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

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

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

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

Interchange the variables.

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

Solve for $y$ .

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

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

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Solve for $y$ .

Multiply both sides of the equation by $- 2$ .

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

Simplify the left side.

Simplify $- 2 (- \frac{y}{2})$ .

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{y}{2}$ into the numerator.

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

Factor $2$ out of $- 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply.

Multiply $- 1$ by $- 1$ .

$1 y = - 2 x^{3}$

Multiply $y$ by $1$ .

$y = - 2 x^{3}$

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

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

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

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

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

Rewrite $- \frac{x}{2}$ as ${({(- 1)}^{3})}^{3} \frac{x}{2}$ .

Rewrite $- 1$ as ${(- 1)}^{3}$ .

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

Rewrite $- 1$ as ${(- 1)}^{3}$ .

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

Pull terms out from under the radical.

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

Raise $- 1$ to the power of $3$ .

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

Rewrite $\sqrt[3]{\frac{x}{2}}$ as $\frac{\sqrt[3]{x}}{\sqrt[3]{2}}$ .

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

Multiply $\frac{\sqrt[3]{x}}{\sqrt[3]{2}}$ by $\frac{{\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{2}}$ .

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

Combine and simplify the denominator.

Multiply $\frac{\sqrt[3]{x}}{\sqrt[3]{2}}$ by $\frac{{\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{2}}$ .

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

Raise $\sqrt[3]{2}$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $2$ .

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

Rewrite ${\sqrt[3]{2}}^{3}$ as $2$ .

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

Simplify the numerator.

Rewrite ${\sqrt[3]{2}}^{2}$ as ${(2^{2})}^{\frac{1}{3}}$ .

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

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

$f^{- 1} (\sqrt[3]{- \frac{x}{2}}) = - 2 {(- \frac{\sqrt[3]{x} \sqrt[3]{4}}{2})}^{3}$

Combine using the product rule for radicals.

$f^{- 1} (\sqrt[3]{- \frac{x}{2}}) = - 2 {(- \frac{\sqrt[3]{x \cdot 4}}{2})}^{3}$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{\sqrt[3]{x \cdot 4}}{2}$ .

$f^{- 1} (\sqrt[3]{- \frac{x}{2}}) = - 2 ({(- 1)}^{3} {(\frac{\sqrt[3]{x \cdot 4}}{2})}^{3})$

Apply the product rule to $\frac{\sqrt[3]{x \cdot 4}}{2}$ .

$f^{- 1} (\sqrt[3]{- \frac{x}{2}}) = - 2 ({(- 1)}^{3} (\frac{{\sqrt[3]{x \cdot 4}}^{3}}{2^{3}}))$

Raise $- 1$ to the power of $3$ .

$f^{- 1} (\sqrt[3]{- \frac{x}{2}}) = - 2 (- \frac{{\sqrt[3]{x \cdot 4}}^{3}}{2^{3}})$

Simplify the numerator.

Rewrite ${\sqrt[3]{x \cdot 4}}^{3}$ as $x \cdot 4$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x \cdot 4}$ as ${(x \cdot 4)}^{\frac{1}{3}}$ .

$f^{- 1} (\sqrt[3]{- \frac{x}{2}}) = - 2 (- \frac{{({(x \cdot 4)}^{\frac{1}{3}})}^{3}}{2^{3}})$

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

$f^{- 1} (\sqrt[3]{- \frac{x}{2}}) = - 2 (- \frac{{(x \cdot 4)}^{\frac{1}{3} \cdot 3}}{2^{3}})$

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

$f^{- 1} (\sqrt[3]{- \frac{x}{2}}) = - 2 (- \frac{{(x \cdot 4)}^{\frac{3}{3}}}{2^{3}})$

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} (\sqrt[3]{- \frac{x}{2}}) = - 2 (- \frac{{(x \cdot 4)}^{\frac{3}{3}}}{2^{3}})$

Rewrite the expression.

$f^{- 1} (\sqrt[3]{- \frac{x}{2}}) = - 2 (- \frac{x \cdot 4}{2^{3}})$

Simplify.

$f^{- 1} (\sqrt[3]{- \frac{x}{2}}) = - 2 (- \frac{x \cdot 4}{2^{3}})$

Move $4$ to the left of $x$ .

$f^{- 1} (\sqrt[3]{- \frac{x}{2}}) = - 2 (- \frac{4 x}{2^{3}})$

Reduce the expression by cancelling the common factors.

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

$f^{- 1} (\sqrt[3]{- \frac{x}{2}}) = - 2 (- \frac{4 x}{8})$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{4 x}{8}$ into the numerator.

$f^{- 1} (\sqrt[3]{- \frac{x}{2}}) = - 2 \frac{- 4 x}{8}$

Factor $2$ out of $- 2$ .

$f^{- 1} (\sqrt[3]{- \frac{x}{2}}) = 2 (- 1) (\frac{- 4 x}{8})$

Factor $2$ out of $8$ .

$f^{- 1} (\sqrt[3]{- \frac{x}{2}}) = 2 \cdot (- 1 \frac{- 4 x}{2 \cdot 4})$

Cancel the common factor.

$f^{- 1} (\sqrt[3]{- \frac{x}{2}}) = 2 \cdot (- 1 \frac{- 4 x}{2 \cdot 4})$

Rewrite the expression.

$f^{- 1} (\sqrt[3]{- \frac{x}{2}}) = - 1 \frac{- 4 x}{4}$

Cancel the common factor of $- 4$ and $4$ .

Factor $4$ out of $- 4 x$ .

$f^{- 1} (\sqrt[3]{- \frac{x}{2}}) = - 1 \frac{4 (- x)}{4}$

Cancel the common factors.

Factor $4$ out of $4$ .

$f^{- 1} (\sqrt[3]{- \frac{x}{2}}) = - 1 \frac{4 (- x)}{4 (1)}$

Cancel the common factor.

$f^{- 1} (\sqrt[3]{- \frac{x}{2}}) = - 1 \frac{4 (- x)}{4 \cdot 1}$

Rewrite the expression.

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

Divide $- x$ by $1$ .

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

Multiply $- 1 (- x)$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{- 2 x^{3}}{2}$ by cancelling the common factors.

Factor $2$ out of $- 2 x^{3}$ .

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

Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- x^{3}$ by $1$ .

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

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

Combine exponents.

Factor out negative.

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

Multiply $- 1$ by $- 1$ .

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

Multiply $x^{3}$ by $1$ .

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

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

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

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

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

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

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