Find the Inverse y=-1/2x^3+6

$y = - \frac{1}{2} x^{3} + 6$

Interchange the variables.

$x = - \frac{1}{2} y^{3} + 6$

Solve for $y$ .

Rewrite the equation as $- \frac{1}{2} y^{3} + 6 = x$ .

$- \frac{1}{2} y^{3} + 6 = x$

Combine $y^{3}$ and $\frac{1}{2}$ .

$- \frac{y^{3}}{2} + 6 = x$

Subtract $6$ from both sides of the equation.

$- \frac{y^{3}}{2} = x - 6$

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

$- 2 (- \frac{y^{3}}{2}) = - 2 (x - 6)$

Simplify both sides of the equation.

Simplify the left side.

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{y^{3}}{2}$ into the numerator.

$- 2 \frac{- y^{3}}{2} = - 2 (x - 6)$

Factor $2$ out of $- 2$ .

$2 (- 1) \frac{- y^{3}}{2} = - 2 (x - 6)$

Cancel the common factor.

$2 \cdot - 1 \frac{- y^{3}}{2} = - 2 (x - 6)$

Rewrite the expression.

$- - y^{3} = - 2 (x - 6)$

Multiply.

Multiply $- 1$ by $- 1$ .

$1 y^{3} = - 2 (x - 6)$

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

$y^{3} = - 2 (x - 6)$

Simplify the right side.

Simplify $- 2 (x - 6)$ .

Apply the distributive property.

$y^{3} = - 2 x - 2 \cdot - 6$

Multiply $- 2$ by $- 6$ .

$y^{3} = - 2 x + 12$

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

$y = \sqrt[3]{- 2 x + 12}$

Factor $2$ out of $- 2 x + 12$ .

Factor $2$ out of $- 2 x$ .

$y = \sqrt[3]{2 (- x) + 12}$

Factor $2$ out of $12$ .

$y = \sqrt[3]{2 (- x) + 2 (6)}$

Factor $2$ out of $2 (- x) + 2 (6)$ .

$y = \sqrt[3]{2 (- x + 6)}$

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

$f^{- 1} (x) = \sqrt[3]{2 (- x + 6)}$

Verify if $f^{- 1} (x) = \sqrt[3]{2 (- x + 6)}$ is the inverse of $f (x) = - \frac{1}{2} x^{3} + 6$ .

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 $f^{- 1} (- \frac{1}{2} x^{3} + 6)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (- \frac{1}{2} x^{3} + 6) = \sqrt[3]{2 (- (- \frac{1}{2} x^{3} + 6) + 6)}$

Combine $x^{3}$ and $\frac{1}{2}$ .

$f^{- 1} (- \frac{1}{2} x^{3} + 6) = \sqrt[3]{2 (- (- \frac{x^{3}}{2} + 6) + 6)}$

Apply the distributive property.

$f^{- 1} (- \frac{1}{2} x^{3} + 6) = \sqrt[3]{2 (\frac{x^{3}}{2} - 1 \cdot 6 + 6)}$

Simplify the expression.

Multiply $- 1$ by $6$ .

$f^{- 1} (- \frac{1}{2} x^{3} + 6) = \sqrt[3]{2 (\frac{x^{3}}{2} - 6 + 6)}$

Add $- 6$ and $6$ .

$f^{- 1} (- \frac{1}{2} x^{3} + 6) = \sqrt[3]{2 (\frac{x^{3}}{2} + 0)}$

Add $\frac{x^{3}}{2}$ and $0$ .

$f^{- 1} (- \frac{1}{2} x^{3} + 6) = \sqrt[3]{2 (\frac{x^{3}}{2})}$

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

$f^{- 1} (- \frac{1}{2} x^{3} + 6) = \sqrt[3]{\frac{2 x^{3}}{2}}$

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.

$f^{- 1} (- \frac{1}{2} x^{3} + 6) = \sqrt[3]{\frac{2 x^{3}}{2}}$

Rewrite the expression.

$f^{- 1} (- \frac{1}{2} x^{3} + 6) = \sqrt[3]{\frac{x^{3}}{1}}$

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

$f^{- 1} (- \frac{1}{2} x^{3} + 6) = \sqrt[3]{x^{3}}$

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

$f^{- 1} (- \frac{1}{2} x^{3} + 6) = x$

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

Set up the composite result function.

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

Evaluate $f (\sqrt[3]{2 (- x + 6)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\sqrt[3]{2 (- x + 6)}) = - \frac{1}{2} \cdot {(\sqrt[3]{2 (- x + 6)})}^{3} + 6$

Simplify each term.

Rewrite ${\sqrt[3]{2 (- x + 6)}}^{3}$ as $2 (- x + 6)$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{2 (- x + 6)}$ as ${(2 (- x + 6))}^{\frac{1}{3}}$ .

$f (\sqrt[3]{2 (- x + 6)}) = - \frac{1}{2} \cdot {({(2 (- x + 6))}^{\frac{1}{3}})}^{3} + 6$

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

$f (\sqrt[3]{2 (- x + 6)}) = - \frac{1}{2} \cdot {(2 (- x + 6))}^{\frac{1}{3} \cdot 3} + 6$

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

$f (\sqrt[3]{2 (- x + 6)}) = - \frac{1}{2} \cdot {(2 (- x + 6))}^{\frac{3}{3}} + 6$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\sqrt[3]{2 (- x + 6)}) = - \frac{1}{2} \cdot {(2 (- x + 6))}^{\frac{3}{3}} + 6$

Rewrite the expression.

$f (\sqrt[3]{2 (- x + 6)}) = - \frac{1}{2} \cdot (2 (- x + 6)) + 6$

Simplify.

$f (\sqrt[3]{2 (- x + 6)}) = - \frac{1}{2} \cdot (2 (- x + 6)) + 6$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$f (\sqrt[3]{2 (- x + 6)}) = \frac{- 1}{2} \cdot (2 (- x + 6)) + 6$

Cancel the common factor.

$f (\sqrt[3]{2 (- x + 6)}) = \frac{- 1}{2} \cdot (2 (- x + 6)) + 6$

Rewrite the expression.

$f (\sqrt[3]{2 (- x + 6)}) = - 1 \cdot (- x + 6) + 6$

Apply the distributive property.

$f (\sqrt[3]{2 (- x + 6)}) = - 1 (- x) - 1 \cdot 6 + 6$

Multiply $- 1 (- x)$ .

Multiply $- 1$ by $- 1$ .

$f (\sqrt[3]{2 (- x + 6)}) = 1 x - 1 \cdot 6 + 6$

Multiply $x$ by $1$ .

$f (\sqrt[3]{2 (- x + 6)}) = x - 1 \cdot 6 + 6$

Multiply $- 1$ by $6$ .

$f (\sqrt[3]{2 (- x + 6)}) = x - 6 + 6$

Combine the opposite terms in $x - 6 + 6$ .

Add $- 6$ and $6$ .

$f (\sqrt[3]{2 (- x + 6)}) = x + 0$

Add $x$ and $0$ .

$f (\sqrt[3]{2 (- x + 6)}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \sqrt[3]{2 (- x + 6)}$ is the inverse of $f (x) = - \frac{1}{2} x^{3} + 6$ .

$f^{- 1} (x) = \sqrt[3]{2 (- x + 6)}$

Find the Inverse y=-1/2x^3+6

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