Precalculus Examples

$f (x) = {(x - 7)}^{3}$ ; find $f^{- 1} (x)$

Step 1

Write $f (x) = {(x - 7)}^{3}$ as an equation.

$y = {(x - 7)}^{3}$

Step 2

Interchange the variables.

$x = {(y - 7)}^{3}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as ${(y - 7)}^{3} = x$ .

${(y - 7)}^{3} = x$

Step 3.2

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

$y - 7 = \sqrt[3]{x}$

Step 3.3

Add $7$ to both sides of the equation.

$y = \sqrt[3]{x} + 7$

Step 4

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

$f^{- 1} (x) = \sqrt[3]{x} + 7$

Step 5

Verify if $f^{- 1} (x) = \sqrt[3]{x} + 7$ is the inverse of $f (x) = {(x - 7)}^{3}$ .

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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 $f^{- 1} ({(x - 7)}^{3})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} ({(x - 7)}^{3}) = \sqrt[3]{{(x - 7)}^{3}} + 7$

Step 5.2.3

Remove parentheses.

$f^{- 1} ({(x - 7)}^{3}) = \sqrt[3]{{(x - 7)}^{3}} + 7$

Step 5.2.4

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

$f^{- 1} ({(x - 7)}^{3}) = x - 7 + 7$

Step 5.2.5

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

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Step 5.2.5.1

Add $- 7$ and $7$ .

$f^{- 1} ({(x - 7)}^{3}) = x + 0$

Step 5.2.5.2

Add $x$ and $0$ .

$f^{- 1} ({(x - 7)}^{3}) = x$

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 $f (\sqrt[3]{x} + 7)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\sqrt[3]{x} + 7) = {((\sqrt[3]{x} + 7) - 7)}^{3}$

Step 5.3.3

Combine the opposite terms in $(\sqrt[3]{x} + 7) - 7$ .

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Step 5.3.3.1

Subtract $7$ from $7$ .

$f (\sqrt[3]{x} + 7) = {(\sqrt[3]{x} + 0)}^{3}$

Step 5.3.3.2

Add $\sqrt[3]{x}$ and $0$ .

$f (\sqrt[3]{x} + 7) = {\sqrt[3]{x}}^{3}$

Step 5.3.4

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

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Step 5.3.4.1

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

$f (\sqrt[3]{x} + 7) = {(x^{\frac{1}{3}})}^{3}$

Step 5.3.4.2

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

$f (\sqrt[3]{x} + 7) = x^{\frac{1}{3} \cdot 3}$

Step 5.3.4.3

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

$f (\sqrt[3]{x} + 7) = x^{\frac{3}{3}}$

Step 5.3.4.4

Cancel the common factor of $3$ .

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Step 5.3.4.4.1

Cancel the common factor.

$f (\sqrt[3]{x} + 7) = x^{\frac{3}{3}}$

Step 5.3.4.4.2

Rewrite the expression.

$f (\sqrt[3]{x} + 7) = x$

Step 5.3.4.5

Simplify.

$f (\sqrt[3]{x} + 7) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \sqrt[3]{x} + 7$ is the inverse of $f (x) = {(x - 7)}^{3}$ .

$f^{- 1} (x) = \sqrt[3]{x} + 7$