Determining whether values are in domain of function

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, then we are going to get a legitimate output f of x. But if for whatever reason f isn't defined at x, or if it gets some kind of undefined state, well then it is, then x would not be in the domain.

So let's try these different values. Is x equal to negative 5 in the domain of f? Let's see what happens if we try to evaluate f of negative five. Well then we 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 3, which is equal to in our numerator we get 0 and our denominator we get negative 8.

Now at first you see the 0, you might get a little bit worried, but it's just a 0 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 and see if you can figure that out. Well f of 0 is going to be equal to, in our numerator we have 0 plus 5. In our denominator we have 0 minus 3. Well, that's just going to get us 5 in the numerator and negative 3 in the denominator. This would just be negative five thirds, 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, try to figure that out. Well, I'll do that up here. f of 3 is going to be equal to what? 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 3 plus 5, and in the denominator we get 3 minus 3.

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 3 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.

All right, so first of all, when x equals negative 3, do we get a legitimate g of x? So let's see g of negative 3. If we try to evaluate, that's going to be the square root of 3 times negative 3, which is equal to the square root of negative 9. Well, with just a principal square root like this, we don't know how to evaluate this, so this is not in the domain.

What about when x equals 0? Well, g of 0 is going to be equal to the square root of 3 times 0, which is equal to the square root of 0, which is equal to 0. So that gave us a legitimate result, so that is in the domain.

And what about g of 2, or x equals 2? What does that give us a legitimate g of 2? Well, g of 2 is going to be equal to the square root of 3 times 2, which is equal to the square root of 6, which is a legitimate output. So x equals 2 is in the domain.

Let's do one last example. So we're told this is 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 1. What's that going to be equal to? Negative 1, every place we see an x, we're going to replace it with a negative 1 minus 5 squared. Well, this is going to be equal to negative 6 squared, negative 6 squared, which is equal to positive 36, which is a very legitimate output. So this is definitely in the domain.

What about 5? So h of 5 is going to be equal to 5 minus 5 squared. Now you might be getting worried because you're seeing a 0 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 5 squared, which is equal to 5 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.