^{\frac{1}{3}} + 3 x^{\frac{1}{3}}$

Step 5.2.4.2

Subtract $6 x^{\frac{1}{3}}$ from $3 x^{\frac{1}{3}}$ .

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

Step 5.2.4.3

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

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Step 5.2.4.3.1

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

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

Step 5.2.4.3.2

Add $x$ and $0$ .

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

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

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

Step 5.4

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

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