WEBVTT
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this problem. Number two of the Stuart Calculus eighth
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edition section in two point eight. Use the given
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graft to estimate the value of each derivative, then
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sketched the graph of F crime. So this is
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a graph of F given as this certain shape.
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And this is a graph for this problem, and
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we're hoping to sketch a graph of F prime or
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the derivative of dysfunction. And we're going to do
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that by finding the derivative for each point from zero
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to seven. X equals zero tech sickle seven and
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then using each of the individual values on DH this
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plot here, tio, plot what our function will
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look like. Okay, so we begin with X
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equals zero. What is the slope? What is
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the derivative at the origin we look at our function
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on? We see that it has a pretty steep
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, positive soap. It goes approximately up little more
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than three. This is with attention. Lang would
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look like, uh so it's not to re over
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one the slope. It probably is more closer to
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four over one or five over one from as to
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how steep the slope of this tangent, Linus.
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So as an estimate. We're going to say that
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the rise over run is for for the slope at
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the origin. Carping at X equals one. What
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? It's a derivative. What is the tension?
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I ll look like tension in line at X equals
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one. It looks like this all right, touches
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the function at X equals one only at one place
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and the line is horizontal, meaning that it has
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a slope of zero at X equals one. Add
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X equals two. We take a look at where
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the function is at X equals two. It's right
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here. Let's say that we went down one and
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over one thats not quite a line that is tangent
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to the function. And so we go down to
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approximately and over one that seems to be a little
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too steep. So when they get one over one
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, it's not steep enough. Negative two or one
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a little too steep. How about halfway halfway between
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negative one Negative too. That seems more appropriate as
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attention line, uh, for this function. So
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half way we can make it a one or negative
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to three over to, and that will be our
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estimate for the Slope X equals two at X equals
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three. What is the derivative? What is this
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stuff of the attention line X equals three. This
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is the point here, crossing that Texas. And
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we see that if we draw attention, line on
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that it is closely associate ID approximately ah, equal
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to a slip of negative one. So that is
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our best estimate. We're going to go with a
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slip of negative one for At X equals three and
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X equals four. We look at where that point
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is. Two. Three, for this is the
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point X equals form. We're trying to estimate what
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potential and my look like we'd go down one and
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we'LL go over one thats not quite steep enough or
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it's a little too steep. So we go over
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to and this seems to be a bit more appropriate
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if we were to draw this tension line here.
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That seems to be a pretty appropriate tension. Lame
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at X equals four. And this slope here is
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down one over to a rise of negative one over
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two. That is a slope of the negative one
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house annexe Eagles fry. What is the derivative of
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F one, two, three, four, five
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This is the point for the function. There's a
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slope world. Well, tryingto as to me by
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going up one in going over one, two.
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Uh, it seems to be a good estimate.
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How about three? I think that is a little
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better. We know that the slope add five as
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it gets closer to the next value. I'm The
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slopes are going from steeper tonight. A Steve.
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So this seems to be a better estimate. Um
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, the sloping arise of negative on and a run
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of three or closer to that than negative one over
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to. So we're going to see that they're slow
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. Estimate at X equals five is negative. One
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over three at X equals two six. This is
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where the function is at X equals six. We
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see that it is it horizontal pendant horizontal line on
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the slope of zero. So six there's a minimum
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and the slope of the tangent line zero. So
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the dirt of zero and finally at X equals seven
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, one, two, three, four, five
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, six, seven. This is where the point
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is. All right, if we go down and
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then over one too steep for the soap over two
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Still seep three, four five at five. We
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imagine it a more approximate tangent. So this is
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what ah, run of five and a rise of
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naked one would look like. So this being a
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slow a rise of one. So this is a
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rise of one and a run of five. That
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is a slope of one over five or one fifth
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. And this seems to be more amore. Appropriate
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estimate for the tension. This liberal potential in an
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X equals seven. So we're going to go with
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one over five. Great. So finally, we're
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going to plant this function in this F prime function
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, and we're going to supply each of these points
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for X equals zero through X equals seven. Starting
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with X equals zero. The point. The value
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of the derivative is for the next point at X
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equals one. The value of the derivative is zero
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for X equals. So we're gonna draw a couple
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things here in a draw, make it one,
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continue the function a little lower, and then draw
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negative too. Okay, being this in mind,
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what is the next point at X equals two.
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The stop is thinking of three over to our negative
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one half that's approximately right here at X equals three
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. The value of the dirt of his negative one
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right about there at X Eagles before the value is
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approximately negative. One half right right there had X
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equals to find the value of the dirt is a
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personal maid of one third. So just a little
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higher closer to the X axis. An ex eagles
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to six, the value the dirt of zero.
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So now across the X axis. And Alex,
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he was the seventh valued. The jury was one
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fifty. So this is what the shape of the
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function will look like. Now we found thie.
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Sure, if it is at each of the points
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. And now we plotted them separately and we join
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them with the single smooth curve to estimate and sketch
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the parent behavior of this function of crime. So
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this is a sketch of the graph of Prime that
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we found it right from the individual irritants. Eight
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different points of the graph of F given here and
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that completes this problem