Find the Inverse 6x^3+4

$6 x^{3} + 4$

Step 1

Interchange the variables.

$x = 6 y^{3} + 4$

Step 2

Solve for $y$ .

Rewrite the equation as $6 y^{3} + 4 = x$ .

$6 y^{3} + 4 = x$

Subtract $4$ from both sides of the equation.

$6 y^{3} = x - 4$

Divide each term in $6 y^{3} = x - 4$ by $6$ and simplify.

Divide each term in $6 y^{3} = x - 4$ by $6$ .

$\frac{6 y^{3}}{6} = \frac{x}{6} + \frac{- 4}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 y^{3}}{6} = \frac{x}{6} + \frac{- 4}{6}$

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

$y^{3} = \frac{x}{6} + \frac{- 4}{6}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $- 4$ and $6$ .

Factor $2$ out of $- 4$ .

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

Cancel the common factors.

Factor $2$ out of $6$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

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

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

Simplify $\sqrt[3]{\frac{x}{6} - \frac{2}{3}}$ .

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

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

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

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

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

Multiply $3$ by $2$ .

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

Combine the numerators over the common denominator.

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

Multiply $- 2$ by $2$ .

$y = \sqrt[3]{\frac{x - 4}{6}}$

Rewrite $\sqrt[3]{\frac{x - 4}{6}}$ as $\frac{\sqrt[3]{x - 4}}{\sqrt[3]{6}}$ .

$y = \frac{\sqrt[3]{x - 4}}{\sqrt[3]{6}}$

Multiply $\frac{\sqrt[3]{x - 4}}{\sqrt[3]{6}}$ by $\frac{{\sqrt[3]{6}}^{2}}{{\sqrt[3]{6}}^{2}}$ .

$y = \frac{\sqrt[3]{x - 4}}{\sqrt[3]{6}} \cdot \frac{{\sqrt[3]{6}}^{2}}{{\sqrt[3]{6}}^{2}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt[3]{x - 4}}{\sqrt[3]{6}}$ by $\frac{{\sqrt[3]{6}}^{2}}{{\sqrt[3]{6}}^{2}}$ .

$y = \frac{\sqrt[3]{x - 4} {\sqrt[3]{6}}^{2}}{\sqrt[3]{6} {\sqrt[3]{6}}^{2}}$

Raise $\sqrt[3]{6}$ to the power of $1$ .

$y = \frac{\sqrt[3]{x - 4} {\sqrt[3]{6}}^{2}}{{\sqrt[3]{6}}^{1} {\sqrt[3]{6}}^{2}}$

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

$y = \frac{\sqrt[3]{x - 4} {\sqrt[3]{6}}^{2}}{{\sqrt[3]{6}}^{1 + 2}}$

Add $1$ and $2$ .

$y = \frac{\sqrt[3]{x - 4} {\sqrt[3]{6}}^{2}}{{\sqrt[3]{6}}^{3}}$

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

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

$y = \frac{\sqrt[3]{x - 4} {\sqrt[3]{6}}^{2}}{{(6^{\frac{1}{3}})}^{3}}$

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

$y = \frac{\sqrt[3]{x - 4} {\sqrt[3]{6}}^{2}}{6^{\frac{1}{3} \cdot 3}}$

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

$y = \frac{\sqrt[3]{x - 4} {\sqrt[3]{6}}^{2}}{6^{\frac{3}{3}}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$y = \frac{\sqrt[3]{x - 4} {\sqrt[3]{6}}^{2}}{6^{\frac{3}{3}}}$

Rewrite the expression.

$y = \frac{\sqrt[3]{x - 4} {\sqrt[3]{6}}^{2}}{6^{1}}$

Evaluate the exponent.

$y = \frac{\sqrt[3]{x - 4} {\sqrt[3]{6}}^{2}}{6}$

Simplify the numerator.

Rewrite ${\sqrt[3]{6}}^{2}$ as $\sqrt[3]{6^{2}}$ .

$y = \frac{\sqrt[3]{x - 4} \sqrt[3]{6^{2}}}{6}$

Raise $6$ to the power of $2$ .

$y = \frac{\sqrt[3]{x - 4} \sqrt[3]{36}}{6}$

Simplify with factoring out.

Combine using the product rule for radicals.

$y = \frac{\sqrt[3]{(x - 4) \cdot 36}}{6}$

Reorder factors in $\frac{\sqrt[3]{(x - 4) \cdot 36}}{6}$ .

$y = \frac{\sqrt[3]{36 (x - 4)}}{6}$

Step 3

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

$f^{- 1} (x) = \frac{\sqrt[3]{36 (x - 4)}}{6}$

Step 4

Verify if $f^{- 1} (x) = \frac{\sqrt[3]{36 (x - 4)}}{6}$ is the inverse of $f (x) = 6 x^{3} + 4$ .

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 $f^{- 1} (6 x^{3} + 4)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (6 x^{3} + 4) = \frac{\sqrt[3]{36 ((6 x^{3} + 4) - 4)}}{6}$

Simplify the numerator.

Subtract $4$ from $4$ .

$f^{- 1} (6 x^{3} + 4) = \frac{\sqrt[3]{36 (6 x^{3} + 0)}}{6}$

Add $6 x^{3}$ and $0$ .

$f^{- 1} (6 x^{3} + 4) = \frac{\sqrt[3]{36 \cdot (6 x^{3})}}{6}$

Multiply $36$ by $6$ .

$f^{- 1} (6 x^{3} + 4) = \frac{\sqrt[3]{216 x^{3}}}{6}$

Rewrite $216 x^{3}$ as ${(6 x)}^{3}$ .

$f^{- 1} (6 x^{3} + 4) = \frac{\sqrt[3]{{(6 x)}^{3}}}{6}$

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

$f^{- 1} (6 x^{3} + 4) = \frac{6 x}{6}$

Cancel the common factor of $6$ .

Cancel the common factor.

$f^{- 1} (6 x^{3} + 4) = \frac{6 x}{6}$

Divide $x$ by $1$ .

$f^{- 1} (6 x^{3} + 4) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{\sqrt[3]{36 (x - 4)}}{6})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\sqrt[3]{36 (x - 4)}}{6}) = 6 {(\frac{\sqrt[3]{36 (x - 4)}}{6})}^{3} + 4$

Simplify each term.

Apply the product rule to $\frac{\sqrt[3]{36 (x - 4)}}{6}$ .

$f (\frac{\sqrt[3]{36 (x - 4)}}{6}) = 6 (\frac{{\sqrt[3]{36 (x - 4)}}^{3}}{6^{3}}) + 4$

Simplify the numerator.

Rewrite ${\sqrt[3]{36 (x - 4)}}^{3}$ as $36 (x - 4)$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{36 (x - 4)}$ as ${(36 (x - 4))}^{\frac{1}{3}}$ .

$f (\frac{\sqrt[3]{36 (x - 4)}}{6}) = 6 (\frac{{({(36 (x - 4))}^{\frac{1}{3}})}^{3}}{6^{3}}) + 4$

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

$f (\frac{\sqrt[3]{36 (x - 4)}}{6}) = 6 (\frac{{(36 (x - 4))}^{\frac{1}{3} \cdot 3}}{6^{3}}) + 4$

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

$f (\frac{\sqrt[3]{36 (x - 4)}}{6}) = 6 (\frac{{(36 (x - 4))}^{\frac{3}{3}}}{6^{3}}) + 4$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\frac{\sqrt[3]{36 (x - 4)}}{6}) = 6 (\frac{{(36 (x - 4))}^{\frac{3}{3}}}{6^{3}}) + 4$

Rewrite the expression.

$f (\frac{\sqrt[3]{36 (x - 4)}}{6}) = 6 (\frac{36 (x - 4)}{6^{3}}) + 4$

Simplify.

$f (\frac{\sqrt[3]{36 (x - 4)}}{6}) = 6 (\frac{36 (x - 4)}{6^{3}}) + 4$

Apply the distributive property.

$f (\frac{\sqrt[3]{36 (x - 4)}}{6}) = 6 (\frac{36 x + 36 \cdot - 4}{6^{3}}) + 4$

Multiply $36$ by $- 4$ .

$f (\frac{\sqrt[3]{36 (x - 4)}}{6}) = 6 (\frac{36 x - 144}{6^{3}}) + 4$

Factor $36$ out of $36 x - 144$ .

Factor $36$ out of $36 x$ .

$f (\frac{\sqrt[3]{36 (x - 4)}}{6}) = 6 (\frac{36 (x) - 144}{6^{3}}) + 4$

Factor $36$ out of $- 144$ .

$f (\frac{\sqrt[3]{36 (x - 4)}}{6}) = 6 (\frac{36 x + 36 \cdot - 4}{6^{3}}) + 4$

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

$f (\frac{\sqrt[3]{36 (x - 4)}}{6}) = 6 (\frac{36 (x - 4)}{6^{3}}) + 4$

Raise $6$ to the power of $3$ .

$f (\frac{\sqrt[3]{36 (x - 4)}}{6}) = 6 (\frac{36 (x - 4)}{216}) + 4$

Cancel the common factor of $6$ .

Factor $6$ out of $216$ .

$f (\frac{\sqrt[3]{36 (x - 4)}}{6}) = 6 (\frac{36 (x - 4)}{6 (36)}) + 4$

Cancel the common factor.

$f (\frac{\sqrt[3]{36 (x - 4)}}{6}) = 6 (\frac{36 (x - 4)}{6 \cdot 36}) + 4$

Rewrite the expression.

$f (\frac{\sqrt[3]{36 (x - 4)}}{6}) = \frac{36 (x - 4)}{36} + 4$

Cancel the common factor of $36$ .

Cancel the common factor.

$f (\frac{\sqrt[3]{36 (x - 4)}}{6}) = \frac{36 (x - 4)}{36} + 4$

Divide $x - 4$ by $1$ .

$f (\frac{\sqrt[3]{36 (x - 4)}}{6}) = x - 4 + 4$

Combine the opposite terms in $x - 4 + 4$ .

Add $- 4$ and $4$ .

$f (\frac{\sqrt[3]{36 (x - 4)}}{6}) = x + 0$

Add $x$ and $0$ .

$f (\frac{\sqrt[3]{36 (x - 4)}}{6}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\sqrt[3]{36 (x - 4)}}{6}$ is the inverse of $f (x) = 6 x^{3} + 4$ .

$f^{- 1} (x) = \frac{\sqrt[3]{36 (x - 4)}}{6}$

Find the Inverse 6x^3+4

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