Find the Domain f^-1(x) = cube root of x+1

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

Move $\sqrt[3]{x + 1}$ to the left side of the equation by subtracting it from both sides.

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

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

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Simplify each term.

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Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Combine $\frac{1}{f}$ and $x$ .

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

To write $- \sqrt[3]{x + 1}$ as a fraction with a common denominator, multiply by $\frac{f}{f}$ .

$\frac{x}{f} - \sqrt[3]{x + 1} \cdot \frac{f}{f} = 0$

Combine $- \sqrt[3]{x + 1}$ and $\frac{f}{f}$ .

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

Combine the numerators over the common denominator.

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

Set the denominator in $\frac{x - \sqrt[3]{x + 1} f}{f}$ equal to $0$ to find where the expression is undefined.

$f = 0$

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, 0) \cup (0, \infty)$

Set-Builder Notation:

${x | x \neq 0}$

Find the Domain f^-1(x) = cube root of x+1

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