Find the Inverse f(x)=4x^3+5

$f (x) = 4 x^{3} + 5$

Replace $f (x)$ with $y$ .

$y = 4 x^{3} + 5$

Interchange the variables.

$x = 4 y^{3} + 5$

Solve for $y$ .

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Rewrite the equation as $4 y^{3} + 5 = x$ .

$4 y^{3} + 5 = x$

Subtract $5$ from both sides of the equation.

$4 y^{3} = x - 5$

Divide each term by $4$ and simplify.

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Divide each term in $4 y^{3} = x - 5$ by $4$ .

$\frac{4 y^{3}}{4} = \frac{x}{4} + \frac{- 5}{4}$

Cancel the common factor of $4$ .

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Cancel the common factor.

$\frac{4 y^{3}}{4} = \frac{x}{4} + \frac{- 5}{4}$

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

$y^{3} = \frac{x}{4} + \frac{- 5}{4}$

Move the negative in front of the fraction.

$y^{3} = \frac{x}{4} - \frac{5}{4}$

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

$y = \sqrt[3]{\frac{x}{4} - \frac{5}{4}}$

Simplify $\sqrt[3]{\frac{x}{4} - \frac{5}{4}}$ .

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Combine the numerators over the common denominator.

$y = \sqrt[3]{\frac{x - 5}{4}}$

Rewrite $\sqrt[3]{\frac{x - 5}{4}}$ as $\frac{\sqrt[3]{x - 5}}{\sqrt[3]{4}}$ .

$y = \frac{\sqrt[3]{x - 5}}{\sqrt[3]{4}}$

Multiply $\frac{\sqrt[3]{x - 5}}{\sqrt[3]{4}}$ by $\frac{{\sqrt[3]{4}}^{2}}{{\sqrt[3]{4}}^{2}}$ .

$y = \frac{\sqrt[3]{x - 5}}{\sqrt[3]{4}} \cdot \frac{{\sqrt[3]{4}}^{2}}{{\sqrt[3]{4}}^{2}}$

Combine and simplify the denominator.

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Multiply $\frac{\sqrt[3]{x - 5}}{\sqrt[3]{4}}$ and $\frac{{\sqrt[3]{4}}^{2}}{{\sqrt[3]{4}}^{2}}$ .

$y = \frac{\sqrt[3]{x - 5} {\sqrt[3]{4}}^{2}}{\sqrt[3]{4} {\sqrt[3]{4}}^{2}}$

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

$y = \frac{\sqrt[3]{x - 5} {\sqrt[3]{4}}^{2}}{{\sqrt[3]{4}}^{1} {\sqrt[3]{4}}^{2}}$

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

$y = \frac{\sqrt[3]{x - 5} {\sqrt[3]{4}}^{2}}{{\sqrt[3]{4}}^{1 + 2}}$

Add $1$ and $2$ .

$y = \frac{\sqrt[3]{x - 5} {\sqrt[3]{4}}^{2}}{{\sqrt[3]{4}}^{3}}$

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

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

$y = \frac{\sqrt[3]{x - 5} {\sqrt[3]{4}}^{2}}{{(4^{\frac{1}{3}})}^{3}}$

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

$y = \frac{\sqrt[3]{x - 5} {\sqrt[3]{4}}^{2}}{4^{\frac{1}{3} \cdot 3}}$

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

$y = \frac{\sqrt[3]{x - 5} {\sqrt[3]{4}}^{2}}{4^{\frac{3}{3}}}$

Cancel the common factor of $3$ .

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Cancel the common factor.

$y = \frac{\sqrt[3]{x - 5} {\sqrt[3]{4}}^{2}}{4^{\frac{3}{3}}}$

Divide $1$ by $1$ .

$y = \frac{\sqrt[3]{x - 5} {\sqrt[3]{4}}^{2}}{4^{1}}$

Evaluate the exponent.

$y = \frac{\sqrt[3]{x - 5} {\sqrt[3]{4}}^{2}}{4}$

Simplify the numerator.

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

$y = \frac{\sqrt[3]{x - 5} \sqrt[3]{4^{2}}}{4}$

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

$y = \frac{\sqrt[3]{x - 5} \sqrt[3]{16}}{4}$

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

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Factor $8$ out of $16$ .

$y = \frac{\sqrt[3]{x - 5} \sqrt[3]{8 (2)}}{4}$

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

$y = \frac{\sqrt[3]{x - 5} \sqrt[3]{2^{3} \cdot 2}}{4}$

Pull terms out from under the radical.

$y = \frac{\sqrt[3]{x - 5} \cdot 2 \sqrt[3]{2}}{4}$

Combine using the product rule for radicals.

$y = \frac{2 \sqrt[3]{(x - 5) \cdot 2}}{4}$

Reduce the expression by cancelling the common factors.

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

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Factor $2$ out of $2 \sqrt[3]{(x - 5) \cdot 2}$ .

$y = \frac{2 (\sqrt[3]{(x - 5) \cdot 2})}{4}$

Cancel the common factors.

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Factor $2$ out of $4$ .

$y = \frac{2 \sqrt[3]{(x - 5) \cdot 2}}{2 \cdot 2}$

Cancel the common factor.

$y = \frac{2 \sqrt[3]{(x - 5) \cdot 2}}{2 \cdot 2}$

Rewrite the expression.

$y = \frac{\sqrt[3]{(x - 5) \cdot 2}}{2}$

Reorder factors in $\frac{\sqrt[3]{(x - 5) \cdot 2}}{2}$ .

$y = \frac{\sqrt[3]{2 (x - 5)}}{2}$

Solve for $y$ and replace with $f^{- 1} (x)$ .

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

$f^{- 1} (x) = \frac{\sqrt[3]{2 (x - 5)}}{2}$

Set up the composite result function.

$g (f (x))$

Evaluate $g (f (x))$ by substituting in the value of $f$ into $g$ .

$g (4 x^{3} + 5) = \frac{\sqrt[3]{2 ((4 x^{3} + 5) - 5)}}{2}$

Simplify the numerator.

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Subtract $5$ from $5$ .

$g (4 x^{3} + 5) = \frac{\sqrt[3]{2 (4 x^{3} + 0)}}{2}$

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

$g (4 x^{3} + 5) = \frac{\sqrt[3]{2 \cdot (4 x^{3})}}{2}$

Multiply $2$ by $4$ .

$g (4 x^{3} + 5) = \frac{\sqrt[3]{8 x^{3}}}{2}$

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

$g (4 x^{3} + 5) = \frac{\sqrt[3]{{(2 x)}^{3}}}{2}$

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

$g (4 x^{3} + 5) = \frac{2 x}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$g (4 x^{3} + 5) = \frac{2 x}{2}$

Divide $x$ by $1$ .

$g (4 x^{3} + 5) = x$

Since $g (f (x)) = x$ , $f^{- 1} (x) = \frac{\sqrt[3]{2 (x - 5)}}{2}$ is the inverse of $f (x) = 4 x^{3} + 5$ .

$f^{- 1} (x) = \frac{\sqrt[3]{2 (x - 5)}}{2}$

Find the Inverse f(x)=4x^3+5

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