 In this video, we provide the solution to question number 15 for practice exam 1 for math 1050. We're given the function f of x equals the square root of 2x plus 1 all over x squared minus 1. We're asked to compute the domain of this function. We need to report it in interval notation. So what are things that can go wrong? Because by the domain convention, we accept the domain to be as large as possible. We accept all real numbers for x, except for those that turn the expression into some non-real number. So we've been trained to watch out for a couple things. So first of all, square roots are problematic. If we take the square root of a negative number, that's not a real number. It's what we call an imaginary number. And so we have to make sure that the radicand, that is the expression inside of the square root, is not negative. So we have to consider the inequality 2x plus 1 is greater than or equal to 0. Equal to 0 is okay. If you take the square root of 0, that's just 0. Nothing non-real about that. So we need to solve this inequality. We can minus 1 from both sides. We get 2x is greater than or equal to negative 1. Divide both sides by 2. We get x is greater than or equal to negative 1 half. So as long as x is greater than negative 1 half, then the numerator, I should say the radicand will be non-negative, I should say. And then the square root of that would then be a real number. So as long as the input is greater than or equal to negative 1, no problems in the numerator. Well, what about the denominator, the x squared right here? One of the problems with denominators is if you divide by 0, right, that gives you a non-real number. So we have to make sure that x squared, excuse me, there, x squared minus 1 is not equal to 0. Well, the easiest way to do that is to solve the equation for 0 and go from there, for which x squared is equal to 1. You can take the square root of both sides. You have to remember when you take the square root, there's actually two square roots of positive and negative 1. And so you end up with plus or minus 1, or specifically 1 and negative 1. And these are the values that x should not be. x cannot be 1 or negative 1. So we have to put these together. Now, notice that negative 1 is less than negative 1 half. So this one already ruled that as right as a possibility. x has to be greater than or equal to negative 1 half, excuse me. So negative 1 is not even in the run at all. But 1 is greater than negative 1. So considering those together, we take the intersection here. We need that x is greater than and or equal to negative 1 half. So we would write something like this bracket, negative 1 half, because negative 1 half is acceptable. Go up to 1. Then we're going to put a parenthesis there, because 1 is not included. If you plug in x equals 1, that makes the denominator go to zero. That's not a real number. We take the union, we jump over to 1 again. So we jumped 1. The parenthesis means that 1 is not part of the domain here. But we include everything up to 1 from the left-hand side and from the right-hand side. And now once you get past 1, there's no more restriction on the domain. So we're going to put infinity right here. We always put a parenthesis on infinity, because infinity is not a real number, hence it's not included. So notice here that you need a parenthesis on infinity. You need a parenthesis on the 1, because it's not included, but you do need a bracket on the negative 1 half, because that is part of the domain. And therefore, the correct answer would be negative 1 half to 1, union 1 to infinity, where 1 is omitted and negative 1 half is included.