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Fen discrepancmeeting t0 discuss the examQuestion 7Let f be the function given by f(r) 2Ic' The graph of f is concave down whenF -...

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Fen discrepancmeeting t0 discuss the examQuestion 7Let f be the function given by f(r) 2Ic' The graph of f is concave down whenF -

Fen discrepanc meeting t0 discuss the exam Question 7 Let f be the function given by f(r) 2Ic' The graph of f is concave down when F -



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Let $F(x)=\int_{1}^{x} f(t) d t,$ where $f$ is the function whose graph
is shown. Where is $F$ concave downward?

Okay for this question Over here, we know we're trying to figure out where is F Conkey downward? In order to do that, we know we're gonna have to calculate the first and the second derivative. Okay, so given this we know after box is intro from wonder Acts of off of T Do you achieve? Remember, we have the fear of the f Prime of axe is lower case of the vax therefore off Double problem Max is lower case off Problem X Therefore, we know the given graph is decreasing on the interval. Negative 1 to 1. Therefore we know it is Kong Cave down on the interval. Negative 1 to 1.

We want to determine where function F is concave up or down where F is equal to X. To the 4/7. We want to follow these two steps listed to solve this problem. First, we're gonna find the second variant of the F F double prime and any zeros or acid votes. These zeros are asuntos break double private act into a series of intervals where in step two, when we when we evaluate F double prime on each of these intervals, we find that F is concave up where F double prime is positive and F is concave down wherever F double prime is negative. So let's go ahead and solve first. Let's find double prime and the zeros of acid totals F double prime or another F prime is 476737 and double prime is negative. 12 or 49 10 7. There's no zeros but there is an acid zero. So we have an intervals from negativity zero and zero to infinity. So we test double prime on each of these intervals. We find that we're both intervals. Double prime is negative, which means that this conclave up nowhere and F is concave down on negative zero and zero to infinity.

We want to identify with a function F. Is conquering up versus down where F. Of X. Is equal to X. To the 4th 7th power. So to answer this question, we need to identify what can cavity means and then proceed to solve. Come gave up versus down needs the sign of ethical prime for X. Is positive or negative. So we have to find ethical prime. That crime is 4/7 1 over X. 2 to 3. Seventh until the problem is negative 12 49 1 over after the 17th. Now we have to check for the partition of critical points where the denominator of F. Or the numerator of F is equal to zero. This is inevitable prime. Not enough. So we get a critical partition point at X equals zero. Thus we check the left and right of zero for the sign of a double prime from negative zero. Such as X equals negative 10. That's what I was positive from X equals 02 infinity. Such as X equals 10. If it's also possible, crime is also positive. Therefore, we have our solution for the con cavity of F. Abbas khan came down nowhere and African came up on the intervals negative infinity to zero, as well as zero to infinity.

Let f be a positive function on a B close interval compare midpoint estimate M N chop suey the rule T. N. And the integral from A to B of F. Given that the graph of F is concave up in part A. And concave town in property. So we're going to write a relationship between these three values the point estimate episode rule and the except value of the integral in each of these two cases. That is when the function the graph of the functions can give up and when it can't get down here, we have a sketch of what happens and and part A. We have a concrete malfunctions. The blue line here we know that if the functions can give up on the interview tab then in the skunk ape up in any serve interval. So we're going to focus the attention on any serve interval from X I two X I plus one. So we have same situation but on the left for a concave all function under right we have the concave down graph. So let's see first the concave up case. We can see here red line china points, P and Q. And that will be the top of trapezoid with base or with heights in any case X one, sorry, excited and excited plus one. So that trap saw it in the case of concave up function is line that top of the trap side lights above the graph of the function that is the red line here lies above the graph the blue line. The graph of the function. And that's an essential property of in concave of function. That is if we join any turbines on the graph, that segment joining the points lies above the graph for the concave down a graph. The situation is the other, the other way around that is The segment joining two points on the graph lies below the graph. In this case we see this red line here joining points, P and Q. Like below the blue line. That is below the graph of the function. Mhm. So we can really see the conclave of case here. The trapezoid P Q X, Y plus one X. I has an area greater than the area under the graph. So it's clear that in the case can give up, we have the T. N. Would be greater than the area under the graph. And we get to notice that we are talking about a positive function. So that's why we can talk about areas here. So we're talking about areas in all all the time here. And that's because we can talk about area under the graph because the function is positive. So the area of the traps oid would stop equal to the red line. There is a line between P and Q in this case here. Thank you, bob has an area greater than the area under the blue line. That is under the graph, which corresponds to the interrupt. It is true that we are looking at a particular or any sub interval. But this relationship, it's going to be the same when we talk about the upside of rule and the whole integral. That is because if we sum all the areas of the trapezoid we get T. N. And if somebody integral between X I and X I plus one over all the sub intervals will get this integral here. So the situation, the relationship we see for any sub interval will be the same. That we will see for the whole numbers and that is because these value here and this Inderal here are obtained by adding all the areas onto the graph and years of the trapezoid over all the sub intervals. Okay, so that is this is true here in this case is clear and the case of concrete down graph, we have the other way around there is a graph under the blue line is greater than the graph under the trapezoid with top PQ which are the Travis side used in the traps of the role. So in this case the integral is greater than the end is the inverse situation we were we saw in they can't give up case we have that. But now we got to see the other method, the mid point you get to see what happens and what in which place it is with respect to T. N. And the interval And here we have another line, green line between the points R and S and that that's where we're going to center our attention now, in fact the midpoint. Remember the midpoint rule is just the step size beam receiver in times the some of the images of the mid points is effort exit variables X 1/2. This point was ever this next point, which is this and so up to f at the last minute point which is equal to this. So it is clear that All right, I'm still here. What we're doing here is adding the areas of rectangles with One side, which is the save interval X. I accept I plus one. And the other side is or height will be these numbers. But what relationship there is between these images here and what we draw here, the green line, that's why we're going to do now. The green line corresponds to attention line to the graph at this point that is at the midpoint. This uh number here, X I bar is the average between the endpoints of the same interval. That is X I plus X. I plus 1/2. The spirit right here, just a midpoint. So at the midpoint we are drawing the tangent line to the graph of the function. This is the point of tendency here in that tangent line. She's this case, the green line has an intersection with the vertical lines through X I N through X plus one. Those points of intersection between the tender line at the midpoint and those vertical lines are the points are here. And as here what we want to show now is that the area is the trapezoid with top equal this red green line. It's exactly the area of the rectangle with based soup interval and hide the much at the mid point. We're going to prove that. And with that we will have automatically the fact that in the case comparable for example the integral that is the area under the graph is greater then mm. N. Because we will have that the area we are going to prove that the area of the Travis oid with top equal this. The segment R. S. Is just equal to the area of the rectangle with with the minus over and there is the length of the soup into a bowl and hide equal the image at the time. And with that when we add up these values here to form an N. We will have a number. We have the some of that will be equal to the sum of the area of the trapezoid with top the green line. And that is less than the area under the graph. So here in this case we will have that and here we have this because the area of the trapezoid we create nine in this case is greater than the area under the graph. We can see clearly that. So the kid thing the only thing we gotta do To have these relationships and this other one is to prove that for any serve interval X I X I plus one. The area of this trapezoid with top line equal R. S. Is exactly equal to the area of direct angle. Using the midpoint formula. That is the part of the lens of the interval X I X I plus one and the image of the midpoint as the high to 30 tingle. So that's what we're going to prove. So first of all we need to write the equation of the tangent line to the graph of the function through the midnight. So tangent line two, the graph of F at the midpoint, the midpoint X I bar of Kenny sub interval IXI plus one too right. The equation of that center line. We need the slope and a point through which the line passes so that we know so Hill slope of this 10-9 we all know is the derivative of F at the point of tendency trees the midpoint because it's a turn in line at the midpoint. That means that the derivative at the midpoint is a slope of the turning mind. Okay, so we have the slope and these tension line passes through the point. It is clear that X. Bar there's a midpoint is the first coordinate of the point of tendency. So us through the point X I. Bar. And the second coordinate is the image of that mid bite. That is this point here is just this point over here. Despite over here, the point of tendency is the first coordinate is the midpoint of the sub interval X. I. Bar. And the second coordinate is the function evaluated at that midnight. We know that because the point is just on the graph of the function. So the second coordinate is the image of the midpoint. It was it's really function if so we know the slope and point through each the land passes. So we can write the equation of the tangent line. It is important to notice that we are working on any 70. So this will will this we will apply to any of these v intervals of the British. So the equation of that line would be okay of this tangent line. Yes. Why -2 coordinate of the point through the line passes. That is f of the midpoint X. I. Bar. And that is equal to the slope of the tenant line, which is a derivative at the midpoint times X -1 Coordinate of the Point Through Which Line Passes. Mhm mm. That is why equal F at the midpoint X bar plus the relative of f at the midpoint exit bar times X minus X. I. Bar. So this line, let's say that we write that as a function. That is ways L of X equal if xie bar plus three motive of F X I bar time, six minutes excited. Bar written as a function to simplify a little bit what we're gonna do now. So this is the line. The equation of the line. The line is stands infinitely too right into the left. But we are interested in those two points are here are here sir. And as here are here as here those points are just intercession of this line. We have obtained and the vertical line through X. I and X. Y plus one. So the point are and S on the figures are the intersection between l've X and the vertical lines through X I N XI plus one respectively. That is our his intersection of L with X. I. And S. Is intersection of X X L X with X. Y plus one. So yeah, the point R is found or second coordinate because the first question is X I. Second coordinator of our he's just a value we're going to obtain. Now that is simply L evaluated and accept exile. That is if we evaluate the line at this value excited, we're going to obtain the image through the line of that value X. I. And that corresponds to decide here. That is the point are and that's either way the line there and we get F X I bar. We are looking here plus the derivative of F X. Bar and now we've gotta replace X here by the value X. I. So times X Y minus X. I. Bar. This is for our and ale of X. I play swan bar. Sorry L. At exactly a swan. What do you think? The image of this value here which he was the you miss correspondent through the line to the point S. So we get equal F. X. I. Bar plus the relative of effort X. Bar times. And now we replace he here X. By this value X. I. Bless want minus X. I. Bar for S. That's it. So now we have the second coordinates of these two points are an S. That's fundamental to have the trap side with top equal the red and the green line. Okay so now let's see what the area. Okay let's see um we have those values and now it is clear that those values corresponds to? Let's see in this fingerless uh get out a little bit of these lines 82. Mhm. See something here. Yeah. T. L evaluated there is the green line of alerted excite for example give us these length here and L. Evaluated at X. I. Plus one. He was teased length here which in this graph are this lens here and these lands here, those are the lens two of the lens of the trapezoid with top equally great. The green line. The other sides of that trap side are the base here which is in fact it's a high to the trump side and yes. Mhm. Is there a bit that's not it didn't work month stay with this. Okay. And the other is just that line here and we know the area of this trapezoid and in this case is this line and he's lying here. We know that the area of these traps are top equal the green line. Is the some of these hide here and decide here This length here in this land here that some of these over two times these lands here, this area of that episode. So it means we get to add up these two images and divide that but to and multiplied by the length of the interval X I. X. Y plus one. So let's put it here. The area of the trapezoid. Let's denote the traps are very carefully here is in the graph that is R. S X I plus one X. I. That's the trap side. R S X I plus one X. I. That is this is an attention. We Used to denote that these are the four notes or corners or points determining the trap side. Okay, so these uh proper side as area equal to. So L. X. I plus L. X. I plus one. Just two lines here and here here and here times that that over two because that's what we call the mean base of the drop side. Times what is the height of the Travis in reality which is the length of the interval X. Y plus X I, exact plus one which is the difference okay this difference we can call it a judge in the case of all the methods we are using that is left employment, right endpoint midpoint trump's order rule. Simpson's rule is The common lands for all seven tales. That is. We have a regular match so we can call that age to simplify notation a little bit. So this area is able to that and this is equal to and now we put the values of X. Y and L. X. Y plus one. That is these two formulas here. So yet the first one is F at X. I bar plus the derivative at X. I bar times X. I hear minus X. I bar. And the other one will be the same expression accepted here. We have exciting plus one. So plus F X bar plus derivative at X bar times XY Plus one. Okay, that's right. Just better this and put all this left a little bit. Okay. Times uh as I said before, X I plus one here instead of excitement because it's a place of excess. We saw in the equation here minus X. I bar All that divided by two and all that times age. Okay, so now we simplify, we have in the numerator here we have if at excite bart two times that is to F X bar. Now we have this term here and this one here and we can say there is a common factor F derivative at X. I bar. So plus after everybody had X. I. Bar times. So we have this first factor here, X I minus X. I bar plus this other factor here. X. Y plus one minus X. I. Bar. It is not necessary to put them in Cyprus. And this is because we have a plus sign in front of it. Okay we have this over to times age. So we have to F at X. I. Bar. Plus there we met you at Exit bar times we have X. I plus X. I plus one In this term here is the same twice. So it's negative two eggs I bar Over two times age. But let's see that this is to F at X. I. Bar plus if derivative at X. I. Bar times X. I bless X. I plus one minus two. Remember Excite bar is X. I Plus eggs I bless 1/2. It means that these two here, it's going to cancel out. And this is equal to to have ex a bar plus the derivative at X. I. Bar Times X. Sightless one exciting, acceptable swan minus. And now we're here we get a bit of parenthesis because we are cancelling the tooth. And this negative affect all discretion here is X. I plus eggs I plus one Over two times H. But see that this expression here, it's the same as this one and they are subtracting it to each other. So this is equal to zero. It means okay let's put it here All this is equal to zero because it's the same expression minus the same expression. So we get to so that notifies all including this product here. So we got to F at X. I. Bar over to times age because all the second turn after the plus year it modifies. So we get this to cancel out again and we get F. X. I. Bar times age. That's just what we wanted. Because this is the terms that appear in the mid point rule. This is a judge the common lens of the serve intervals you might say over it. And this is a measure at the midpoint that is X. I. Bar in the case of the interval X. I accept this one. So for any sub interval, the area of the trapezoid with top line equal the green line in the graph is equal to the area of the rectangle. With besides age. That is the interval excites sexual swan and the image at the midpoint of that internet. If we add up those areas of the travel side, we get the midpoint rule. So for that reason having shown that that those areas are the same. It is easy to see in this graph that the relationship we have is this the case that can give up and this in the case of concrete down. So it's just a reverse situation in both cases. So we can see that um yeah, the situation we have in one serve interval will say which is the situation globally in the interval maybe because where we get to add all the areas of the trapezoid to have Each of the Method Travis is a rule and one rule and we get to have all the integral or the areas under the graph over each some interval to have the area under the graph. On the whole interval A. B. Yeah. Yeah. Okay. So that's the proof and I think it's it's not complicated but it's just a geometrical observation of between these three areas. That's what's important. And then to prove the area of the rectangle used in the Melbourne rule is the same area of the trapezoid form by using the tangent line to the graph through the midpoint and comprised between the vertical lines. Two rigs I accept plus one and the interval itself x i is I plus one


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