Find the Inverse h(x) = cube root of x-3

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

Step 1

Write $h (x) = \sqrt[3]{x - 3}$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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

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

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$y - 3 = x^{3}$

Add $3$ to both sides of the equation.

$y = x^{3} + 3$

Step 4

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

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

Step 5

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

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

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

Set up the composite result function.

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

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

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

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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

Add $- 3$ and $3$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Subtract $3$ from $3$ .

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

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

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

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

$h (x^{3} + 3) = x$

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

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

Find the Inverse h(x) = cube root of x-3

Related Questions

Support LCMGCF.com by disabling your adblocker.