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Identifying relative minimum and maximum values
Sal analyzes graphs of functions to find relative extremum points. Created by Sal Khan.
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In order to be a relative or local maximum or minimum, doesn't a function need to be continuous? in the interval?(10 votes)
A functionƒ, defined on a setS, is said to have a relative maximum at a pointcinSif there is some open intervalIcontainingcsuch thatƒ(x) ≤ ƒ(c)for allxwhich lie inI ∩ S.
The concept of relative minimum is similarly defined by reversing the inequality. These definitions does not assume anything about the nature ofƒatc, and in particular,ƒneed not be continuous atc.(16 votes)
ACCORDING TO EXTREME VALUE THEOREM , graph should be continuous it should not be discontinuous... while in the above video the graph is discontinuous so we cannot find the maximum or minimum value from the graph. am i right according to it ?(6 votes)
The extreme value theorem is for absolute maxima/minima. With a graph with discontinuous, that means that you might not find an absolute maximum/minimum, but you can still find relative maximums and minimums.(13 votes)
isnt f(c) an absolute min(3 votes)
The hand-drawn graph in this video isn't precise in the sense that we have exact x and y values at each point, it's merely a demonstration of how to identify relative max and mins. However, judging by the way the graph is drawn, one could argue that f(c) is also the absolute min of the graph shown in the video.
Keep in mind that a relative min or max can also be the absolute min or max as well. Re-watch the extreme value theorem video if this doesn't make sense and pay close attention to how Saul defines the intervals for a relative min/max and just think of expanding that interval to encompass the entire graph for an absolute min or max.(11 votes)
How can f(c) be a local minimum? I thought that to be a local minimum, the derivative of the left side had to be negative and the derivative of the right is positive? Also, how can endpoints (generally) be local minimum/maximum, by this definition? We don't know what the right/left side derivative is. Thanks in advance.(5 votes)
First rule in differentiation:
The function must be continuous.
The functions in the video are not continuous, therefore they cannot be evaluated by applying derivative techniques.(5 votes)
In my text book it says that end points can not be considered as local min or max.
But here we do consider them! Why ?!(2 votes)
Please read your textbook more carefully. It should say that the endpoints of an OPEN interval cannot be considered as max or min. The reason is simple: on an open interval, the endpoints are OUTSIDE the domain and thus not part of the function's definition. Thus, they are not max, min nor any other part of the function.
However, on a CLOSED interval, the endpoints of a differentiable function are ALWAYS local extrema. This is because if the function is increasing at it reaches the endpoint, then the endpoint must be the maximum in that region. If the function is decreasing as it reaches the endpoint, then the endpoint must be the minimum point in the region. Thus, if a function is differentiable, then the endpoints of a CLOSED interval (but not an open interval) must be local max or min.
If your textbook says that the endpoints of a differentiable function a CLOSED interval are not local extrema, then the book is wrong.
Special situation note: if the function is stationary, this is it is completely horizontal, as it reaches the closed interval endpoint, then it has the unusual property of being both a max and a min at the same time, as would all points on the horizontal region of the function. This is rather unusual and you probably won't be asked about it in class.(6 votes)
If you were given an equation and asked to find the relative minimum and maximum, would you graph the function or is there an algebraic way? I also know you can do it by graphing calculator, but sometimes those aren't allowed.(2 votes)- In calculus, you'll have to learn to identify the extrema (that is the general term for either max or min) by taking the first derivative.
The extrema of a continuous function can only lie at one of these places:
1. Where the first derivative equals zero.
2. Where the first derivative fails to exist.
3. The endpoints of a closed interval.
However, there is no certainty that the function actually has extrema at the first two situations, you do have to check. The endpoints of a closed interval are always relative extrema.
So, finding an extremum graphically or getting an approximate value with a graphing calculator will not be sufficient. You will be expected to compute the exact values using differential calculus and typically without a graph.(5 votes)
In one of the review questions in the next quiz section we are asked to find the absolute minima of y=(-3x^2 +24x +10) so I found the two critical points x=0 and x=8. The function's value at x=0 is 10. At the included endpoint x=12 it is also 10. Considering that x=12 is an endpoint and did not come up as a critical point, I decided it could not be an absolute minimum, but this was incorrect. Could someone explain the error in my thinking. Do we not use the critical point method anymore?...we were just using that a few videos ago... Thank you!(3 votes)- When we're asked for the absolute min or max we have to find the most extreme values within the given range. Sometimes these occur at critical points, so we always have to check those points. We also have to check the endpoints of the range, though, and if we find values that exceed the ones we found at the critical points, we use the endpoint values instead. We aren't abandoning the critical point method, but we're answering a question that requires us to consider whether there are other values in the range that are more extreme. In this case, I believe the correct value where x = 12 is less than 10.(2 votes)
- what is the difference between finding "all values at which x has a local minimum" and "all local minimum values of x"(2 votes)
- Both should mean the same thing. Quite frankly, both are awkwardly phrased, as if we were trying to find the minimum value of x (which, of course, is pointless). A much better phrasing would be "all values of x for which the graph has a local minimum".(4 votes)
- I think in order for c to be a local min there has to be a local max just to the left of c. you said in another video that for a local min to exist f' must be negative as it approaches that point.(3 votes)
- For a local min (or max) to exist at a point, f' must be positive just to one side of that point and negative on the other and f must be continuous.
Consider a parabola y=x^2, this has a local min at x=0 and no local max anywhere.(1 vote)
would f(d) be the absolute maximum for this graph?(1 vote)- Yes, it looks like it. Any absolute maximum is also a relative/local one by definition, it's just that the open interval around it that you use to define it extends farther. Since the value of f(x) when x=d looks greatest across the interval, that would make f(d) the absolute maximum.
However, as Jesse has pointed out in another comment, you would have to make an assumption that the interval where you're saying f(d) is the absolute maximum—which Sal hasn't explicitly defined—starts and ends where the orange curve drawn starts and ends. Otherwise you might get a very negative value somewhere to the left and a very positive value "in the dark," to the right, potentially exceeding f(d).(4 votes)
Video transcript
So what we want
to think about is, at what x values does our
function here in orange-- let me make this clear. This is the graph of
y is equal to f of x. At what x values, and we
have some choices here, which of these x values
I should say, does f of x hit relative maximum values
or relative minimum values? And I encourage you to pause
the video and think about it and classify whether we hit
a relative maximum value or a relative minimum
value at each of these x's. So let's first
look at x equals a. f of a is right over here. This is f of a. And I can pretty easily
construct an open interval around a so that f of x, if
x is in this open interval, is going to be, is definitely
going to be less than or equal to f of a. f of x in
that interval is definitely, they're all lower
values of f of a. So this right over here, and
you can even see it visually, this is kind of the classic
relative maximum value that we've gotten to. Now what about this? If this was filled in, if
we were continuous here, this would be pretty obviously
a relative minimum point. But this does something
interesting, it jumps up. And so this right
over here, let's see, this is the
value of f of b. That is f of b right over here. This is a little bit
counterintuitive. But I actually can construct
an open interval around b. I can actually construct
an open interval around b where the value of f
of x, if it's in that interval, is a less than or
equal to f of b. So f of b right over here is
also a relative maximum value. Now what about c
right over here? Well if this was just at
the bottom of a kind of, if it would look like e, e is
your classic relative minimum point. But c, look at
this discontinuity. What's going on here? But we just have to
think about, well can we construct an open
interval around c where f of c is-- this is f of c right
over here-- where f of c is less than or equal to
the x's in, is less than or equal to f of x for the
x's in that open interval. Well let's see, in this open
interval the way I've drawn it, the f of x's are here
and they are over here. So it looks like
f of x is always greater than or equal to f of c. So that by that definition,
by the definition of a relative
minimum point, this makes it, or relative
minimum value. So that actually is a
relative minimum value. Now we get over here to d. And really, by the
same argument that we used for b, that is
also at d our function takes on another
relative maximum point. And then e, when
x is equal to e, this is the function
hitting what could really be considered a classic
relative minimum point. We can easily
construct an interval where you take any x
in that interval, f of x is going to be greater
than or equal to f of e. So this is a relative
minimum value as well.