AP®︎/College Calculus AB
Course: AP®︎/College Calculus AB > Unit 5Lesson 2: Extreme value theorem, global versus local extrema, and critical points
Extreme value theorem
The Extreme value theorem states that if a function is continuous on a closed interval [a,b], then the function must have a maximum and a minimum on the interval. This makes sense: when a function is continuous you can draw its graph without lifting the pencil, so you must hit a high point and a low point on that interval. Created by Sal Khan.
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- He is using continuous, as opposed to discreet, correct? I still don't understand the continuous aspect of the statement and why it is important. The closed aspect makes sense, but not he continuous part.(12 votes)
- The fact that a graph is continuous makes a big difference when doing calculus. Let's say you had the function x^2. A, you know that function is something you can draw without lifting your pen, and b, you know that function's smooth and doesn't have any kinks or pointy edges in it. So that function moves into this elite class of functions that we call continuous and differentiable; continuous because you can draw it without stopping and differentiable because you can take the derivative at every point on the curve. I'm not sure how far along you are in calculus, but as you progress, you'll learn to find the slope of a curve, just like how you learned to find the slope of a line. However, you can't find the slope if the graph is shaped funny, like a heart or a star, or happens to have a jump in it, like a set of steps. The heart and star may be continuous, but they are not differentiable, because they have a pointy edge at which you cannot calculate the slope. The steps are neither continuous nor differentiable, because you obviously can't draw them without lifting your pen, and they have breaks in them, so you can't find the slopes at every point. Hope this helps you understand it a li'l bit better! ;)(73 votes)
- Is there a series/practice problem set on math logic and reading it? I'd like to get used to reading backwards E's and "members of" signs through some good ol' KA practice.(19 votes)
- This (free) course(or the accompanying book) should get you where you want to be:
- So if the intervals is not CLOSED meaning the endpoints are not included, there are NO ABSOLUTE MAX OR MIN?(12 votes)
- Example of why the interval needs to be closed for the theorem to be conclusive: does the identity function have a minimum or maximum on the open interval
(0, 1)? That is, if we let
ƒ(x) = xfor
xin the open interval
(0, 1), does
ƒhave a maximum or minimum?
(Answer: no, neither.)
However, if we define
ƒon the closed interval
[0, 1], then
ƒhas a minimum at
0and a maximum at
However, some functions do have maxima and / or minima on open intervals. For instance, let
ƒ(x) = 1 - x²for
xin the open interval
(-1, 1). Then
ƒhas a maximum at
ƒhas no minimum.
In short: the extreme value theorem only applies when the function is continuous on an interval that is both closed and bounded (well, strictly speaking, the function need not be defined on an interval, its domain need only be compact.)(17 votes)
- What about a function with a slope of 0, like f(x)=3? There are no min or max points, unless all points are simultaneously min and max points. Doesn't that definition break down?(13 votes)
- When dealing with constant functions, every point is both a maximum and a minimum.(13 votes)
- 6:00, is it a bit convenient for us when the discontinuity points (holes) coincidentally overlap the extreme points (maxima & minima)? If those two points were filled in AND another hole was created somewhere between, would the function still have the extreme points?(13 votes)
- The extreme value theorem states that a function that is continuous over a closed interval is guaranteed to have a maximum or minimum value over a closed interval. The same cannot be said for functions that do not satisfy these conditions, although it is possible to find or construct such functions that have a maximum and minimum over a closed interval. In the beginning, Sal drew an arbitrary continuous function over a closed interval to visually confirm that a continuous function does indeed have a maximum point and minimum point inside a closed interval. He then proceeded to show why the conditions (that f must be continuous over a closed interval) are necessary in order for the conclusion(that f have a max and min) to follow. He purposely choose to overlap the discontinuity points with the relative extrema in order to show an example of a discontinuous function that does not technically have a max or min. If Sal placed the discontinuous points elsewhere on the graph, then the function would have maximum and minimum points, since f would be defined at those points.
Similarly, Sal chose to depict a linear function over a half interval in order to show an example of a continuous function that does not have relative extrema over such an interval. Again, the moral of the story behind the Extreme Value Theorem is that while a function continuous over a closed interval must have a maximum and a minimum value, a function that does not fit this description may or may not reach its peak values.
I hope this helps.(14 votes)
- Wouldn't this also hold for [-∞, ∞]?(4 votes)
- No. The Extreme value theorem requires a closed interval. You cannot have a closed bound of ±∞ because ∞ is never a value that can actually be reached. Thus, a bound of infinity must be an open bound.
So, it is (−∞, +∞), it cannot be [−∞, +∞]. Thus, the requirements for the Extreme value theorem are not met, so it does not apply.(14 votes)
- At about5:47, Sal starts talking about how you can't ever add enough 9s onto 4 to get the exact x value of the maximum. But, if you keep putting on more and more 9s to infinity, you get 4.999... on forever. If I'm not mistaken, 0.999... = 1, so wouldn't the maximum be still at x = (4+0.999...) = (4+1) = 5?(6 votes)
- There is no such number as infinity, so you cannot actually add an infinite number of 9s. So, this is a limit, not an actual finite sum.(8 votes)
- Here's an edge case: What if my interval is a single point i.e. [1,1] is there still an extreme maximum and minimum?(4 votes)
- A degenerate interval has only one real member. The range for any continuous, differentiable function on that interval would have only one value, which would be the maximum and minimum value.
Does this have any practical value?(5 votes)
- When he says "there are an" at1:11he implies that there are only one maximum value. That isn't true because you can have multiple values of maximum and minimum. right?(3 votes)
- Whereas there can be multiple relative extrema, there can be at most one absolute maximum and at most one absolute minimum.(5 votes)
- Aren't there some curves that are continuous but don't have absolute maxima and minima? I'm thinking here of a Koch curve, which I realize isn't a function, but I thought there might be a function that has similar properties.
I would think that the extreme value theorem would need a function that is continuous and differentiable over [a, b].(3 votes)
- There exist continuous functions on open intervals that don't reach min/max values, because they are unbounded in the interval (read: they can tend to -∞, ∞, or both on this open interval). If you are dealing with continuous functions on a closed interval, on the other hand, you know they reach max/minimums. That is, it is quite easy to show that in this kind of setup, the function is necessarily bounded by a certain K ∈ ℝ and that the supremum/infimum of the value-set is in fact a value of f(x) with x ∈ the (closed) interval.(4 votes)
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.