WEBVTT
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we will sketch the graph of the function given by
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this expression, the piecewise function the first part or
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piece is two X plus one. For x Greater
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than or equal to zero and less than one.
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And the second piece is 4-2 eggs. If
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x is greater than or equal to one and less
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than or equal to three. So it's a piecewise
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function with two pieces. Both pieces are straight lines
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. This case they will be segments. And the
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domain of the function defined that way is zero three
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. Close interval, close both. And points in
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the first piece series included but one is not included
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and in the second piece both one and 3 and
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include are included. So let's uh draw first to
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line two explosive one for X greater than or equal
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to and less than one At zero. The line
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to Express one has the value one. So and
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it is included. Yeah. So we have this
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value here. This red dot here. Okay then
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at one it is not included but if we calculate
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the value is one of the piece to express one
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we get three. That means that at one we
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will have divided three but we indicate is not in
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good with this open circle here and then we draw
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the line Between those two points. Let's see if
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like you fitted well here almost there. So you're
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here this is a line it's a segment that because
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um x is uh in the interval ceo one including
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Sedona including one and now the second part is second
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piece straight line again And for X Equal one,
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the images 4-2 which is two. Yeah,
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so to at once or the values too. So
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this and it is included in the graph. That
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is these points here. Okay. Say said we
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are including it in the graph and for X equals
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three, we got four minus two times three.
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That is four minus six. That is negative two
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. So three we have negative negative two and it's
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included in the gravels. We have to join the
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two lines and we see that we have to cross
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the X axis. That too because at X equal
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to we get zero. So we draw the line
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something like this then we um yeah feed better.
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So there there's a line is a little bit longer
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than we need and that's it. So this is
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a line and we see it cross exactly at Mexico's
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two because the image of two following the second piece
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is 4-4 is considered. So we have this
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uh this continuous function here and we can see that
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we have an absolute minimum value over here at this
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point. It's the lowest point of the graph and
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is included in the graph. That's why it is
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the lowest point. So f has an absolute minimum
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value of negative two. In that value. Of
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course at X equal three. Mhm. Respect to
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local minimum values. We don't have local minimum values
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because for example, this point here we have now
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function to the left of that point. Could not
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be a local minimum At this point. one here
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It's not a local minimum because if we take any
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interval around one Small interpret that is, we are
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close to one. There are all their images that
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are smaller of the image at one which is to
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to the right, we have a smaller values and
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to the left we have greater values in these other
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parts of the functions. So one is not a
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local minimum and there are no more possibilities of local
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minimum. So have has no local minimum and we
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respect to the maximum values. There is no absolute
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maximum because the highest point should be this fine but
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it's not included. So there is always created values
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that uh get closer to the image at uh three
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at the value three but never reaching it. So
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it has no absolute maximum but it don't have also
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local maximum because the same reason before there is at
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any point that we choose on the graph inside the
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domain, not the end points. There are always
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images that are greater and smaller than the image at
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the point. So it has no absolute or local
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maximum values. So the only thing that this function
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has is an absolute minimum value-2 at a sequel
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three at the writing point of the domain. Yeah
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. And so this is the result in this problem
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. Mhm