Find the Inverse (x-2)^3-1

${(x - 2)}^{3} - 1$

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

Add $1$ to both sides of the equation.

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

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

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

Add $2$ to both sides of the equation.

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

Step 3

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

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

Step 4

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

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

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

Combine the opposite terms in $\sqrt[3]{({(x - 2)}^{3} - 1) + 1} + 2$ .

Add $- 1$ and $1$ .

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

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

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

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

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

Combine the opposite terms in $x - 2 + 2$ .

Add $- 2$ and $2$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Combine the opposite terms in ${((\sqrt[3]{x + 1} + 2) - 2)}^{3} - 1$ .

Subtract $2$ from $2$ .

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

Add $\sqrt[3]{x + 1}$ and $0$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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

Subtract $1$ from $1$ .

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

Add $x$ and $0$ .

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

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

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

Find the Inverse (x-2)^3-1

Related Questions

Support LCMGCF.com by disabling your adblocker.