Find the Inverse -2x^3+9

$- 2 x^{3} + 9$

Step 1

Interchange the variables.

$x = - 2 y^{3} + 9$

Step 2

Solve for $y$ .

Rewrite the equation as $- 2 y^{3} + 9 = x$ .

$- 2 y^{3} + 9 = x$

Subtract $9$ from both sides of the equation.

$- 2 y^{3} = x - 9$

Divide each term in $- 2 y^{3} = x - 9$ by $- 2$ and simplify.

Divide each term in $- 2 y^{3} = x - 9$ by $- 2$ .

$\frac{- 2 y^{3}}{- 2} = \frac{x}{- 2} + \frac{- 9}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 y^{3}}{- 2} = \frac{x}{- 2} + \frac{- 9}{- 2}$

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

$y^{3} = \frac{x}{- 2} + \frac{- 9}{- 2}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$y^{3} = - \frac{x}{2} + \frac{- 9}{- 2}$

Dividing two negative values results in a positive value.

$y^{3} = - \frac{x}{2} + \frac{9}{2}$

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

$y = \sqrt[3]{- \frac{x}{2} + \frac{9}{2}}$

Simplify $\sqrt[3]{- \frac{x}{2} + \frac{9}{2}}$ .

Combine the numerators over the common denominator.

$y = \sqrt[3]{\frac{- x + 9}{2}}$

Rewrite $\sqrt[3]{\frac{- x + 9}{2}}$ as $\frac{\sqrt[3]{- x + 9}}{\sqrt[3]{2}}$ .

$y = \frac{\sqrt[3]{- x + 9}}{\sqrt[3]{2}}$

Multiply $\frac{\sqrt[3]{- x + 9}}{\sqrt[3]{2}}$ by $\frac{{\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{2}}$ .

$y = \frac{\sqrt[3]{- x + 9}}{\sqrt[3]{2}} \cdot \frac{{\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{2}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt[3]{- x + 9}}{\sqrt[3]{2}}$ by $\frac{{\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{2}}$ .

$y = \frac{\sqrt[3]{- x + 9} {\sqrt[3]{2}}^{2}}{\sqrt[3]{2} {\sqrt[3]{2}}^{2}}$

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

$y = \frac{\sqrt[3]{- x + 9} {\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{1} {\sqrt[3]{2}}^{2}}$

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

$y = \frac{\sqrt[3]{- x + 9} {\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{1 + 2}}$

Add $1$ and $2$ .

$y = \frac{\sqrt[3]{- x + 9} {\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{3}}$

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

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

$y = \frac{\sqrt[3]{- x + 9} {\sqrt[3]{2}}^{2}}{{(2^{\frac{1}{3}})}^{3}}$

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

$y = \frac{\sqrt[3]{- x + 9} {\sqrt[3]{2}}^{2}}{2^{\frac{1}{3} \cdot 3}}$

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

$y = \frac{\sqrt[3]{- x + 9} {\sqrt[3]{2}}^{2}}{2^{\frac{3}{3}}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$y = \frac{\sqrt[3]{- x + 9} {\sqrt[3]{2}}^{2}}{2^{\frac{3}{3}}}$

Rewrite the expression.

$y = \frac{\sqrt[3]{- x + 9} {\sqrt[3]{2}}^{2}}{2^{1}}$

Evaluate the exponent.

$y = \frac{\sqrt[3]{- x + 9} {\sqrt[3]{2}}^{2}}{2}$

Simplify the numerator.

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

$y = \frac{\sqrt[3]{- x + 9} \sqrt[3]{2^{2}}}{2}$

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

$y = \frac{\sqrt[3]{- x + 9} \sqrt[3]{4}}{2}$

Simplify with factoring out.

Combine using the product rule for radicals.

$y = \frac{\sqrt[3]{(- x + 9) \cdot 4}}{2}$

Reorder factors in $\frac{\sqrt[3]{(- x + 9) \cdot 4}}{2}$ .

$y = \frac{\sqrt[3]{4 (- x + 9)}}{2}$

Step 3

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

$f^{- 1} (x) = \frac{\sqrt[3]{4 (- x + 9)}}{2}$

Step 4

Verify if $f^{- 1} (x) = \frac{\sqrt[3]{4 (- x + 9)}}{2}$ is the inverse of $f (x) = - 2 x^{3} + 9$ .

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 $f^{- 1} (- 2 x^{3} + 9)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (- 2 x^{3} + 9) = \frac{\sqrt[3]{4 (- (- 2 x^{3} + 9) + 9)}}{2}$

Simplify the numerator.

Apply the distributive property.

$f^{- 1} (- 2 x^{3} + 9) = \frac{\sqrt[3]{4 (- (- 2 x^{3}) - 1 \cdot 9 + 9)}}{2}$

Multiply $- 2$ by $- 1$ .

$f^{- 1} (- 2 x^{3} + 9) = \frac{\sqrt[3]{4 (2 x^{3} - 1 \cdot 9 + 9)}}{2}$

Multiply $- 1$ by $9$ .

$f^{- 1} (- 2 x^{3} + 9) = \frac{\sqrt[3]{4 (2 x^{3} - 9 + 9)}}{2}$

Add $- 9$ and $9$ .

$f^{- 1} (- 2 x^{3} + 9) = \frac{\sqrt[3]{4 (2 x^{3} + 0)}}{2}$

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

$f^{- 1} (- 2 x^{3} + 9) = \frac{\sqrt[3]{4 \cdot (2 x^{3})}}{2}$

Multiply $4$ by $2$ .

$f^{- 1} (- 2 x^{3} + 9) = \frac{\sqrt[3]{8 x^{3}}}{2}$

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

$f^{- 1} (- 2 x^{3} + 9) = \frac{\sqrt[3]{{(2 x)}^{3}}}{2}$

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

$f^{- 1} (- 2 x^{3} + 9) = \frac{2 x}{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (- 2 x^{3} + 9) = \frac{2 x}{2}$

Divide $x$ by $1$ .

$f^{- 1} (- 2 x^{3} + 9) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{\sqrt[3]{4 (- x + 9)}}{2})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = - 2 {(\frac{\sqrt[3]{4 (- x + 9)}}{2})}^{3} + 9$

Simplify each term.

Apply the product rule to $\frac{\sqrt[3]{4 (- x + 9)}}{2}$ .

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = - 2 \frac{{\sqrt[3]{4 (- x + 9)}}^{3}}{2^{3}} + 9$

Simplify the numerator.

Rewrite ${\sqrt[3]{4 (- x + 9)}}^{3}$ as $4 (- x + 9)$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{4 (- x + 9)}$ as ${(4 (- x + 9))}^{\frac{1}{3}}$ .

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = - 2 \frac{{({(4 (- x + 9))}^{\frac{1}{3}})}^{3}}{2^{3}} + 9$

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

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = - 2 \frac{{(4 (- x + 9))}^{\frac{1}{3} \cdot 3}}{2^{3}} + 9$

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

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = - 2 \frac{{(4 (- x + 9))}^{\frac{3}{3}}}{2^{3}} + 9$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = - 2 \frac{{(4 (- x + 9))}^{\frac{3}{3}}}{2^{3}} + 9$

Rewrite the expression.

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = - 2 \frac{4 (- x + 9)}{2^{3}} + 9$

Simplify.

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = - 2 \frac{4 (- x + 9)}{2^{3}} + 9$

Apply the distributive property.

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = - 2 \frac{4 (- x) + 4 \cdot 9}{2^{3}} + 9$

Multiply $- 1$ by $4$ .

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = - 2 \frac{- 4 x + 4 \cdot 9}{2^{3}} + 9$

Multiply $4$ by $9$ .

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = - 2 \frac{- 4 x + 36}{2^{3}} + 9$

Factor $4$ out of $- 4 x + 36$ .

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

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = - 2 \frac{4 (- x) + 36}{2^{3}} + 9$

Factor $4$ out of $36$ .

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = - 2 \frac{4 (- x) + 4 (9)}{2^{3}} + 9$

Factor $4$ out of $4 (- x) + 4 (9)$ .

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = - 2 \frac{4 (- x + 9)}{2^{3}} + 9$

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

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = - 2 \frac{4 (- x + 9)}{8} + 9$

Cancel the common factor of $2$ .

Factor $2$ out of $- 2$ .

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = 2 (- 1) (\frac{4 (- x + 9)}{8}) + 9$

Factor $2$ out of $8$ .

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = 2 \cdot (- 1 \frac{4 (- x + 9)}{2 \cdot 4}) + 9$

Cancel the common factor.

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = 2 \cdot (- 1 \frac{4 (- x + 9)}{2 \cdot 4}) + 9$

Rewrite the expression.

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = - 1 \frac{4 (- x + 9)}{4} + 9$

Cancel the common factor of $4$ .

Cancel the common factor.

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = - 1 \frac{4 (- x + 9)}{4} + 9$

Divide $- x + 9$ by $1$ .

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = - 1 (- x + 9) + 9$

Apply the distributive property.

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = - 1 (- x) - 1 \cdot 9 + 9$

Multiply $- 1 (- x)$ .

Multiply $- 1$ by $- 1$ .

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = 1 x - 1 \cdot 9 + 9$

Multiply $x$ by $1$ .

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = x - 1 \cdot 9 + 9$

Multiply $- 1$ by $9$ .

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = x - 9 + 9$

Combine the opposite terms in $x - 9 + 9$ .

Add $- 9$ and $9$ .

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = x + 0$

Add $x$ and $0$ .

$f (\frac{\sqrt[3]{4 (- x + 9)}}{2}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\sqrt[3]{4 (- x + 9)}}{2}$ is the inverse of $f (x) = - 2 x^{3} + 9$ .

$f^{- 1} (x) = \frac{\sqrt[3]{4 (- x + 9)}}{2}$

Find the Inverse -2x^3+9

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