Find the Domain and Range f^-1(x) = square root of x+3

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

Set the radicand in $\sqrt{x + 3}$ greater than or equal to $0$ to find where the expression is defined.

$x + 3 \geq 0$

Subtract $3$ from both sides of the inequality.

$x \geq - 3$

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

Interval Notation:

$[- 3, \infty)$

Set-Builder Notation:

${x | x \geq - 3}$

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

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${y | y \geq 0}$

Determine the domain and range.

Domain: $[- 3, \infty), {x | x \geq - 3}$

Range: $[0, \infty), {y | y \geq 0}$