Find the Inverse f(x)=(2x+1)^3-4

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

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

$y = {(2 x + 1)}^{3} - 4$

Interchange the variables.

$x = {(2 y + 1)}^{3} - 4$

Solve for $y$ .

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Rewrite the equation as ${(2 y + 1)}^{3} - 4 = x$ .

${(2 y + 1)}^{3} - 4 = x$

Add $4$ to both sides of the equation.

${(2 y + 1)}^{3} = x + 4$

Take the cube root of each side of the equation to set up the solution for $y$

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

Remove the perfect root factor $2 y + 1$ under the radical to solve for $y$ .

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

Subtract $1$ from both sides of the equation.

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

Divide each term by $2$ and simplify.

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Divide each term in $2 y = \sqrt[3]{x + 4} - 1$ by $2$ .

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

Divide $y$ by $1$ .

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

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.

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

Set up the composite result function.

$g (f (x))$

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

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

Simplify the numerator.

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Use the Binomial Theorem.

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

Simplify each term.

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Apply the product rule to $2 x$ .

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

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

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

Apply the product rule to $2 x$ .

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

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

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

Multiply $4$ by $3$ .

$g ({(2 x + 1)}^{3} - 4) = \frac{\sqrt[3]{8 x^{3} + 12 x^{2} \cdot 1 + 3 (2 x) \cdot 1^{2} + 1^{3} - 4 + 4} - 1}{2}$

Multiply $12$ by $1$ .

$g ({(2 x + 1)}^{3} - 4) = \frac{\sqrt[3]{8 x^{3} + 12 x^{2} + 3 (2 x) \cdot 1^{2} + 1^{3} - 4 + 4} - 1}{2}$

Multiply $2$ by $3$ .

$g ({(2 x + 1)}^{3} - 4) = \frac{\sqrt[3]{8 x^{3} + 12 x^{2} + 6 x \cdot 1^{2} + 1^{3} - 4 + 4} - 1}{2}$

One to any power is one.

$g ({(2 x + 1)}^{3} - 4) = \frac{\sqrt[3]{8 x^{3} + 12 x^{2} + 6 x \cdot 1 + 1^{3} - 4 + 4} - 1}{2}$

Multiply $6$ by $1$ .

$g ({(2 x + 1)}^{3} - 4) = \frac{\sqrt[3]{8 x^{3} + 12 x^{2} + 6 x + 1^{3} - 4 + 4} - 1}{2}$

One to any power is one.

$g ({(2 x + 1)}^{3} - 4) = \frac{\sqrt[3]{8 x^{3} + 12 x^{2} + 6 x + 1 - 4 + 4} - 1}{2}$

Subtract $4$ from $1$ .

$g ({(2 x + 1)}^{3} - 4) = \frac{\sqrt[3]{8 x^{3} + 12 x^{2} + 6 x - 3 + 4} - 1}{2}$

Add $- 3$ and $4$ .

$g ({(2 x + 1)}^{3} - 4) = \frac{\sqrt[3]{8 x^{3} + 12 x^{2} + 6 x + 1} - 1}{2}$

Rewrite $8 x^{3} + 12 x^{2} + 6 x + 1$ in a factored form.

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Regroup terms.

$g ({(2 x + 1)}^{3} - 4) = \frac{\sqrt[3]{8 x^{3} + 1 + 12 x^{2} + 6 x} - 1}{2}$

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

$g ({(2 x + 1)}^{3} - 4) = \frac{\sqrt[3]{{(2 x)}^{3} + 1 + 12 x^{2} + 6 x} - 1}{2}$

Rewrite $1$ as $1^{3}$ .

$g ({(2 x + 1)}^{3} - 4) = \frac{\sqrt[3]{{(2 x)}^{3} + 1^{3} + 12 x^{2} + 6 x} - 1}{2}$

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = 2 x$ and $b = 1$ .

$g ({(2 x + 1)}^{3} - 4) = \frac{\sqrt[3]{(2 x + 1) ({(2 x)}^{2} - 2 x \cdot 1 + 1^{2}) + 12 x^{2} + 6 x} - 1}{2}$

Simplify.

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Apply the product rule to $2 x$ .

$g ({(2 x + 1)}^{3} - 4) = \frac{\sqrt[3]{(2 x + 1) (2^{2} x^{2} - 2 x \cdot 1 + 1^{2}) + 12 x^{2} + 6 x} - 1}{2}$

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

$g ({(2 x + 1)}^{3} - 4) = \frac{\sqrt[3]{(2 x + 1) (4 x^{2} - 2 x \cdot 1 + 1^{2}) + 12 x^{2} + 6 x} - 1}{2}$

Multiply $2$ by $- 1$ .

$g ({(2 x + 1)}^{3} - 4) = \frac{\sqrt[3]{(2 x + 1) (4 x^{2} - 2 x \cdot 1 + 1^{2}) + 12 x^{2} + 6 x} - 1}{2}$

Multiply $- 2$ by $1$ .

$g ({(2 x + 1)}^{3} - 4) = \frac{\sqrt[3]{(2 x + 1) (4 x^{2} - 2 x + 1^{2}) + 12 x^{2} + 6 x} - 1}{2}$

One to any power is one.

$g ({(2 x + 1)}^{3} - 4) = \frac{\sqrt[3]{(2 x + 1) (4 x^{2} - 2 x + 1) + 12 x^{2} + 6 x} - 1}{2}$

Factor $6 x$ out of $12 x^{2} + 6 x$ .

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Factor $6 x$ out of $12 x^{2}$ .

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

Factor $6 x$ out of $6 x$ .

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

Factor $6 x$ out of $6 x (2 x) + 6 x (1)$ .

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

Factor $2 x + 1$ out of $(2 x + 1) (4 x^{2} - 2 x + 1) + 6 x (2 x + 1)$ .

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Factor $2 x + 1$ out of $6 x (2 x + 1)$ .

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

Factor $2 x + 1$ out of $(2 x + 1) (4 x^{2} - 2 x + 1) + (2 x + 1) (6 x)$ .

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

Add $- 2 x$ and $6 x$ .

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

Factor using the perfect square rule.

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Rewrite $4 x^{2}$ as ${(2 x)}^{2}$ .

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

Rewrite $1$ as $1^{2}$ .

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

Check the middle term by multiplying $2 a b$ and compare this result with the middle term in the original expression.

$2 a b = 2 \cdot (2 x) \cdot 1$

Simplify.

$2 a b = 4 x$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 2 x$ and $b = 1$ .

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

Multiply $2 x + 1$ by ${(2 x + 1)}^{2}$ by adding the exponents.

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Multiply $2 x + 1$ by ${(2 x + 1)}^{2}$ .

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Raise $2 x + 1$ to the power of $1$ .

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

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

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

Add $1$ and $2$ .

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

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

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

Subtract $1$ from $1$ .

$g ({(2 x + 1)}^{3} - 4) = \frac{2 x + 0}{2}$

Add $2 x$ and $0$ .

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

Cancel the common factor of $2$ .

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Cancel the common factor.

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

Divide $x$ by $1$ .

$g ({(2 x + 1)}^{3} - 4) = x$

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

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

Find the Inverse f(x)=(2x+1)^3-4

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