Determining the domain of a function
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- [Instructor] We're asked to determine for each x-value whether it is in the domain of f or not. And they have our definition of f of x up here. So pause this video and see if you can work through this before we do it together. All right, so just as a bit of a review, if x is in the domain of our function, that means that if we input our x into our function, when we are going to get a legitimate output f of x. But if for whatever reason f isn't defined at x or it gets some kind of undefined state, well, then x would not be in the domain. So let's try these different values. Is x equal to negative five in the domain of f? Well, let's see what happens if we try to evaluate f of negative five. Well, then in the numerator, we get negative five plus five. Every place where we see an x, we replace it with a negative five. So it's negative five plus five, over negative five minus three, which is equal to in our numerator, we get zero, and in our denominator, we get negative eight. Now, at first you see the zero, and you might get a little bit worried, but it's just a zero in the numerator, so this whole thing just evaluates to a zero, which is a completely legitimate output. So x equals negative five is in the domain. What about x equals zero? Is that in the domain? Pause the video. See if you can figure that out. Well, f of zero is going to be equal to in our numerator, we have zero plus five, and in our denominator, we have zero minus three. Well, that's just going to get us five in the numerator and negative three in the denominator. This would just be negative 5/3. But this is a completely legitimate output. So the function is defined at x equals zero, so it's in the domain for sure. Now what about x equals three? Pause the video and try to figure that out. Well, I'll do that up here. f of three is going to be equal to what? And you might already see some warning signs as to what's going to happen here in the denominator, but I'll just evaluate the whole thing. In the numerator, we get three plus five. In the denominator, we get three minus three. So this is going to be equal to eight over zero. Now what is eight divided by zero? Well, we don't know. This is one of those fascinating things in mathematics. We haven't defined what happens when something is divided by zero. So three is not in the domain. The function is not defined there, not in domain. Let's do another example. Determine for each x-value whether it is in the domain of g or not. So pause this video and try to work through all three of these. So first of all, when x equals negative three, do we get a legitimate g of x? So let's see. g of negative three, if we try to evaluate this, that's going to be the square root of three times negative three, which is equal to the square root of negative nine. Well, with just a principle square root like this, we don't know how to evaluate this. So this is not in the domain. What about when x equals zero? Well, g of zero is going to be equal to the square root of three times zero, which is equal to the square root of zero, which is equal to zero, so that gave us a legitimate result. So that is in the domain. Now what about g of two, or x equals two? Does that give us a legitimate g of two? Well, g of two is going to be equal to the square root of three times two, which is equal to the square root of six which is a legitimate output. So x equals two is in the domain. Let's do one last example. So we're told, this h of x right over here, and once again, we have to figure out whether these x-values are in the domain or not. Pause this video and see if you can work through that. All right, well, let's just first think about h of negative one. What's that going to be equal to? Negative one, every place we see an x, we're going to replace it with a negative one, minus five, squared. Well, this is going to be equal to negative six squared, negative six squared, which is equal to positive 36, which is a very legitimate output, and so this is definitely in the domain. What about five? So h of five is going to be equal to five minus five squared. Now you might be getting worried 'cause you're seeing a zero here, but it's not like we're trying to divide by zero. We're just squaring zero, which is completely legitimate. So zero squared is just a zero, and so h of five is very much defined. So this is in the domain. Now what about h of 10? Well, h of 10 is going to be equal to 10 minus five squared, which is equal to five squared, which is equal to 25. Once again, it's a very legitimate output. So the function is definitely defined for x equals 10, and we're done.