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### Course: Calculus 1>Unit 5

Lesson 2: Extreme value theorem 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.

## Want to join the conversation?

• 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?
• When dealing with constant functions, every point is both a maximum and a minimum.
• 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.
• So if the intervals is not CLOSED meaning the endpoints are not included, there are NO ABSOLUTE MAX OR MIN?
• 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) = x for x in the open interval (0, 1), does ƒ have a maximum or minimum?

However, if we define ƒ on the closed interval [0, 1], then ƒ has a minimum at 0 and a maximum at 1.

However, some functions do have maxima and / or minima on open intervals. For instance, let ƒ(x) = 1 - x² for x in the open interval (-1, 1). Then ƒ has a maximum at 0, but ƒ has 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.)
• , 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?
• 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.
• 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.
• Actually continuous functions are not opposed to discreet. As an example, any function defined on only the integers is continuous. If you want to understand why continuity matters first remember: if f(x) is continuous on [a,b] then whenever the limit of f(x) exists as x approaches x' then it is equivalent to f(x'). Now what does this mean? Well lets avoid that and think about the negation of this statement or for a moment: there is a x' where the limit of f(x) as x approaches x' is not equal to f(x'). This is exactly what is happening after @. The maximum would be where the hole is but it is not. So what we see is that the Extreme Value Theorem can only promise us absolute maximums/minimums if the function is continuous on [a,b]. It is important to note that it only promises us this. So you can have a non continuous function that does have a global maximum or minimum.
• Wouldn't this also hold for [-∞, ∞]?
• 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.
• At about , 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?
• 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.
• 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?
• 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?