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

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

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

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

Interchange the variables.

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

Solve for $y$ .

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

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

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

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

Add $4$ to both sides of the equation.

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

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

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

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

Remove parentheses.

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

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

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

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

Add $- 4$ and $4$ .

$f^{- 1} ({(x - 4)}^{3}) = x + 0$

Add $x$ and $0$ .

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

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 $f$ .

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

Combine the opposite terms in $(\sqrt[3]{x} + 4) - 4$ .

Subtract $4$ from $4$ .

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

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

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

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

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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

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

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

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