Find the Inverse 3x^3-1

$3 x^{3} - 1$

Step 1

Interchange the variables.

$x = 3 y^{3} - 1$

Step 2

Solve for $y$ .

Rewrite the equation as $3 y^{3} - 1 = x$ .

$3 y^{3} - 1 = x$

Add $1$ to both sides of the equation.

$3 y^{3} = x + 1$

Divide each term in $3 y^{3} = x + 1$ by $3$ and simplify.

Divide each term in $3 y^{3} = x + 1$ by $3$ .

$\frac{3 y^{3}}{3} = \frac{x}{3} + \frac{1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y^{3}}{3} = \frac{x}{3} + \frac{1}{3}$

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

$y^{3} = \frac{x}{3} + \frac{1}{3}$

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

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

Simplify $\sqrt[3]{\frac{x}{3} + \frac{1}{3}}$ .

Combine the numerators over the common denominator.

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

Rewrite $\sqrt[3]{\frac{x + 1}{3}}$ as $\frac{\sqrt[3]{x + 1}}{\sqrt[3]{3}}$ .

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

Multiply $\frac{\sqrt[3]{x + 1}}{\sqrt[3]{3}}$ by $\frac{{\sqrt[3]{3}}^{2}}{{\sqrt[3]{3}}^{2}}$ .

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

Combine and simplify the denominator.

Multiply $\frac{\sqrt[3]{x + 1}}{\sqrt[3]{3}}$ by $\frac{{\sqrt[3]{3}}^{2}}{{\sqrt[3]{3}}^{2}}$ .

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

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

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

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

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

Add $1$ and $2$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

Simplify the numerator.

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

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

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

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

Simplify with factoring out.

Combine using the product rule for radicals.

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

Reorder factors in $\frac{\sqrt[3]{(x + 1) \cdot 9}}{3}$ .

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

Step 3

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

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

Step 4

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

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

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

Simplify the numerator.

Add $- 1$ and $1$ .

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

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

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

Multiply $9$ by $3$ .

$f^{- 1} (3 x^{3} - 1) = \frac{\sqrt[3]{27 x^{3}}}{3}$

Rewrite $27 x^{3}$ as ${(3 x)}^{3}$ .

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Apply the product rule to $\frac{\sqrt[3]{9 (x + 1)}}{3}$ .

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

Simplify the numerator.

Rewrite ${\sqrt[3]{9 (x + 1)}}^{3}$ as $9 (x + 1)$ .

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Apply the distributive property.

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

Multiply $9$ by $1$ .

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

Factor $9$ out of $9 x + 9$ .

Factor $9$ out of $9 x$ .

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

Factor $9$ out of $9$ .

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

Factor $9$ out of $9 (x) + 9 (1)$ .

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

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

$f (\frac{\sqrt[3]{9 (x + 1)}}{3}) = 3 (\frac{9 (x + 1)}{27}) - 1$

Cancel the common factor of $3$ .

Factor $3$ out of $27$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $9$ .

Cancel the common factor.

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

Divide $x + 1$ by $1$ .

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

Combine the opposite terms in $x + 1 - 1$ .

Subtract $1$ from $1$ .

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

Add $x$ and $0$ .

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

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

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

Find the Inverse 3x^3-1

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