Find the Inverse g(x) = cube root of x+1+2

$g (x) = \sqrt[3]{x + 1 + 2}$

Write $g (x) = \sqrt[3]{x + 1 + 2}$ as an equation.

$y = \sqrt[3]{x + 1 + 2}$

Interchange the variables.

$x = \sqrt[3]{y + 1 + 2}$

Solve for $y$ .

Rewrite the equation as $\sqrt[3]{y + 1 + 2} = x$ .

$\sqrt[3]{y + 1 + 2} = x$

To remove the radical on the left side of the equation, cube both sides of the equation.

${\sqrt[3]{y + 1 + 2}}^{3} = x^{3}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{y + 1 + 2}$ as ${(y + 1 + 2)}^{\frac{1}{3}}$ .

${({(y + 1 + 2)}^{\frac{1}{3}})}^{3} = x^{3}$

Simplify the left side.

Simplify ${({(y + 1 + 2)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(y + 1 + 2)}^{\frac{1}{3}})}^{3}$ .

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

${(y + 1 + 2)}^{\frac{1}{3} \cdot 3} = x^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(y + 1 + 2)}^{\frac{1}{3} \cdot 3} = x^{3}$

Rewrite the expression.

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

Add $1$ and $2$ .

${(y + 3)}^{1} = x^{3}$

Simplify.

$y + 3 = x^{3}$

Subtract $3$ from both sides of the equation.

$y = x^{3} - 3$

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

$g^{- 1} (x) = x^{3} - 3$

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

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Add $1$ and $2$ .

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

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 ${(x + 3)}^{\frac{1}{3}}$ .

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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

Subtract $3$ from $3$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Add $- 3$ and $1$ .

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

Add $- 2$ and $2$ .

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

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

$g (x^{3} - 3) = \sqrt[3]{x^{3}}$

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

$g (x^{3} - 3) = x$

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

$g^{- 1} (x) = x^{3} - 3$

Find the Inverse g(x) = cube root of x+1+2

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