Find the Inverse cube root of 6x-7

$\sqrt[3]{6 x - 7}$

Step 1

Interchange the variables.

$x = \sqrt[3]{6 y - 7}$

Step 2

Solve for $y$ .

Rewrite the equation as $\sqrt[3]{6 y - 7} = x$ .

$\sqrt[3]{6 y - 7} = x$

To remove the radical on the left side of the equation, cube both sides of the equation.

${\sqrt[3]{6 y - 7}}^{3} = x^{3}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{6 y - 7}$ as ${(6 y - 7)}^{\frac{1}{3}}$ .

${({(6 y - 7)}^{\frac{1}{3}})}^{3} = x^{3}$

Simplify the left side.

Simplify ${({(6 y - 7)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(6 y - 7)}^{\frac{1}{3}})}^{3}$ .

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

${(6 y - 7)}^{\frac{1}{3} \cdot 3} = x^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(6 y - 7)}^{\frac{1}{3} \cdot 3} = x^{3}$

Rewrite the expression.

${(6 y - 7)}^{1} = x^{3}$

Simplify.

$6 y - 7 = x^{3}$

Solve for $y$ .

Add $7$ to both sides of the equation.

$6 y = x^{3} + 7$

Divide each term in $6 y = x^{3} + 7$ by $6$ and simplify.

Divide each term in $6 y = x^{3} + 7$ by $6$ .

$\frac{6 y}{6} = \frac{x^{3}}{6} + \frac{7}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 y}{6} = \frac{x^{3}}{6} + \frac{7}{6}$

Divide $y$ by $1$ .

$y = \frac{x^{3}}{6} + \frac{7}{6}$

Step 3

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

$f^{- 1} (x) = \frac{x^{3}}{6} + \frac{7}{6}$

Step 4

Verify if $f^{- 1} (x) = \frac{x^{3}}{6} + \frac{7}{6}$ is the inverse of $f (x) = \sqrt[3]{6 x - 7}$ .

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 $f^{- 1} (\sqrt[3]{6 x - 7})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\sqrt[3]{6 x - 7}) = \frac{{(\sqrt[3]{6 x - 7})}^{3}}{6} + \frac{7}{6}$

Combine the numerators over the common denominator.

$f^{- 1} (\sqrt[3]{6 x - 7}) = \frac{{(\sqrt[3]{6 x - 7})}^{3} + 7}{6}$

Rewrite ${\sqrt[3]{6 x - 7}}^{3}$ as $6 x - 7$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{6 x - 7}$ as ${(6 x - 7)}^{\frac{1}{3}}$ .

$f^{- 1} (\sqrt[3]{6 x - 7}) = \frac{{({(6 x - 7)}^{\frac{1}{3}})}^{3} + 7}{6}$

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

$f^{- 1} (\sqrt[3]{6 x - 7}) = \frac{{(6 x - 7)}^{\frac{1}{3} \cdot 3} + 7}{6}$

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

$f^{- 1} (\sqrt[3]{6 x - 7}) = \frac{{(6 x - 7)}^{\frac{3}{3}} + 7}{6}$

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} (\sqrt[3]{6 x - 7}) = \frac{{(6 x - 7)}^{\frac{3}{3}} + 7}{6}$

Rewrite the expression.

$f^{- 1} (\sqrt[3]{6 x - 7}) = \frac{(6 x - 7) + 7}{6}$

Simplify.

$f^{- 1} (\sqrt[3]{6 x - 7}) = \frac{6 x - 7 + 7}{6}$

Simplify terms.

Combine the opposite terms in $6 x - 7 + 7$ .

Add $- 7$ and $7$ .

$f^{- 1} (\sqrt[3]{6 x - 7}) = \frac{6 x + 0}{6}$

Add $6 x$ and $0$ .

$f^{- 1} (\sqrt[3]{6 x - 7}) = \frac{6 x}{6}$

Cancel the common factor of $6$ .

Cancel the common factor.

$f^{- 1} (\sqrt[3]{6 x - 7}) = \frac{6 x}{6}$

Divide $x$ by $1$ .

$f^{- 1} (\sqrt[3]{6 x - 7}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{x^{3}}{6} + \frac{7}{6})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x^{3}}{6} + \frac{7}{6}) = \sqrt[3]{6 (\frac{x^{3}}{6} + \frac{7}{6}) - 7}$

Apply the distributive property.

$f (\frac{x^{3}}{6} + \frac{7}{6}) = \sqrt[3]{6 (\frac{x^{3}}{6}) + 6 (\frac{7}{6}) - 7}$

Cancel the common factor of $6$ .

Cancel the common factor.

$f (\frac{x^{3}}{6} + \frac{7}{6}) = \sqrt[3]{6 (\frac{x^{3}}{6}) + 6 (\frac{7}{6}) - 7}$

Rewrite the expression.

$f (\frac{x^{3}}{6} + \frac{7}{6}) = \sqrt[3]{x^{3} + 6 (\frac{7}{6}) - 7}$

Cancel the common factor of $6$ .

Cancel the common factor.

$f (\frac{x^{3}}{6} + \frac{7}{6}) = \sqrt[3]{x^{3} + 6 (\frac{7}{6}) - 7}$

Rewrite the expression.

$f (\frac{x^{3}}{6} + \frac{7}{6}) = \sqrt[3]{x^{3} + 7 - 7}$

Simplify by subtracting numbers.

Subtract $7$ from $7$ .

$f (\frac{x^{3}}{6} + \frac{7}{6}) = \sqrt[3]{x^{3} + 0}$

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

$f (\frac{x^{3}}{6} + \frac{7}{6}) = \sqrt[3]{x^{3}}$

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

$f (\frac{x^{3}}{6} + \frac{7}{6}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x^{3}}{6} + \frac{7}{6}$ is the inverse of $f (x) = \sqrt[3]{6 x - 7}$ .

$f^{- 1} (x) = \frac{x^{3}}{6} + \frac{7}{6}$

Find the Inverse cube root of 6x-7

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