Precalculus Examples

$f (n) = 2 {(n - 2)}^{3}$

Step 1

Write $f (n) = 2 {(n - 2)}^{3}$ as an equation.

$y = 2 {(n - 2)}^{3}$

Step 2

Interchange the variables.

$n = 2 {(y - 2)}^{3}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $2 {(y - 2)}^{3} = n$ .

$2 {(y - 2)}^{3} = n$

Step 3.2

Divide each term in $2 {(y - 2)}^{3} = n$ by $2$ and simplify.

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Step 3.2.1

Divide each term in $2 {(y - 2)}^{3} = n$ by $2$ .

$\frac{2 {(y - 2)}^{3}}{2} = \frac{n}{2}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $2$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{2 {(y - 2)}^{3}}{2} = \frac{n}{2}$

Step 3.2.2.1.2

Divide ${(y - 2)}^{3}$ by $1$ .

${(y - 2)}^{3} = \frac{n}{2}$

Step 3.3

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y - 2 = \sqrt[3]{\frac{n}{2}}$

Step 3.4

Simplify $\sqrt[3]{\frac{n}{2}}$ .

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Step 3.4.1

Rewrite $\sqrt[3]{\frac{n}{2}}$ as $\frac{\sqrt[3]{n}}{\sqrt[3]{2}}$ .

$y - 2 = \frac{\sqrt[3]{n}}{\sqrt[3]{2}}$

Step 3.4.2

Multiply $\frac{\sqrt[3]{n}}{\sqrt[3]{2}}$ by $\frac{{\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{2}}$ .

$y - 2 = \frac{\sqrt[3]{n}}{\sqrt[3]{2}} \cdot \frac{{\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{2}}$

Step 3.4.3

Combine and simplify the denominator.

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Step 3.4.3.1

Multiply $\frac{\sqrt[3]{n}}{\sqrt[3]{2}}$ by $\frac{{\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{2}}$ .

$y - 2 = \frac{\sqrt[3]{n} {\sqrt[3]{2}}^{2}}{\sqrt[3]{2} {\sqrt[3]{2}}^{2}}$

Step 3.4.3.2

Raise $\sqrt[3]{2}$ to the power of $1$ .

$y - 2 = \frac{\sqrt[3]{n} {\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{1} {\sqrt[3]{2}}^{2}}$

Step 3.4.3.3

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

$y - 2 = \frac{\sqrt[3]{n} {\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{1 + 2}}$

Step 3.4.3.4

Add $1$ and $2$ .

$y - 2 = \frac{\sqrt[3]{n} {\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{3}}$

Step 3.4.3.5

Rewrite ${\sqrt[3]{2}}^{3}$ as $2$ .

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Step 3.4.3.5.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{2}$ as $2^{\frac{1}{3}}$ .

$y - 2 = \frac{\sqrt[3]{n} {\sqrt[3]{2}}^{2}}{{(2^{\frac{1}{3}})}^{3}}$

Step 3.4.3.5.2

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

$y - 2 = \frac{\sqrt[3]{n} {\sqrt[3]{2}}^{2}}{2^{\frac{1}{3} \cdot 3}}$

Step 3.4.3.5.3

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

$y - 2 = \frac{\sqrt[3]{n} {\sqrt[3]{2}}^{2}}{2^{\frac{3}{3}}}$

Step 3.4.3.5.4

Cancel the common factor of $3$ .

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Step 3.4.3.5.4.1

Cancel the common factor.

$y - 2 = \frac{\sqrt[3]{n} {\sqrt[3]{2}}^{2}}{2^{\frac{3}{3}}}$

Step 3.4.3.5.4.2

Rewrite the expression.

$y - 2 = \frac{\sqrt[3]{n} {\sqrt[3]{2}}^{2}}{2^{1}}$

Step 3.4.3.5.5

Evaluate the exponent.

$y - 2 = \frac{\sqrt[3]{n} {\sqrt[3]{2}}^{2}}{2}$

Step 3.4.4

Simplify the numerator.

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Step 3.4.4.1

Rewrite ${\sqrt[3]{2}}^{2}$ as $\sqrt[3]{2^{2}}$ .

$y - 2 = \frac{\sqrt[3]{n} \sqrt[3]{2^{2}}}{2}$

Step 3.4.4.2

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

$y - 2 = \frac{\sqrt[3]{n} \sqrt[3]{4}}{2}$

Step 3.4.5

Simplify with factoring out.

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Step 3.4.5.1

Combine using the product rule for radicals.

$y - 2 = \frac{\sqrt[3]{n \cdot 4}}{2}$

Step 3.4.5.2

Reorder factors in $\frac{\sqrt[3]{n \cdot 4}}{2}$ .

$y - 2 = \frac{\sqrt[3]{4 n}}{2}$

Step 3.5

Add $2$ to both sides of the equation.

$y = \frac{\sqrt[3]{4 n}}{2} + 2$

Step 4

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

$f^{- 1} (n) = \frac{\sqrt[3]{4 n}}{2} + 2$

Step 5

Verify if $f^{- 1} (n) = \frac{\sqrt[3]{4 n}}{2} + 2$ is the inverse of $f (n) = 2 {(n - 2)}^{3}$ .

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Step 5.1

To verify the inverse, check if $f^{- 1} (f (n)) = n$ and $f (f^{- 1} (n)) = n$ .

Step 5.2

Evaluate $f^{- 1} (f (n))$ .

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Step 5.2.1

Set up the composite result function.

$f^{- 1} (f (n))$

Step 5.2.2

Evaluate $f^{- 1} (2 {(n - 2)}^{3})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (2 {(n - 2)}^{3}) = \frac{\sqrt[3]{4 (2 {(n - 2)}^{3})}}{2} + 2$

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

Multiply $4$ by $2$ .

$f^{- 1} (2 {(n - 2)}^{3}) = \frac{\sqrt[3]{8 {(n - 2)}^{3}}}{2} + 2$

Step 5.2.3.1.2

Rewrite $8 {(n - 2)}^{3}$ as ${(2 (n - 2))}^{3}$ .

$f^{- 1} (2 {(n - 2)}^{3}) = \frac{\sqrt[3]{{(2 (n - 2))}^{3}}}{2} + 2$

Step 5.2.3.1.3

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

$f^{- 1} (2 {(n - 2)}^{3}) = \frac{2 (n - 2)}{2} + 2$

Step 5.2.3.1.4

Apply the distributive property.

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

Step 5.2.3.1.5

Multiply $2$ by $- 2$ .

$f^{- 1} (2 {(n - 2)}^{3}) = \frac{2 n - 4}{2} + 2$

Step 5.2.3.1.6

Factor $2$ out of $2 n - 4$ .

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Step 5.2.3.1.6.1

Factor $2$ out of $2 n$ .

$f^{- 1} (2 {(n - 2)}^{3}) = \frac{2 (n) - 4}{2} + 2$

Step 5.2.3.1.6.2

Factor $2$ out of $- 4$ .

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

Step 5.2.3.1.6.3

Factor $2$ out of $2 n + 2 \cdot - 2$ .

$f^{- 1} (2 {(n - 2)}^{3}) = \frac{2 (n - 2)}{2} + 2$

Step 5.2.3.2

Cancel the common factor of $2$ .

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Step 5.2.3.2.1

Cancel the common factor.

$f^{- 1} (2 {(n - 2)}^{3}) = \frac{2 (n - 2)}{2} + 2$

Step 5.2.3.2.2

Divide $n - 2$ by $1$ .

$f^{- 1} (2 {(n - 2)}^{3}) = n - 2 + 2$

Step 5.2.4

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

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Step 5.2.4.1

Add $- 2$ and $2$ .

$f^{- 1} (2 {(n - 2)}^{3}) = n + 0$

Step 5.2.4.2

Add $n$ and $0$ .

$f^{- 1} (2 {(n - 2)}^{3}) = n$

Step 5.3

Evaluate $f (f^{- 1} (n))$ .

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Step 5.3.1

Set up the composite result function.

$f (f^{- 1} (n))$

Step 5.3.2

Evaluate $f (\frac{\sqrt[3]{4 n}}{2} + 2)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\sqrt[3]{4 n}}{2} + 2) = 2 {((\frac{\sqrt[3]{4 n}}{2} + 2) - 2)}^{3}$

Step 5.3.3

Combine the opposite terms in $(\frac{\sqrt[3]{4 n}}{2} + 2) - 2$ .

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Step 5.3.3.1

Subtract $2$ from $2$ .

$f (\frac{\sqrt[3]{4 n}}{2} + 2) = 2 {(\frac{\sqrt[3]{4 n}}{2} + 0)}^{3}$

Step 5.3.3.2

Add $\frac{\sqrt[3]{4 n}}{2}$ and $0$ .

$f (\frac{\sqrt[3]{4 n}}{2} + 2) = 2 {(\frac{\sqrt[3]{4 n}}{2})}^{3}$

Step 5.3.4

Apply the product rule to $\frac{\sqrt[3]{4 n}}{2}$ .

$f (\frac{\sqrt[3]{4 n}}{2} + 2) = 2 (\frac{{\sqrt[3]{4 n}}^{3}}{2^{3}})$

Step 5.3.5

Rewrite ${\sqrt[3]{4 n}}^{3}$ as $4 n$ .

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Step 5.3.5.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{4 n}$ as ${(4 n)}^{\frac{1}{3}}$ .

$f (\frac{\sqrt[3]{4 n}}{2} + 2) = 2 (\frac{{({(4 n)}^{\frac{1}{3}})}^{3}}{2^{3}})$

Step 5.3.5.2

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

$f (\frac{\sqrt[3]{4 n}}{2} + 2) = 2 (\frac{{(4 n)}^{\frac{1}{3} \cdot 3}}{2^{3}})$

Step 5.3.5.3

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

$f (\frac{\sqrt[3]{4 n}}{2} + 2) = 2 (\frac{{(4 n)}^{\frac{3}{3}}}{2^{3}})$

Step 5.3.5.4

Cancel the common factor of $3$ .

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Step 5.3.5.4.1

Cancel the common factor.

$f (\frac{\sqrt[3]{4 n}}{2} + 2) = 2 (\frac{{(4 n)}^{\frac{3}{3}}}{2^{3}})$

Step 5.3.5.4.2

Rewrite the expression.

$f (\frac{\sqrt[3]{4 n}}{2} + 2) = 2 (\frac{4 n}{2^{3}})$

Step 5.3.5.5

Simplify.

$f (\frac{\sqrt[3]{4 n}}{2} + 2) = 2 (\frac{4 n}{2^{3}})$

Step 5.3.6

Raise $2$ to the power of $3$ .

$f (\frac{\sqrt[3]{4 n}}{2} + 2) = 2 (\frac{4 n}{8})$

Step 5.3.7

Cancel the common factor of $2$ .

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Step 5.3.7.1

Factor $2$ out of $8$ .

$f (\frac{\sqrt[3]{4 n}}{2} + 2) = 2 (\frac{4 n}{2 (4)})$

Step 5.3.7.2

Cancel the common factor.

$f (\frac{\sqrt[3]{4 n}}{2} + 2) = 2 (\frac{4 n}{2 \cdot 4})$

Step 5.3.7.3

Rewrite the expression.

$f (\frac{\sqrt[3]{4 n}}{2} + 2) = \frac{4 n}{4}$

Step 5.3.8

Cancel the common factor of $4$ .

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Step 5.3.8.1

Cancel the common factor.

$f (\frac{\sqrt[3]{4 n}}{2} + 2) = \frac{4 n}{4}$

Step 5.3.8.2

Divide $n$ by $1$ .

$f (\frac{\sqrt[3]{4 n}}{2} + 2) = n$

Step 5.4

Since $f^{- 1} (f (n)) = n$ and $f (f^{- 1} (n)) = n$ , then $f^{- 1} (n) = \frac{\sqrt[3]{4 n}}{2} + 2$ is the inverse of $f (n) = 2 {(n - 2)}^{3}$ .

$f^{- 1} (n) = \frac{\sqrt[3]{4 n}}{2} + 2$