Find the Inverse f^-1(x)=(x^3)/2

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

Step 1

Write $f^{- 1} (x) = \frac{x^{3}}{2}$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{y^{3}}{2} = x$ .

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

Multiply both sides of the equation by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$y^{3} = 2 x$

Take the cube root of both sides of the equation to eliminate the exponent on the left side.

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

Step 4

Replace $y$ with $f (x)$ to show the final answer.

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

Step 5

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

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

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

Set up the composite result function.

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

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

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

Combine $2$ and $\frac{x^{3}}{2}$ .

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

Reduce the expression by cancelling the common factors.

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

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

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

Set up the composite result function.

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

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

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

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

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

Find the Inverse f^-1(x)=(x^3)/2

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