 there's one important remark I want to make here. A critical point must be in the domain of the function. Let me give you an example to help illustrate that. Consider the function f of x equals x over x minus 4. By the usual quotient rule, one can determine that the derivative of f, which you can see right here, f prime would be negative 4 over x minus 4. So where is this function equal to 0? Well, because it's a fraction, the only way it could be equal to 0 is if the numerator is 0, which that can't equal 0 because it's negative 4. The other concern is, well, what makes the denominator go to 0? That would make the derivative undefined. We see that the critical number, there's a single critical number, which you're going to get, is going to be x equals 4. But you'll notice that if I look at f of 4, not f prime at 4, if you look at f of 4, you plug it into here, you end up with 4 over 0, which of course is undefined. It does not exist. The original function is not defined at x equals 4. Technically speaking, x equals 4 is not a critical number because it's not a point in the domain of the original function. But if you were to draw a sign chart going on here at x equals 4, you'll notice that if you pick a number larger than 4, let's take, for example, say like 10 or something. If you plugged in 10 here, 10 minus 4 squared is 6, squared is 36. You get negative 4 over 36. That's going to be a negative number. So we can plug that in right there. If we pick something less than 4, let's take like, for example, 0. 0 minus 4 is negative 4 squared. It's positive 16. You get negative 4 over positive 16. That's again going to be negative. So you look at that, it's like, oh, there's no sign change. What does that mean about the graph? Well, that just means, so the derivative didn't switch its signs. The function f would be if it's decreasing, and then it's decreasing again, okay? If we were just to sketch the picture of f of x real quick here, some things to note is it does have a vertical asymptote at x equals 4. It has a horizontal asymptote at x equals 1. And if we fill in the rest of the picture, we'll get something like this, which is kind of a curious thing to note here. This function is always decreasing, but there are parts of the graph that are higher up than other ones later on. And that's because it's always decreasing on its interval. Because you have this vertical asymptote at x equals 4, it can actually sort of like go through a teleport. You know, I think this happens in movies and video games sometimes where if you fall through a portal, then you can come up from above, like in Thor Ragnarok, Dr. Strange says that the Loki falls for 30 minutes because of this type of phenomenon, right? He's always falling, but because of the wraparound feature, you can keep on going around in a loop. So the reason I bring up this example here is that x equals 4 is not a critical number per se, but it is important to still consider what happens at x equals 4 because the function, it doesn't have a local maximum at x equals 4. It has a vertical asymptote. There's no extremum there. But we should still mention that the intervals of increase and decreasing occur from negative infinity to 4, union 4 to infinity, because it's not increasing at 4. And we can do examples like this all over the place, like another quick example. If we take y equals 1 over x squared, for example, the graph looks something like this. Same basic idea. We don't have, we don't have a local maximum at x equals 0 because it's undefined there. The derivative of this function, of course, would be negative 2 over x cubed, for which if we were to draw the sign chart, we get something like your critical number 0. It's going to look like you, when you're to the right of 0, you're going to get positive, excuse me, negative. And when you're to the left of 0, you're going to get positive. So it goes from increasing to decreasing, but that doesn't mean you have a local maximum because you're outside the domain. So you have to consider these numbers outside the domain, but if you're outside the domain of the original function, then you cannot be a critical number, but the function could switch its monotonicity at that point.