Find the domain of 𝑓 of 𝑥 equals the square root of 𝑥 minus one over the square root of nine minus 𝑥 minus the square root of 𝑥 minus three.
Remember, the domain of a function is the set of possible values that we can input into that function. In this case, we have a function that involves two potential issues. Firstly, we want to make sure that the denominator of the function is never equal to zero. So when we’re finding the domain of the function, we’ll disregard any values of 𝑥 that cause this to happen. Next, when we’re working with square root functions, we need to make sure that the expressions inside the square root are nonnegative. In other words, 𝑥 minus one must be greater than or equal to zero as must nine minus 𝑥 and 𝑥 minus three.
So because it’s a little simpler, we’ll just begin by looking at the expressions inside each of our square root signs, that is, 𝑥 minus one, nine minus 𝑥, and 𝑥 minus three. We’ll solve each of these inequalities in turn. Our first, we’ll add one to both sides. That gives us 𝑥 is greater than or equal to one. In our second, we’ll add 𝑥 to both sides. So nine is greater than or equal to 𝑥 or alternatively 𝑥 is less than or equal to nine. Finally, we’ll add three to both sides of our third equation. That gives us 𝑥 is greater than or equal to three. There is some overlap between these three regions. But we’ll consider that when we’ve looked at the denominator of our fraction.
We know that we want to make sure the denominator of this fraction is not equal to zero. So what we’ll do is we’ll solve by setting equal to zero and find the values of 𝑥 that we’re going to need to disregard. That is, the square root of nine minus 𝑥 minus the square root of 𝑥 minus three equals zero. Let’s begin by adding the square root of 𝑥 minus three to both sides. That gives us the square root of nine minus 𝑥 equals the square root of 𝑥 minus three. Now, we approach the next step with great care. We’re going to square both sides of this equation. We would usually need to check our solutions at the end and ensure we haven’t created any extra. However, since we have square roots on both sides, both sides must in fact be positive. So we’ll never get something strange like negative one equals one, which would be squared to give one equals one.
So squaring both sides, and we get nine minus 𝑥 equals 𝑥 minus three. Then, we add 𝑥 and three to both sides, and we get 12 equals two 𝑥, which when we divide through by two, we see that 𝑥 is equal to six. And so we have all of our criteria. We know 𝑥 must be greater than one, less than nine, greater than three, and not equal to six. We might spot firstly that when 𝑥 is greater than or equal to three, it absolutely also has to be greater than one. Since our domain is the intersection of all these possible criteria, we disregard this and we just stick with 𝑥 is greater than or equal to three.
So combining all of these, we see that 𝑥 is greater than or equal to three, less than or equal to nine, but it’s also not equal to six. Let’s rewrite this using set notation. When we do, we find that the domain of our function 𝑓 of 𝑥 is the set of values in the closed interval from three to nine minus the set containing the element six.