Find the Inverse x^3+6

$x^{3} + 6$

Step 1

Interchange the variables.

$x = y^{3} + 6$

Step 2

Solve for $y$ .

Rewrite the equation as $y^{3} + 6 = x$ .

$y^{3} + 6 = x$

Subtract $6$ from both sides of the equation.

$y^{3} = x - 6$

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

$y = \sqrt[3]{x - 6}$

Step 3

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

$f^{- 1} (x) = \sqrt[3]{x - 6}$

Step 4

Verify if $f^{- 1} (x) = \sqrt[3]{x - 6}$ is the inverse of $f (x) = x^{3} + 6$ .

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

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

Set up the composite result function.

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

Evaluate $f^{- 1} (x^{3} + 6)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (x^{3} + 6) = \sqrt[3]{(x^{3} + 6) - 6}$

Subtract $6$ from $6$ .

$f^{- 1} (x^{3} + 6) = \sqrt[3]{x^{3} + 0}$

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

$f^{- 1} (x^{3} + 6) = \sqrt[3]{x^{3}}$

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

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

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

Set up the composite result function.

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

Evaluate $f (\sqrt[3]{x - 6})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\sqrt[3]{x - 6}) = {(\sqrt[3]{x - 6})}^{3} + 6$

Rewrite ${\sqrt[3]{x - 6}}^{3}$ as $x - 6$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x - 6}$ as ${(x - 6)}^{\frac{1}{3}}$ .

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

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

$f (\sqrt[3]{x - 6}) = {(x - 6)}^{\frac{1}{3} \cdot 3} + 6$

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

$f (\sqrt[3]{x - 6}) = {(x - 6)}^{\frac{3}{3}} + 6$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\sqrt[3]{x - 6}) = {(x - 6)}^{\frac{3}{3}} + 6$

Rewrite the expression.

$f (\sqrt[3]{x - 6}) = (x - 6) + 6$

Simplify.

$f (\sqrt[3]{x - 6}) = x - 6 + 6$

Combine the opposite terms in $x - 6 + 6$ .

Add $- 6$ and $6$ .

$f (\sqrt[3]{x - 6}) = x + 0$

Add $x$ and $0$ .

$f (\sqrt[3]{x - 6}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \sqrt[3]{x - 6}$ is the inverse of $f (x) = x^{3} + 6$ .

$f^{- 1} (x) = \sqrt[3]{x - 6}$

Find the Inverse x^3+6

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