Find the Inverse f(x)=x^3-4

$f (x) = x^{3} - 4$

Write $f (x) = x^{3} - 4$ as an equation.

$y = x^{3} - 4$

Interchange the variables.

$x = y^{3} - 4$

Solve for $y$ .

Rewrite the equation as $y^{3} - 4 = x$ .

$y^{3} - 4 = x$

Add $4$ to both sides of the equation.

$y^{3} = x + 4$

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

$y = \sqrt[3]{x + 4}$

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

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

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

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 $f^{- 1} (x^{3} - 4)$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Add $- 4$ and $4$ .

$f^{- 1} (x^{3} - 4) = \sqrt[3]{x^{3} + 0}$

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

$f^{- 1} (x^{3} - 4) = \sqrt[3]{x^{3}}$

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

$f^{- 1} (x^{3} - 4) = x$

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

Set up the composite result function.

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

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

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

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

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

$f (\sqrt[3]{x + 4}) = (x + 4) - 4$

Simplify.

$f (\sqrt[3]{x + 4}) = x + 4 - 4$

Combine the opposite terms in $x + 4 - 4$ .

Subtract $4$ from $4$ .

$f (\sqrt[3]{x + 4}) = x + 0$

Add $x$ and $0$ .

$f (\sqrt[3]{x + 4}) = x$

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

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

Find the Inverse f(x)=x^3-4

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