Find the Domain f(x) = square root of x+6 if -6<=x<=-2; 0-x if x>-2

$f (x) = {\begin{cases} \sqrt{x + 6} & - 6 \leq x \leq - 2 \\ 0 - x & x > - 2 \end{cases}$

Step 1

Find the domain for $\sqrt{x + 6}$ .

$[- 6, \infty)$

Step 2

Compare the domain of $\sqrt{x + 6}$ with the restriction $- 6 \leq x \leq - 2$ to see where this piece of the function is defined.

$[- 6, - 2]$

Step 3

Since the domain of $0 - x$ is all real numbers, the domain of this piece of the function is its restriction, $x > - 2$ .

$(- 2, \infty)$

Step 4

Find the domain by taking the union of all of the intervals where the function is defined.

$[- 6, \infty)$

Step 5