Find the Inverse 2x^3-8

$2 x^{3} - 8$

Step 1

Interchange the variables.

$x = 2 y^{3} - 8$

Step 2

Solve for $y$ .

Rewrite the equation as $2 y^{3} - 8 = x$ .

$2 y^{3} - 8 = x$

Add $8$ to both sides of the equation.

$2 y^{3} = x + 8$

Divide each term in $2 y^{3} = x + 8$ by $2$ and simplify.

Divide each term in $2 y^{3} = x + 8$ by $2$ .

$\frac{2 y^{3}}{2} = \frac{x}{2} + \frac{8}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y^{3}}{2} = \frac{x}{2} + \frac{8}{2}$

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

$y^{3} = \frac{x}{2} + \frac{8}{2}$

Simplify the right side.

Divide $8$ by $2$ .

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

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

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

Simplify $\sqrt[3]{\frac{x}{2} + 4}$ .

To write $4$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Combine $4$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

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

Multiply $4$ by $2$ .

$y = \sqrt[3]{\frac{x + 8}{2}}$

Rewrite $\sqrt[3]{\frac{x + 8}{2}}$ as $\frac{\sqrt[3]{x + 8}}{\sqrt[3]{2}}$ .

$y = \frac{\sqrt[3]{x + 8}}{\sqrt[3]{2}}$

Multiply $\frac{\sqrt[3]{x + 8}}{\sqrt[3]{2}}$ by $\frac{{\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{2}}$ .

$y = \frac{\sqrt[3]{x + 8}}{\sqrt[3]{2}} \cdot \frac{{\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{2}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt[3]{x + 8}}{\sqrt[3]{2}}$ by $\frac{{\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{2}}$ .

$y = \frac{\sqrt[3]{x + 8} {\sqrt[3]{2}}^{2}}{\sqrt[3]{2} {\sqrt[3]{2}}^{2}}$

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

$y = \frac{\sqrt[3]{x + 8} {\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{1} {\sqrt[3]{2}}^{2}}$

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

$y = \frac{\sqrt[3]{x + 8} {\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{1 + 2}}$

Add $1$ and $2$ .

$y = \frac{\sqrt[3]{x + 8} {\sqrt[3]{2}}^{2}}{{\sqrt[3]{2}}^{3}}$

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

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

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

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

$y = \frac{\sqrt[3]{x + 8} {\sqrt[3]{2}}^{2}}{2^{\frac{1}{3} \cdot 3}}$

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

$y = \frac{\sqrt[3]{x + 8} {\sqrt[3]{2}}^{2}}{2^{\frac{3}{3}}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$y = \frac{\sqrt[3]{x + 8} {\sqrt[3]{2}}^{2}}{2^{\frac{3}{3}}}$

Rewrite the expression.

$y = \frac{\sqrt[3]{x + 8} {\sqrt[3]{2}}^{2}}{2^{1}}$

Evaluate the exponent.

$y = \frac{\sqrt[3]{x + 8} {\sqrt[3]{2}}^{2}}{2}$

Simplify the numerator.

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

$y = \frac{\sqrt[3]{x + 8} \sqrt[3]{2^{2}}}{2}$

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

$y = \frac{\sqrt[3]{x + 8} \sqrt[3]{4}}{2}$

Simplify with factoring out.

Combine using the product rule for radicals.

$y = \frac{\sqrt[3]{(x + 8) \cdot 4}}{2}$

Reorder factors in $\frac{\sqrt[3]{(x + 8) \cdot 4}}{2}$ .

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

Step 3

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

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

Step 4

Verify if $f^{- 1} (x) = \frac{\sqrt[3]{4 (x + 8)}}{2}$ is the inverse of $f (x) = 2 x^{3} - 8$ .

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 $f^{- 1} (2 x^{3} - 8)$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Simplify the numerator.

Add $- 8$ and $8$ .

$f^{- 1} (2 x^{3} - 8) = \frac{\sqrt[3]{4 (2 x^{3} + 0)}}{2}$

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

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

Multiply $4$ by $2$ .

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

$f^{- 1} (2 x^{3} - 8) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{\sqrt[3]{4 (x + 8)}}{2})$ by substituting in the value of $f^{- 1}$ into $f$ .

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

Simplify each term.

Apply the product rule to $\frac{\sqrt[3]{4 (x + 8)}}{2}$ .

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

Simplify the numerator.

Rewrite ${\sqrt[3]{4 (x + 8)}}^{3}$ as $4 (x + 8)$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{4 (x + 8)}$ as ${(4 (x + 8))}^{\frac{1}{3}}$ .

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Apply the distributive property.

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

Multiply $4$ by $8$ .

$f (\frac{\sqrt[3]{4 (x + 8)}}{2}) = 2 (\frac{4 x + 32}{2^{3}}) - 8$

Factor $4$ out of $4 x + 32$ .

Factor $4$ out of $4 x$ .

$f (\frac{\sqrt[3]{4 (x + 8)}}{2}) = 2 (\frac{4 (x) + 32}{2^{3}}) - 8$

Factor $4$ out of $32$ .

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

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

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

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

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

Cancel the common factor of $2$ .

Factor $2$ out of $8$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Divide $x + 8$ by $1$ .

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

Combine the opposite terms in $x + 8 - 8$ .

Subtract $8$ from $8$ .

$f (\frac{\sqrt[3]{4 (x + 8)}}{2}) = x + 0$

Add $x$ and $0$ .

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

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\sqrt[3]{4 (x + 8)}}{2}$ is the inverse of $f (x) = 2 x^{3} - 8$ .

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

Find the Inverse 2x^3-8

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