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

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

Set the radicand in $\sqrt{x + 1}$ greater than or equal to $0$ to find where the expression is defined.

$x + 1 \geq 0$

Subtract $1$ from both sides of the inequality.

$x \geq - 1$

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

Interval Notation:

$[- 1, \infty)$

Set-Builder Notation:

${x | x \geq - 1}$

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: $[- 1, \infty), {x | x \geq - 1}$

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