 In this video, we wanna present the solution to question number 14 for practice exam number one for math 12-10. In this question, we're asked to compute the domain of the function f of x, which equals the square root of two x plus one all over x square minus one. And we need to report our answer in interval notation. Well, when it comes to these functions, when we're trying to find the domain, the domain convention tells us that we're gonna accept, unless told otherwise, which we weren't, we're gonna accept all real numbers which make the expression into a well-defined And there's basically three problems we have to look for. We have to look for taking square roots of negatives. There is a square root in our function, so we do have to watch out for that. We need to look out for division by zero, which there is a denominator here, so we need to look out, watch out for that. We also need to look out for logs of zero or logs of a negative. There's no logarithms in this function, so we don't have to worry about that one so much. But we have to worry about taking square root of a negative and dividing by zero, all right? And so those are the only real domain issues we have to worry about when it comes to these calculus functions here. So when you look at the square root, right, we need to make sure, when you're examining the square root, the radicand, that is the two x plus one, that needs to be non-negative. It can, being equal to zero is okay. So we have to solve the inequality. Two x plus one is greater than equal to zero. We subtract one from both sides. We get two x is greater than equal to negative one. We divide by two. We get x needs to be greater than equal to negative one half. So that's what the domain of the numerator requires. But we also have to make sure the denominator never goes to zero. So when we solve the equation x squared minus one equals zero, we get x squared equals one. We take the square root and see x equals plus or minus one. These are actually forbidden values. X should not equal these. And so we need to put this stuff together. It's like, okay, x needs to be greater than or equal to negative one. Excuse me, greater than equal to negative one half. Negative one is less than that. But also not equal to plus or minus one. So when you put those together, when you look at the intersection of these domains, we're gonna get negative one half to one union one to infinity. This would be the domain of our function. We can only accept things greater than negative one half. So we get this. We have to intersect that with the domain here. The domain here was negative infinity to negative one union, negative one to one union one to infinity. You'll notice that this domain makes no mention of negative one, like jumping over it because negative one's less than negative one half. And so this one already told us we weren't ever gonna be close to negative one in the domain.