Precalculus Examples

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

Tap for more steps...

Step 2.1

Rewrite the equation as $\sqrt[3]{y - 7} = x$ .

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

Step 2.2

To remove the radical on the left side of the equation, cube both sides of the equation.

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

Step 2.3

Simplify each side of the equation.

Tap for more steps...

Step 2.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{y - 7}$ as ${(y - 7)}^{\frac{1}{3}}$ .

${({(y - 7)}^{\frac{1}{3}})}^{3} = x^{3}$

Step 2.3.2

Simplify the left side.

Tap for more steps...

Step 2.3.2.1

Simplify ${({(y - 7)}^{\frac{1}{3}})}^{3}$ .

Tap for more steps...

Step 2.3.2.1.1

Multiply the exponents in ${({(y - 7)}^{\frac{1}{3}})}^{3}$ .

Tap for more steps...

Step 2.3.2.1.1.1

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

${(y - 7)}^{\frac{1}{3} \cdot 3} = x^{3}$

Step 2.3.2.1.1.2

Cancel the common factor of $3$ .

Tap for more steps...

Step 2.3.2.1.1.2.1

Cancel the common factor.

${(y - 7)}^{\frac{1}{3} \cdot 3} = x^{3}$

Step 2.3.2.1.1.2.2

Rewrite the expression.

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

Step 2.3.2.1.2

Simplify.

$y - 7 = x^{3}$

Step 2.4

Add $7$ to both sides of the equation.

$y = x^{3} + 7$

Step 3

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

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

Step 4

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

Tap for more steps...

Step 4.1

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

Step 4.2

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

Tap for more steps...

Step 4.2.1

Set up the composite result function.

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

Step 4.2.2

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

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

Step 4.2.3

Rewrite ${\sqrt[3]{x - 7}}^{3}$ as $x - 7$ .

Tap for more steps...

Step 4.2.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x - 7}$ as ${(x - 7)}^{\frac{1}{3}}$ .

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

Step 4.2.3.2

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

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

Step 4.2.3.3

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

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

Step 4.2.3.4

Cancel the common factor of $3$ .

Tap for more steps...

Step 4.2.3.4.1

Cancel the common factor.

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

Step 4.2.3.4.2

Rewrite the expression.

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

Step 4.2.3.5

Simplify.

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

Step 4.2.4

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

Tap for more steps...

Step 4.2.4.1

Add $- 7$ and $7$ .

$f^{- 1} (\sqrt[3]{x - 7}) = x + 0$

Step 4.2.4.2

Add $x$ and $0$ .

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

Step 4.3

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

Tap for more steps...

Step 4.3.1

Set up the composite result function.

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

Step 4.3.2

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

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

Step 4.3.3

Subtract $7$ from $7$ .

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

Step 4.3.4

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

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

Step 4.3.5

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

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

Step 4.4

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

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