Find the Inverse f(x)=1/2 cube root of x

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

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

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

Interchange the variables.

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

Solve for $y$ .

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

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

Multiply both sides of the equation by $2$ .

$2 (\frac{1}{2} \cdot \sqrt[3]{y}) = 2 x$

Simplify the left side.

Simplify $2 (\frac{1}{2} \cdot \sqrt[3]{y})$ .

Combine $\frac{1}{2}$ and $\sqrt[3]{y}$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

Apply the product rule to $2 x$ .

$y = 2^{3} x^{3}$

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

$y = 8 x^{3}$

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

$f^{- 1} (x) = 8 x^{3}$

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

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

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

Combine $\frac{1}{2}$ and $\sqrt[3]{x}$ .

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

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

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

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

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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

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

Cancel the common factor of $8$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

Set up the composite result function.

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

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

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

Remove parentheses.

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

Rewrite $8 x^{3}$ as ${(2 x)}^{3}$ .

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

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

$f (8 x^{3}) = \frac{1}{2} \cdot (2 x)$

Cancel the common factor of $2$ .

Factor $2$ out of $2 x$ .

$f (8 x^{3}) = \frac{1}{2} \cdot (2 (x))$

Cancel the common factor.

$f (8 x^{3}) = \frac{1}{2} \cdot (2 x)$

Rewrite the expression.

$f (8 x^{3}) = x$

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

$f^{- 1} (x) = 8 x^{3}$

Find the Inverse f(x)=1/2 cube root of x

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