Extreme value theorem and critical points
Extreme value theorem
So we'll now think about the extreme value theorem. Which we'll see is a bit of common sense. But in all of these theorems it's always fun to think about the edge cases. Why is it laid out the way it is? And that might give us a little bit more intuition about it. So the extreme value theorem says if we have some function that is continuous over a closed interval, let's say the closed interval from a to b. And when we say a closed interval, that means we include the end points a and b. That is we have these brackets here instead of parentheses. Then there will be an absolute maximum value for f and an absolute minimum value for f. So then that means there exists-- this is the logical symbol for there exists-- there exists an absolute maximum value of f over interval and absolute minimum value of f over the interval. So let's think about that a little bit. And this probably is pretty intuitive for you. You're probably saying, well why did they even have to write a theorem here? And why do we even have to have this continuity there? And we'll see in a second why the continuity actually matters. So this is my x-axis, that's my y-axis. And let's draw the interval. So the interval is from a to b. So let's say this is a and this is b right over here. Let's say that this right over here is f of a. So that is f of a. And let's say this right over here is f of b. So this value right over here is f of b. And let's say the function does something like this. Let's say the function does something like this over the interval. And I'm just drawing something somewhat arbitrary right over here. So I've drawn a continuous function. I really didn't have to pick up my pen as I drew this right over here. And so you can see at least the way this continuous function that I've drawn, it's clear that there's an absolute maximum and absolute minimum point over this interval. The absolute minimum point, well it seems like we hit it right over here, when x is, let's say this is x is c. And this is f of c right over there. And it looks like we had our absolute maximum point over the interval right over there when x is, let's say this is x is equal to d. And this right over here is f of d. So another way to say this statement right over here if f is continuous over the interval, we could say there exists a c and d that are in the interval. So they're members of the set that are in the interval such that-- and I'm just using the logical notation here. Such that f c is less than or equal to f of x, which is less than or equal to f of d for all x in the interval. Just like that. So in this case you're saying, look, we hit our minimum value when x is equal to c. That's that right over here. Our maximum value when x is equal to d. And for all the other Xs in the interval we are between those two values. Now one thing, we could draw other continuous functions. And once again I'm not doing a proof of the extreme value theorem. But just to make you familiar with it and why it's stated the way it is. And you could draw a bunch of functions here that are continuous over this closed interval. Here our maximum point happens right when we hit b. And our minimum point happens at a. For a flat function we could put any point as a maximum or the minimum point. And we'll see that this would actually be true. But let's dig a little bit deeper as to why f needs to be continuous, and why this needs to be a closed interval. So first let's think about why does f need to be continuous? Well I can easily construct a function that is not continuous over a closed interval where it is hard to articulate a minimum or a maximum point. And I encourage you, actually pause this video and try to construct that function on your own. Try to construct a non-continuous function over a closed interval where it would be very difficult or you can't really pick out an absolute minimum or an absolute maximum value over that interval. Well let's see, let me draw a graph here. So let's say that this right over here is my interval. Let's say that's a, that's b. Let's say our function did something like this. Let's say our function did something right where you would have expected to have a maximum value let's say the function is not defined. And right where you would have expected to have a minimum value, the function is not defined. And so right over here you could say, well look, the function is clearly approaching, as x approaches this value right over here, the function is clearly approaching this limit. But that limit can't be the maxima because the function never gets to that. So you could say, well let's a little closer here. Maybe this number right over here is 5. So you could say, maybe the maximum is 4.9. Then you could get your x even closer to this value and make your y be 4.99, or 4.999. You could keep adding another 9. So there is no maximum value. Similarly here, on the minimum. Let me draw it a little bit so it looks more like a minimum. There is-- you can get closer and closer to it, but there's no minimum. Let's say that this value right over here is 1. So you could get to 1.1, or 1.01, or 1.0001. And so you could keep drawing some 0s between the two 1s but there's no absolute minimum value there. Now let's think about why it being a closed interval matters. Why you have to include your endpoints as kind of candidates for your maximum and minimum values over the interval. Well let's imagine that it was an open interval. Let's imagine open interval. And sometimes, if we want to be particular, we could make this is the closed interval right of here in brackets. And if we wanted to do an open interval right over here, that's a and that's b. And let's just pick very simple function, let's say a function like this. So right over here, if a were in our interval, it looks like we hit our minimum value at a. f of a would have been our minimum value. And f of b looks like it would have been our maximum value. But we're not including a and b in the interval. This is an open interval so you can keep getting closer, and closer, and closer, to b and keep getting higher, and higher, and higher values without ever quite getting to be. Because once again we're not including the point b. Similarly, you could get closer, and closer, and closer, to a and get smaller, and smaller values. But a is not included in your set under consideration. So f of a cannot be your minimum value. So that on one level, it's kind of a very intuitive, almost obvious theorem. But on the other hand, it is nice to know why they had to say continuous and why they had to say a closed interval like this.