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## Differential Calculus

### Course: Differential Calculus>Unit 5

Lesson 5: Absolute (global) extrema

# Finding absolute extrema on a closed interval

Sal finds the absolute maximum value of f(x)=8ln(x)-x² over the interval [1,4]. Created by Sal Khan.

## Want to join the conversation?

• Is there any method to identify that critical point `x = 2` is maxima or minima, instead of replacing 1, 2 and 4 into `f(x)`? •  You can see whether `x=2` is a local maximum or minimum by using either the First Derivative Test (testing whether `f'(x)` changes sign at `x=2`) or the Second Derivative Test (determining whether `f"(2)` is positive or negative). However, neither of these will tell you whether `f(2)` is an absolute maximum or minimum on the closed interval [1, 4], which is what the Extreme Value Theorem is talking about here. You need to actually compare the values of the function at the critical numbers and at the endpoints to determine which is the highest and lowest on the interval. Hope that helps!
• Did we really have to check the values of the function at the endpoints to see if they are greater than the local maximum f(2)? I mean, since we didn't find any other x value in the interval [1,4] where f(x) is either 0 or undefined and since the function is continuous, isn't that enough to know for sure that neither of the endpoints has a greater value than f(2)? Because if at least one of them did, shouldn't the function have a local minimum in order to increase to a greater value than f(2)? And if it did, we would have already found that local minimum, but we didn't. • Once we prove that f(2) is the local maximum by taking derivatives of intervals before and after it, and that there are no other critical points, then you are right, I don't see any other information needed to prove that f(2) is also the absolute maximum over the domain. However, he didn't prove it was the local maximum before proving it was the absolute maximum by comparison to the endpoints.

Hopefully this helps someone 4 years later.
• Is there any proof of the extreme value theorem • So could we say that the absolute(global) minimum is f(4)? I re-watched the video right at the end but i don't think he said it. • My question is if any question has a critical value is "infinity". now, we ignore the critical value or put that value in original function to find max. or mini.?? • A critical point cannot have a value of infinity. There may be a critical point because the first derivative diverges toward infinity, but in such a case the first derivative fails to exist at that point.

If the original function is defined at a point and its first derivative fails to exist at that point, then you would proceed to see whether it is an extremum in the usual way -- seeing if the first derivative changes signs by comparing the first derivative to just before vs. just after to see if there is a sign change OR by plugging in the critical point into the original function and then comparing that to points arbitrarily close to it on either side.
• I'm not sure if someone else asked this or not, but wouldn't -1^2 = 1 not -1?
(1 vote) • How to determine minima or maxima (absolute) when there are asymptotes? • If there are vertical or oblique asymptotes, the chances are that you don't have an absolute max and/or min. Specifically, if there is anywhere the function heads toward positive infinity, you don't have an absolute max. If there is anywhere the function heads toward negative infinity, you don't have an absolute min.

If you have a specific problem in mind, please post that so we can examine the specifics.
• Sal you said at that ln(2) (.693147...) is going to be some fraction. And that fraction, if greater than 1/2 will result in 8*ln(2) being more than 4. While ln(2) * 8 is greater than 4 (5.54518), I wanted to follow your reasoning and got lost at this point.

Being more than have will eventually give you a negative number when added with -4. • Let's say I'm looking at a graph with two maximums and two minimums (four points). So the absolute max/min would be the highest/lowest value out of the four points on a graph, and the relative or local max/min would be the other two points on the graph?
(1 vote) • No, that is not correct.

First the plural of "maximum" is "maxima" and the plural of "minimum" is "minima".
There is not always an absolute max/min, but if there is it will be one of the local max/min. To find out here is what to do:
If you have a closed interval, then the endpoints are automatically local max/min. If you have an open interval the endpoints are never max/min (because they are not in the domain).
Every max/min is a local max/min.
Identify the largest local max and the least local min. If and only if there is no other point in the domain greater than the greatest local max, then that local max is also an absolute max. Likewise, if and only if there is no other point in the domain that is less than the least local min, then that min is also an absolute min.

If there is any point in the domain greater than the greatest local max, then there is no absolute max. Likewise, if there is any point in the domain less than the least local min, then there is no absolute min.

Note, if you have two or more max that with equal y values and which are greater than all other points in the domain, then both (or however many there are) are absolute max. Similar for there being more than one absolute min. Consider the case of the sine function. Every local min is -1 and every local max is +1 AND there are no points where sine is greater than +1 nor less than -1. Thus, with the sine function all of the local maxima are absolute maxima and all of the local minima are absolute minima. So, sine has infinitely many absolute maxima and infinitely many absolute minima. 