Find the Inverse f(x) = cube root of x-1+4

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

Step 1

Write $f (x) = \sqrt[3]{x - 1 + 4}$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

Rewrite the equation as $\sqrt[3]{y - 1 + 4} = x$ .

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

To remove the radical on the left side of the equation, cube both sides of the equation.

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

Simplify each side of the equation.

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

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

Simplify the left side.

Simplify ${({(y - 1 + 4)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(y - 1 + 4)}^{\frac{1}{3}})}^{3}$ .

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

${(y - 1 + 4)}^{\frac{1}{3} \cdot 3} = x^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(y - 1 + 4)}^{\frac{1}{3} \cdot 3} = x^{3}$

Rewrite the expression.

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

Add $- 1$ and $4$ .

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

Simplify.

$y + 3 = x^{3}$

Subtract $3$ from both sides of the equation.

$y = x^{3} - 3$

Step 4

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

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

Step 5

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

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

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

Simplify each term.

Add $- 1$ and $4$ .

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

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

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

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

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Combine the opposite terms in $x + 3 - 3$ .

Subtract $3$ from $3$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Subtract $1$ from $- 3$ .

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

Add $- 4$ and $4$ .

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

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

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

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

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

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

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

Find the Inverse f(x) = cube root of x-1+4

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