Find the Inverse F(x)=x^3-8

$F (x) = x^{3} - 8$

Step 1

Write $F (x) = x^{3} - 8$ as an equation.

$y = x^{3} - 8$

Step 2

Interchange the variables.

$x = y^{3} - 8$

Step 3

Solve for $y$ .

Rewrite the equation as $y^{3} - 8 = x$ .

$y^{3} - 8 = x$

Add $8$ to both sides of the equation.

$y^{3} = x + 8$

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

$y = \sqrt[3]{x + 8}$

Step 4

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

$F^{- 1} (x) = \sqrt[3]{x + 8}$

Step 5

Verify if $F^{- 1} (x) = \sqrt[3]{x + 8}$ is the inverse of $F (x) = x^{3} - 8$ .

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 $F^{- 1} (x^{3} - 8)$ by substituting in the value of $F$ into $F^{- 1}$ .

$F^{- 1} (x^{3} - 8) = \sqrt[3]{(x^{3} - 8) + 8}$

Add $- 8$ and $8$ .

$F^{- 1} (x^{3} - 8) = \sqrt[3]{x^{3} + 0}$

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

$F^{- 1} (x^{3} - 8) = \sqrt[3]{x^{3}}$

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

$F^{- 1} (x^{3} - 8) = x$

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

Set up the composite result function.

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

Evaluate $F (\sqrt[3]{x + 8})$ by substituting in the value of $F^{- 1}$ into $F$ .

$F (\sqrt[3]{x + 8}) = {(\sqrt[3]{x + 8})}^{3} - 8$

Rewrite ${\sqrt[3]{x + 8}}^{3}$ as $x + 8$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x + 8}$ as ${(x + 8)}^{\frac{1}{3}}$ .

$F (\sqrt[3]{x + 8}) = {({(x + 8)}^{\frac{1}{3}})}^{3} - 8$

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

$F (\sqrt[3]{x + 8}) = {(x + 8)}^{\frac{1}{3} \cdot 3} - 8$

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

$F (\sqrt[3]{x + 8}) = {(x + 8)}^{\frac{3}{3}} - 8$

Cancel the common factor of $3$ .

Cancel the common factor.

$F (\sqrt[3]{x + 8}) = {(x + 8)}^{\frac{3}{3}} - 8$

Rewrite the expression.

$F (\sqrt[3]{x + 8}) = (x + 8) - 8$

Simplify.

$F (\sqrt[3]{x + 8}) = x + 8 - 8$

Combine the opposite terms in $x + 8 - 8$ .

Subtract $8$ from $8$ .

$F (\sqrt[3]{x + 8}) = x + 0$

Add $x$ and $0$ .

$F (\sqrt[3]{x + 8}) = x$

Since $F^{- 1} (F (x)) = x$ and $F (F^{- 1} (x)) = x$ , then $F^{- 1} (x) = \sqrt[3]{x + 8}$ is the inverse of $F (x) = x^{3} - 8$ .

$F^{- 1} (x) = \sqrt[3]{x + 8}$

Find the Inverse F(x)=x^3-8

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