Find the Domain f^-1(x) = square root of 14+x

$f^{- 1} (x) = \sqrt{14 + x}$

Step 1

Set the radicand in $\sqrt{14 + x}$ greater than or equal to $0$ to find where the expression is defined.

$14 + x \geq 0$

Step 2

Subtract $14$ from both sides of the inequality.

$x \geq - 14$

Step 3

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

Interval Notation:

$[- 14, \infty)$

Set-Builder Notation:

${x | x \geq - 14}$

Step 4