The domain is the set of values of x that can be substituted into a functional equation. Negative numbers cannot be permitted under a square root radical, because you can't multiply a negative by itself, which is a negative) and get a negative, because you'll always get a positive,So what is under that radical cannot be negative. So we cannot substitute any number for x and get an answer that will make a negative number. So we set it zero, so it won't be negative. We can divide through both sides by 3 without reversing the inequality because we are dividing through by a positive number, not by a negative number. (If we had divided both sides by a negative number we would have had to reverse the symbol of imequality). Now factor: The critical values are found by setting the left side equal to zero. 2-x=0 has solution x=2 2+x=0 has solution x=-2 Therefore these are the critical numbers. We mark them on a number line with open circles "o" (but we may close them later). -------------o---------------o--------- -5 -4 -3 -2 -1 0 +1 +2 +3 +4 We pick a test value left of -2, say -3 since it's the easiest. We substitute -3 in to see if it's true or false. That's false so we do not shade that part of the number line, so we still just have this: -------------o---------------o--------- -5 -4 -3 -2 -1 0 1 2 3 4 Next we pick a test value between -2 and 2, say 0 since it's the easiest. We substitute 0 in to see if it's true or false. That's true so we shade that part of the number line, and now we have this: -------------o===============o--------- -5 -4 -3 -2 -1 0 +1 +2 +3 +4 Next we pick a test value right of +2, say +3 since it's the easiest. We substitute +3 in to see if it's true or false. That's false so we do not shade that part of the number line, so we still just have this: -------------o===============o--------- -5 -4 -3 -2 -1 0 +1 +2 +3 +4 Next we test the critical points themselves, -2 and +2 Testing -2 This is true so we darken the circle at -2 -------------@===============o--------- -5 -4 -3 -2 -1 0 +1 +2 +3 +4 Testing +2 This is true too so we darken the circle at +2 -------------@===============@--------- -5 -4 -3 -2 -1 0 +1 +2 +3 +4 In interval notation this is [-2, 2] Edwin