Find the Inverse (x+2)^3-3

${(x + 2)}^{3} - 3$

Step 1

Interchange the variables.

$x = {(y + 2)}^{3} - 3$

Step 2

Solve for $y$ .

Rewrite the equation as ${(y + 2)}^{3} - 3 = x$ .

${(y + 2)}^{3} - 3 = x$

Add $3$ to both sides of the equation.

${(y + 2)}^{3} = x + 3$

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

$y + 2 = \sqrt[3]{x + 3}$

Subtract $2$ from both sides of the equation.

$y = \sqrt[3]{x + 3} - 2$

Step 3

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

$f^{- 1} (x) = \sqrt[3]{x + 3} - 2$

Step 4

Verify if $f^{- 1} (x) = \sqrt[3]{x + 3} - 2$ is the inverse of $f (x) = {(x + 2)}^{3} - 3$ .

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 $f^{- 1} ({(x + 2)}^{3} - 3)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} ({(x + 2)}^{3} - 3) = \sqrt[3]{({(x + 2)}^{3} - 3) + 3} - 2$

Combine the opposite terms in $\sqrt[3]{({(x + 2)}^{3} - 3) + 3} - 2$ .

Add $- 3$ and $3$ .

$f^{- 1} ({(x + 2)}^{3} - 3) = \sqrt[3]{{(x + 2)}^{3} + 0} - 2$

Add ${(x + 2)}^{3}$ and $0$ .

$f^{- 1} ({(x + 2)}^{3} - 3) = \sqrt[3]{{(x + 2)}^{3}} - 2$

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

$f^{- 1} ({(x + 2)}^{3} - 3) = x + 2 - 2$

Combine the opposite terms in $x + 2 - 2$ .

Subtract $2$ from $2$ .

$f^{- 1} ({(x + 2)}^{3} - 3) = x + 0$

Add $x$ and $0$ .

$f^{- 1} ({(x + 2)}^{3} - 3) = x$

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

Set up the composite result function.

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

Evaluate $f (\sqrt[3]{x + 3} - 2)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\sqrt[3]{x + 3} - 2) = {((\sqrt[3]{x + 3} - 2) + 2)}^{3} - 3$

Combine the opposite terms in ${((\sqrt[3]{x + 3} - 2) + 2)}^{3} - 3$ .

Add $- 2$ and $2$ .

$f (\sqrt[3]{x + 3} - 2) = {(\sqrt[3]{x + 3} + 0)}^{3} - 3$

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

$f (\sqrt[3]{x + 3} - 2) = {\sqrt[3]{x + 3}}^{3} - 3$

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

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

$f (\sqrt[3]{x + 3} - 2) = {({(x + 3)}^{\frac{1}{3}})}^{3} - 3$

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

$f (\sqrt[3]{x + 3} - 2) = {(x + 3)}^{\frac{1}{3} \cdot 3} - 3$

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

$f (\sqrt[3]{x + 3} - 2) = {(x + 3)}^{\frac{3}{3}} - 3$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\sqrt[3]{x + 3} - 2) = {(x + 3)}^{\frac{3}{3}} - 3$

Rewrite the expression.

$f (\sqrt[3]{x + 3} - 2) = (x + 3) - 3$

Simplify.

$f (\sqrt[3]{x + 3} - 2) = x + 3 - 3$

Combine the opposite terms in $x + 3 - 3$ .

Subtract $3$ from $3$ .

$f (\sqrt[3]{x + 3} - 2) = x + 0$

Add $x$ and $0$ .

$f (\sqrt[3]{x + 3} - 2) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \sqrt[3]{x + 3} - 2$ is the inverse of $f (x) = {(x + 2)}^{3} - 3$ .

$f^{- 1} (x) = \sqrt[3]{x + 3} - 2$

Find the Inverse (x+2)^3-3

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