Find the Inverse 8x^3-6

$8 x^{3} - 6$

Step 1

Interchange the variables.

$x = 8 y^{3} - 6$

Step 2

Solve for $y$ .

Rewrite the equation as $8 y^{3} - 6 = x$ .

$8 y^{3} - 6 = x$

Add $6$ to both sides of the equation.

$8 y^{3} = x + 6$

Divide each term in $8 y^{3} = x + 6$ by $8$ and simplify.

Divide each term in $8 y^{3} = x + 6$ by $8$ .

$\frac{8 y^{3}}{8} = \frac{x}{8} + \frac{6}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 y^{3}}{8} = \frac{x}{8} + \frac{6}{8}$

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

$y^{3} = \frac{x}{8} + \frac{6}{8}$

Simplify the right side.

Cancel the common factor of $6$ and $8$ .

Factor $2$ out of $6$ .

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

Cancel the common factors.

Factor $2$ out of $8$ .

$y^{3} = \frac{x}{8} + \frac{2 \cdot 3}{2 \cdot 4}$

Cancel the common factor.

$y^{3} = \frac{x}{8} + \frac{2 \cdot 3}{2 \cdot 4}$

Rewrite the expression.

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

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

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

Simplify $\sqrt[3]{\frac{x}{8} + \frac{3}{4}}$ .

To write $\frac{3}{4}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Write each expression with a common denominator of $8$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{3}{4}$ by $\frac{2}{2}$ .

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

Multiply $4$ by $2$ .

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

Combine the numerators over the common denominator.

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

Multiply $3$ by $2$ .

$y = \sqrt[3]{\frac{x + 6}{8}}$

Rewrite $\frac{x + 6}{8}$ as ${(\frac{1}{2})}^{3} (x + 6)$ .

Factor the perfect power $1^{3}$ out of $x + 6$ .

$y = \sqrt[3]{\frac{1^{3} (x + 6)}{8}}$

Factor the perfect power $2^{3}$ out of $8$ .

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

Rearrange the fraction $\frac{1^{3} (x + 6)}{2^{3} \cdot 1}$ .

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

Pull terms out from under the radical.

$y = \frac{1}{2} \sqrt[3]{x + 6}$

Combine $\frac{1}{2}$ and $\sqrt[3]{x + 6}$ .

$y = \frac{\sqrt[3]{x + 6}}{2}$

Step 3

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

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

Step 4

Verify if $f^{- 1} (x) = \frac{\sqrt[3]{x + 6}}{2}$ is the inverse of $f (x) = 8 x^{3} - 6$ .

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 $f^{- 1} (8 x^{3} - 6)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (8 x^{3} - 6) = \frac{\sqrt[3]{(8 x^{3} - 6) + 6}}{2}$

Simplify the numerator.

Add $- 6$ and $6$ .

$f^{- 1} (8 x^{3} - 6) = \frac{\sqrt[3]{8 x^{3} + 0}}{2}$

Add $8 x^{3}$ and $0$ .

$f^{- 1} (8 x^{3} - 6) = \frac{\sqrt[3]{8 x^{3}}}{2}$

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

$f^{- 1} (8 x^{3} - 6) = \frac{\sqrt[3]{{(2 x)}^{3}}}{2}$

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

$f^{- 1} (8 x^{3} - 6) = \frac{2 x}{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (8 x^{3} - 6) = \frac{2 x}{2}$

Divide $x$ by $1$ .

$f^{- 1} (8 x^{3} - 6) = x$

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

Set up the composite result function.

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

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

$f (\frac{\sqrt[3]{x + 6}}{2}) = 8 {(\frac{\sqrt[3]{x + 6}}{2})}^{3} - 6$

Simplify each term.

Apply the product rule to $\frac{\sqrt[3]{x + 6}}{2}$ .

$f (\frac{\sqrt[3]{x + 6}}{2}) = 8 (\frac{{\sqrt[3]{x + 6}}^{3}}{2^{3}}) - 6$

Rewrite ${\sqrt[3]{x + 6}}^{3}$ as $x + 6$ .

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

$f (\frac{\sqrt[3]{x + 6}}{2}) = 8 (\frac{{({(x + 6)}^{\frac{1}{3}})}^{3}}{2^{3}}) - 6$

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

$f (\frac{\sqrt[3]{x + 6}}{2}) = 8 (\frac{{(x + 6)}^{\frac{1}{3} \cdot 3}}{2^{3}}) - 6$

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

$f (\frac{\sqrt[3]{x + 6}}{2}) = 8 (\frac{{(x + 6)}^{\frac{3}{3}}}{2^{3}}) - 6$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\frac{\sqrt[3]{x + 6}}{2}) = 8 (\frac{{(x + 6)}^{\frac{3}{3}}}{2^{3}}) - 6$

Rewrite the expression.

$f (\frac{\sqrt[3]{x + 6}}{2}) = 8 (\frac{x + 6}{2^{3}}) - 6$

Simplify.

$f (\frac{\sqrt[3]{x + 6}}{2}) = 8 (\frac{x + 6}{2^{3}}) - 6$

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

$f (\frac{\sqrt[3]{x + 6}}{2}) = 8 (\frac{x + 6}{8}) - 6$

Cancel the common factor of $8$ .

Cancel the common factor.

$f (\frac{\sqrt[3]{x + 6}}{2}) = 8 (\frac{x + 6}{8}) - 6$

Rewrite the expression.

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

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

Subtract $6$ from $6$ .

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

Add $x$ and $0$ .

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

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\sqrt[3]{x + 6}}{2}$ is the inverse of $f (x) = 8 x^{3} - 6$ .

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

Find the Inverse 8x^3-6

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