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

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

Step 1

Write $f (x) = {(2 x + 1)}^{3} - 4$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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

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

Subtract $1$ from both sides of the equation.

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

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Combine the numerators over the common denominator.

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

Step 4

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

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

Step 5

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

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 $f^{- 1} ({(2 x + 1)}^{3} - 4)$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Simplify the numerator.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{{(2 x + 1)}^{3} - 4 + 4}$ as ${({(2 x + 1)}^{3} - 4 + 4)}^{\frac{1}{3}}$ .

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

Combine the opposite terms in ${(2 x + 1)}^{3} - 4 + 4$ .

Add $- 4$ and $4$ .

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

Add ${(2 x + 1)}^{3}$ and $0$ .

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

Multiply the exponents in ${({(2 x + 1)}^{3})}^{\frac{1}{3}}$ .

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Subtract $1$ from $1$ .

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

Add $2 x$ and $0$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

Evaluate $f (\frac{\sqrt[3]{x + 4} - 1}{2})$ by substituting in the value of $f^{- 1}$ into $f$ .

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

Simplify each term.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Combine the opposite terms in $\sqrt[3]{x + 4} - 1 + 1$ .

Add $- 1$ and $1$ .

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

Add $\sqrt[3]{x + 4}$ and $0$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Combine the opposite terms in $x + 4 - 4$ .

Subtract $4$ from $4$ .

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

Add $x$ and $0$ .

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

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