WEBVTT
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we want to find a bar a sketch of a
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graph of a function that has to local maximum,
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one local minimum and no absolute minimum barbie. We
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sketch the graph of functions that have three local minimum
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. two local maximum and seven critical numbers. Let's
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start with a that is a function that has to
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local maximum, one local minimum. And now as
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the minimum seems to be easy in this case because
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we have something like this where we have let's Let's
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see two local maximum. So we have this point
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here over here is a local maximum and this point
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over here is the local maximum. So we have
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to local maximum only in this graph we have then
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one local minimum and this one is here one local
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minimum, there is only one and now we want
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no absolute minimum. And in this case let's see
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that the lowest point in the graphic which should be
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the absolute minimum should happen here. But the point
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has been removed just the image of this value here
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as fame put this here up and for that reason
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we uh have a graph of the function that do
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not does not obtain it, that was value.
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Yeah. So uh we can say that in this
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craft there is no absolute minimum, there is some
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sort of maximum but now as the minimum as we
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want it. So two local maxima. Ah those
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points here and here and here one local minimum you
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at this point sorry and no absolute minimum because we
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have these behavior here on the lowest point should be
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here but we have removed the image and put it
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on, put it above in order to not attain
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the absolute minimum because uh something similar if we don't
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want an absolute maximum that is we can't put the
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highest point on the graph to the right and removed
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that point of the graph and put it up below
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. That's uh the similar case or no absolute maximum
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. So we have these three property on this graph
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. Okay, now we go for barbie which should
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be a little bit more interesting. And we see
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the following, we have three local minimum to local
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maximum. It seems that there are already there five
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critical numbers. We need two more to have seven
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curriculum. So we have thrown this and as we
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can see here we have Uh let's do it with
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the statement three local minimum. So we have here
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local minimum here local minimum here and local minimum here
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. So we have three local minimum. Mhm And
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we wanted uh two local maximum and we have them
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up here and here it could be differential. There
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is no need that is be like this. But
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this way, but you can have a perfectly differential
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function here and here and each of these local maximum
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. So to local maximum at these two points as
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we said And now we have these five points here
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are critical numbers because there we have either There are
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derivative in which case to be zero because they are
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local extremes or there are these are points where there
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is no derivative like this in this example and they
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are also critical numbers because they are in the domain
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and function and the derivative that does not exceed that
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exists at those points. So it means that We
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have already with this property here. We have five
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critical numbers. So we need two more and the
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idea is to make the graphic discontinuity with the chump
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discontinued like this with this graph, we don't have
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neither local minimum nor nor local maximum at these two
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points because if we take any interval around those points
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, there is always the images are greater than the
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image of the point and lower the the image of
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the point. So these two points are now extreme
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by local stream values but but they are critical numbers
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because they are in the domain of the function.
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We have put images to those points in this case
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, we have put the image up here and here
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at this point, we have put it up here
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here in this case here and here in this case
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here. So the points are in the domain or
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the values of the numbers and there is no derivative
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. So they are critical numbers. So we had
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these two anomalies near the 10 points of the domain
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in order to Add two more critical numbers but which
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are not local maximum nor local minimum. So that's
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the idea behind that. So I take this out
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a bit, so this is an example and solving
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for P. And we had this a little bit
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easier in for A. And that's it.