Find the Domain and Range f^-1(x) = cube root of x+2

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

Step 1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 2

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${y | y \in ℝ}$

Step 3

Determine the domain and range.

Domain: $(- \infty, \infty), {x | x \in ℝ}$

Range: $(- \infty, \infty), {y | y \in ℝ}$

Step 4

Find the Domain and Range f^-1(x) = cube root of x+2

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