Find the Inverse f(x)=4x^3-1

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

Write $f (x) = 4 x^{3} - 1$ as an equation.

$y = 4 x^{3} - 1$

Interchange the variables.

$x = 4 y^{3} - 1$

Solve for $y$ .

Rewrite the equation as $4 y^{3} - 1 = x$ .

$4 y^{3} - 1 = x$

Add $1$ to both sides of the equation.

$4 y^{3} = x + 1$

Divide each term in $4 y^{3} = x + 1$ by $4$ and simplify.

Divide each term in $4 y^{3} = x + 1$ by $4$ .

$\frac{4 y^{3}}{4} = \frac{x}{4} + \frac{1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y^{3}}{4} = \frac{x}{4} + \frac{1}{4}$

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

$y^{3} = \frac{x}{4} + \frac{1}{4}$

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

$y = \sqrt[3]{\frac{x}{4} + \frac{1}{4}}$

Simplify $\sqrt[3]{\frac{x}{4} + \frac{1}{4}}$ .

Combine the numerators over the common denominator.

$y = \sqrt[3]{\frac{x + 1}{4}}$

Rewrite $\sqrt[3]{\frac{x + 1}{4}}$ as $\frac{\sqrt[3]{x + 1}}{\sqrt[3]{4}}$ .

$y = \frac{\sqrt[3]{x + 1}}{\sqrt[3]{4}}$

Multiply $\frac{\sqrt[3]{x + 1}}{\sqrt[3]{4}}$ by $\frac{{\sqrt[3]{4}}^{2}}{{\sqrt[3]{4}}^{2}}$ .

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

Combine and simplify the denominator.

Multiply $\frac{\sqrt[3]{x + 1}}{\sqrt[3]{4}}$ by $\frac{{\sqrt[3]{4}}^{2}}{{\sqrt[3]{4}}^{2}}$ .

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

Raise $\sqrt[3]{4}$ to the power of $1$ .

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

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

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

Add $1$ and $2$ .

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

Rewrite ${\sqrt[3]{4}}^{3}$ as $4$ .

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

Simplify the numerator.

Rewrite ${\sqrt[3]{4}}^{2}$ as ${(4^{2})}^{\frac{1}{3}}$ .

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

Raise $4$ to the power of $2$ .

$y = \frac{\sqrt[3]{x + 1} \sqrt[3]{16}}{4}$

Rewrite $16$ as $2^{3} \cdot 2$ .

Factor $8$ out of $16$ .

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

Rewrite $8$ as $2^{3}$ .

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

Pull terms out from under the radical.

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

Combine using the product rule for radicals.

$y = \frac{2 \sqrt[3]{(x + 1) \cdot 2}}{4}$

Reduce the expression by cancelling the common factors.

Cancel the common factor of $2$ and $4$ .

Factor $2$ out of $2 \sqrt[3]{(x + 1) \cdot 2}$ .

$y = \frac{2 (\sqrt[3]{(x + 1) \cdot 2})}{4}$

Cancel the common factors.

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Reorder factors in $\frac{\sqrt[3]{(x + 1) \cdot 2}}{2}$ .

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

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

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

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

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

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

Simplify the numerator.

Add $- 1$ and $1$ .

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

Add $4 x^{3}$ and $0$ .

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

Multiply $2$ by $4$ .

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

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

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

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

$f^{- 1} (4 x^{3} - 1) = \frac{2 x}{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (4 x^{3} - 1) = \frac{2 x}{2}$

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Apply the product rule to $\frac{\sqrt[3]{2 (x + 1)}}{2}$ .

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

Simplify the numerator.

Rewrite ${\sqrt[3]{2 (x + 1)}}^{3}$ as $2 (x + 1)$ .

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Apply the distributive property.

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

Multiply $2$ by $1$ .

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

Factor $2$ out of $2 x + 2$ .

Factor $2$ out of $2 x$ .

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

Factor $2$ out of $2$ .

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

Factor $2$ out of $2 (x) + 2 (1)$ .

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

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

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

Cancel the common factor of $4$ .

Factor $4$ out of $8$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x + 1$ by $1$ .

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

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

Subtract $1$ from $1$ .

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

Add $x$ and $0$ .

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

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

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

Find the Inverse f(x)=4x^3-1

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