Find the Inverse y=1/4x^3-2

$y = \frac{1}{4} x^{3} - 2$

Step 1

Interchange the variables.

$x = \frac{1}{4} y^{3} - 2$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{1}{4} y^{3} - 2 = x$ .

$\frac{1}{4} y^{3} - 2 = x$

Combine $\frac{1}{4}$ and $y^{3}$ .

$\frac{y^{3}}{4} - 2 = x$

Add $2$ to both sides of the equation.

$\frac{y^{3}}{4} = x + 2$

Multiply both sides of the equation by $4$ .

$4 \frac{y^{3}}{4} = 4 (x + 2)$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$4 \frac{y^{3}}{4} = 4 (x + 2)$

Rewrite the expression.

$y^{3} = 4 (x + 2)$

Simplify the right side.

Simplify $4 (x + 2)$ .

Apply the distributive property.

$y^{3} = 4 x + 4 \cdot 2$

Multiply $4$ by $2$ .

$y^{3} = 4 x + 8$

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

$y = \sqrt[3]{4 x + 8}$

Factor $4$ out of $4 x + 8$ .

Factor $4$ out of $4 x$ .

$y = \sqrt[3]{4 (x) + 8}$

Factor $4$ out of $8$ .

$y = \sqrt[3]{4 x + 4 \cdot 2}$

Factor $4$ out of $4 x + 4 \cdot 2$ .

$y = \sqrt[3]{4 (x + 2)}$

Step 3

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

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

Step 4

Verify if $f^{- 1} (x) = \sqrt[3]{4 (x + 2)}$ is the inverse of $f (x) = \frac{1}{4} x^{3} - 2$ .

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 $f^{- 1} (\frac{1}{4} x^{3} - 2)$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Combine $\frac{1}{4}$ and $x^{3}$ .

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

Simplify by adding numbers.

Add $- 2$ and $2$ .

$f^{- 1} (\frac{1}{4} x^{3} - 2) = \sqrt[3]{4 (\frac{x^{3}}{4} + 0)}$

Add $\frac{x^{3}}{4}$ and $0$ .

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

Combine $4$ and $\frac{x^{3}}{4}$ .

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

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{4 x^{3}}{4}$ by cancelling the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

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

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

Set up the composite result function.

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

Evaluate $f (\sqrt[3]{4 (x + 2)})$ by substituting in the value of $f^{- 1}$ into $f$ .

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

Simplify each term.

Rewrite ${\sqrt[3]{4 (x + 2)}}^{3}$ as $4 (x + 2)$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{4 (x + 2)}$ as ${(4 (x + 2))}^{\frac{1}{3}}$ .

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

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

Combine the opposite terms in $x + 2 - 2$ .

Subtract $2$ from $2$ .

$f (\sqrt[3]{4 (x + 2)}) = x + 0$

Add $x$ and $0$ .

$f (\sqrt[3]{4 (x + 2)}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \sqrt[3]{4 (x + 2)}$ is the inverse of $f (x) = \frac{1}{4} x^{3} - 2$ .

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

Find the Inverse y=1/4x^3-2

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