Find the Inverse y=2 cube root of x-5

$y = 2 \sqrt[3]{x - 5}$

Step 1

Interchange the variables.

$x = 2 \sqrt[3]{y - 5}$

Step 2

Solve for $y$ .

Rewrite the equation as $2 \sqrt[3]{y - 5} = x$ .

$2 \sqrt[3]{y - 5} = x$

Divide each term in $2 \sqrt[3]{y - 5} = x$ by $2$ and simplify.

Divide each term in $2 \sqrt[3]{y - 5} = x$ by $2$ .

$\frac{2 \sqrt[3]{y - 5}}{2} = \frac{x}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 \sqrt[3]{y - 5}}{2} = \frac{x}{2}$

Divide $\sqrt[3]{y - 5}$ by $1$ .

$\sqrt[3]{y - 5} = \frac{x}{2}$

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

${\sqrt[3]{y - 5}}^{3} = {(\frac{x}{2})}^{3}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{y - 5}$ as ${(y - 5)}^{\frac{1}{3}}$ .

${({(y - 5)}^{\frac{1}{3}})}^{3} = {(\frac{x}{2})}^{3}$

Simplify the left side.

Simplify ${({(y - 5)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(y - 5)}^{\frac{1}{3}})}^{3}$ .

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

${(y - 5)}^{\frac{1}{3} \cdot 3} = {(\frac{x}{2})}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(y - 5)}^{\frac{1}{3} \cdot 3} = {(\frac{x}{2})}^{3}$

Rewrite the expression.

${(y - 5)}^{1} = {(\frac{x}{2})}^{3}$

Simplify.

$y - 5 = {(\frac{x}{2})}^{3}$

Simplify the right side.

Simplify ${(\frac{x}{2})}^{3}$ .

Apply the product rule to $\frac{x}{2}$ .

$y - 5 = \frac{x^{3}}{2^{3}}$

Raise $2$ to the power of $3$ .

$y - 5 = \frac{x^{3}}{8}$

Add $5$ to both sides of the equation.

$y = \frac{x^{3}}{8} + 5$

Step 3

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

$f^{- 1} (x) = \frac{x^{3}}{8} + 5$

Step 4

Verify if $f^{- 1} (x) = \frac{x^{3}}{8} + 5$ is the inverse of $f (x) = 2 \sqrt[3]{x - 5}$ .

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 $f^{- 1} (2 \sqrt[3]{x - 5})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (2 \sqrt[3]{x - 5}) = \frac{{(2 \sqrt[3]{x - 5})}^{3}}{8} + 5$

Simplify each term.

Simplify the numerator.

Apply the product rule to $2 \sqrt[3]{x - 5}$ .

$f^{- 1} (2 \sqrt[3]{x - 5}) = \frac{2^{3} {\sqrt[3]{x - 5}}^{3}}{8} + 5$

Raise $2$ to the power of $3$ .

$f^{- 1} (2 \sqrt[3]{x - 5}) = \frac{8 {\sqrt[3]{x - 5}}^{3}}{8} + 5$

Rewrite ${\sqrt[3]{x - 5}}^{3}$ as $x - 5$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x - 5}$ as ${(x - 5)}^{\frac{1}{3}}$ .

$f^{- 1} (2 \sqrt[3]{x - 5}) = \frac{8 {({(x - 5)}^{\frac{1}{3}})}^{3}}{8} + 5$

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

$f^{- 1} (2 \sqrt[3]{x - 5}) = \frac{8 {(x - 5)}^{\frac{1}{3} \cdot 3}}{8} + 5$

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

$f^{- 1} (2 \sqrt[3]{x - 5}) = \frac{8 {(x - 5)}^{\frac{3}{3}}}{8} + 5$

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} (2 \sqrt[3]{x - 5}) = \frac{8 {(x - 5)}^{\frac{3}{3}}}{8} + 5$

Rewrite the expression.

$f^{- 1} (2 \sqrt[3]{x - 5}) = \frac{8 (x - 5)}{8} + 5$

Simplify.

$f^{- 1} (2 \sqrt[3]{x - 5}) = \frac{8 (x - 5)}{8} + 5$

Cancel the common factor of $8$ .

Cancel the common factor.

$f^{- 1} (2 \sqrt[3]{x - 5}) = \frac{8 (x - 5)}{8} + 5$

Divide $x - 5$ by $1$ .

$f^{- 1} (2 \sqrt[3]{x - 5}) = x - 5 + 5$

Combine the opposite terms in $x - 5 + 5$ .

Add $- 5$ and $5$ .

$f^{- 1} (2 \sqrt[3]{x - 5}) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (2 \sqrt[3]{x - 5}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{x^{3}}{8} + 5)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x^{3}}{8} + 5) = 2 \sqrt[3]{(\frac{x^{3}}{8} + 5) - 5}$

Subtract $5$ from $5$ .

$f (\frac{x^{3}}{8} + 5) = 2 \sqrt[3]{\frac{x^{3}}{8} + 0}$

Add $\frac{x^{3}}{8}$ and $0$ .

$f (\frac{x^{3}}{8} + 5) = 2 \sqrt[3]{\frac{x^{3}}{8}}$

Rewrite $8$ as $2^{3}$ .

$f (\frac{x^{3}}{8} + 5) = 2 \sqrt[3]{\frac{x^{3}}{2^{3}}}$

Rewrite $\frac{x^{3}}{2^{3}}$ as ${(\frac{x}{2})}^{3}$ .

$f (\frac{x^{3}}{8} + 5) = 2 \sqrt[3]{{(\frac{x}{2})}^{3}}$

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

$f (\frac{x^{3}}{8} + 5) = 2 (\frac{x}{2})$

Cancel the common factor of $2$ .

Cancel the common factor.

$f (\frac{x^{3}}{8} + 5) = 2 (\frac{x}{2})$

Rewrite the expression.

$f (\frac{x^{3}}{8} + 5) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x^{3}}{8} + 5$ is the inverse of $f (x) = 2 \sqrt[3]{x - 5}$ .

$f^{- 1} (x) = \frac{x^{3}}{8} + 5$

Find the Inverse y=2 cube root of x-5

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