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Question: Given the Least Upper Bound Property (definition 1.1.3 below), Proposition 2.1.10 below, the Bolzano-Weierstrass Theorem (Theorem 2.3.8 below), the Min-Ma...

Question

Question: Given the Least Upper Bound Property (definition 1.1.3 below), Proposition 2.1.10 below, the Bolzano-Weierstrass Theorem (Theorem 2.3.8 below), the Min-Max Theorem (Theorem 3.3.2 below), Definition 4.2.1, Lemma 4.2.2,and Rolle's Theorem (Theorem 4.2.3 below). Describe how these are connected to obtain complete proof of the Mean Value Theorem (Theorem 4.2.4 below) and prove the theorem_ Note will give thumbs up for cetailed answer. can also breakthese into separate questions necess

Question: Given the Least Upper Bound Property (definition 1.1.3 below), Proposition 2.1.10 below, the Bolzano-Weierstrass Theorem (Theorem 2.3.8 below), the Min-Max Theorem (Theorem 3.3.2 below), Definition 4.2.1, Lemma 4.2.2,and Rolle's Theorem (Theorem 4.2.3 below). Describe how these are connected to obtain complete proof of the Mean Value Theorem (Theorem 4.2.4 below) and prove the theorem_ Note will give thumbs up for cetailed answer. can also breakthese into separate questions necessany. Delinition An ordend sel has the least-upper-bound propenty if every nonempty subset that is bounded aboie has least upper bound, that SupP exists in $ Froposition LIO. miondrole SCGMece {xa} bounded if and only if it convergen: Furuermon i {X, onafon increasing bounded. Ihen sup{Xa eN} If {saa} montolone decreasing (d bounded then lim inf{Xn : n N} Theorem 2}.8 (Bolzano_Weierstrass . Suppose sequence {xn} of real numbers bounded Then there exists convergent subsequence {Xrs} Theorem 3 .2 (Minimum-maximum theorem) Let f: [0.b] Rbe continuous function Then achieves both an absolute minit (ld &n absolute Paxirm on [a.b]: Delinition +1. Let $ Rb a sel and let f : Rbe a funelion. The function is said MaVd 4relative #IcLLIDHANI at € = if thete exists = 0 such that for allx where have flx) < flc)- The definition of relative minimum AmToCOl Lemna 422 Lct f:l4.b] utriclon diferemtiable &l € (4.b). ard relulut naanneelee reltive maximum Ut €: Then Theonm 42} (Rolle). Let f : [a.b] + Rbe cOntimuons function differentiable on (4.b) Such ahat f(a) f(6). Then there exists & € (a.b) such thar f"(c) Theorem 424 (Mean value theorem). Let f : lu.b] Rbe FOHIomNuuclon diferentiable (0,b} Then there exists Fau GCWc f() f(e) f"(cl(b



Answers

Let f(x) = 4?x2.
Consider a rectangle with vertices(0,0),(x,0),(x,y), and(0,y),where the vertex (x,y) lies on the curve y=f(x),0?x?2(see the picture at right).Part I.(a)
Express the perimeter P of the rectangle as a function of x,P(x). (b)
Find the absolute maximum and absolute minimum of P(x)on the interval0?x?2.(c)
Find the dimensions of the rectangle with the largest perimeter.Part II.(a)
Express the area A of the rectangle as a function of x,A(x).
(b)
Find the absolute maximum and absolute minimum ofA(x)on the interval0?x?2.(c)
Find the dimensions of the rectangle with the largest area.

Yeah. So in this problem we need to sure that any bounded decreasing sequence of real numbers converges. So we are given about a decreasing sequence. We named that A. N. And we need to show that A. N. Can gorgeous to what limit? We don't care about this. Okay, so let's start with the proof, send the sequence in is bounded in particular. It does bounded below that is the set A. And such that and comes from natural numbers is a bounded subset five. Have we known that are has completeness property also known as and you'll be property that is least upper bound property. And this is equal into G. L. B. Property also known as greatest lower bound property, basically this means, but if any subset of R is bounded below, then it must have an infamy. Um So in fema of the scent and coming from and coming from national numbers is some real number. Let's call this L. So our claim is like the sequence N. Can we just to L now, since L is the greatest No, it bound after said of elements of sequences of the terms of sequences. So for any absolutely. Didn't see role as minus epsilon, which is strictly less than N. I'm sorry, L plus epsilon, which is strictly created. An L cannot be by lower bound. So there exists some. And in particular there exists some term A N. That is less than L plus of silence. But since in is a decreasing sequence we have a N less than equal to capital and for every and greater than or equal to capital. And so in particular we have and less than L plus a silent for every angry to an end no sense. And is can feel them. We have less than equal to a N. For every n natural numbers. In particular, l minus epsilon, which is strictly less than L, would also be less than a N for every angrier than capital. And so we have this inequality right here and this inequality right here and we combined them to find that for every epsilon created in zero, there exists a national number and such side for every end. After that comes after that natural number added minus upside and is less than a. In which is less than L. Plus of silent. That means A. In the sequence A in Canada. Just to L. This completes option for this term.

In this problem, we are given to subsets of our name, the S. And T. And we have the property that given us and T. S. Is going to be less than or equal to T. For every S. And S. And T. N. T. So we're first supposed to show that S. Is bounded from above. So in order to do that, we just pick a tea from tea. So it's like T. One B. From T. Then for every S. And S. S. Is less than or equal to D. One us founded above. Similarly for bounded below. For S. Let her T. Sorry, let S. One B. And S. Then for every T. And T. S. One is less than or equal to T. Us founded. Hello. We're now supposed to show approve that's the supreme um of S. Is less than the infamous um of T. So let's do this by contradiction. So suppose this is not the case. That's a pretty mom of S strictly greater than inform um of T. And this implies there exists some S. In S. Such that the S. Is greater than the infamous um of T. Furthermore there exists a T. Into such that the in form um of T. Is less than or equal to T. It's less than S. Because otherwise if that wasn't the case S. Would be the in form. Um Right. And we have picked that S. Is greater than the infamous. Um Yeah so that means there exists T. And S. Such. That T. Is less than S. But this is a contradiction because all asses are less than three. Quality us. The in form um O. T. Is less than or equal to the supreme um of us. So then we're supposed to come up with a couple of examples where we have S. Intersect T. Is not the empty set. And we can simply put this as well. S. B. 0 to 1 and that T be 1 to 2. All right. Every element of S. Is less than a record of every element of T. And the intersection it's not the empty set. We want. Now an example where the supreme um of S. Equals the in form um of tea but we want s intersex T. To be the empty set. So how can we do this? Well we can take our other example and just open it up. Sp 0 to 1. The open interval and let T. V. One two. The closed interval open interval. So the in form um of T. Is one, the supreme um of S. Is one and those are equal.

Okay, so the question says the graph of the function is given, we have to find all the local maximum and local minimum values of the function and we have to find the values that acts in which they occur. And we also have to find the intervals in which a function is decreasing and in which a function is increasing. So basically the local maximum, if you guys don't know. So on the left side side aftereffects is increasing and on the right aftereffects, Yeah, it's decreasing. So basically it's increasing on the left, decreasing on the right. Um Okay, decreasing on the right. And local men is the opposite. It's your decreasing on the left side and increasing on the right side. So let's find the local minimums and local maximums. Uh They are point coordinates. They are just and we we just have to say local men had X equals blank. Local max at X equals blank. So, and you can have more than one local minimums and you can have more than one local maximums. So uh I thi local men next to two. So as you can see the X. Values they increased by two. So at the coordinate for comma negative three. There's a look there's a local minimum it's a local men. So here's one of the coordinates negative four three. Is it 43 or negative 43? It's 43 I want to make that positive. So at that coordinate that papa that is a coordinate that's a local men and there's also local men for in the negatives. Um I believe it is negative six comma negative five. So negative six com on negative five. So those are the two local men's. Now let's go for the local maximums. Mhm. So local max. You have zero comma one that's one of them. And you also have 2468 10 comma three 10 comma three. Um Yeah so let me just check my answers. So here 01 10 3 negative six negative five and four negative three. That is what I had except I accidentally put 43 instead of four negative three. My mistake. Okay now we have to find the intervals in which the function is increasing and on which the function is decreasing. So we have to use interval negotiation here. So I'm just gonna put eye and see and the function is increasing at the intervals um negative six negative six comma Okay zero just X values and union. And where else is increasing? Four comma 10 four comma 10. So these are all using parentheses because these do not involve endpoints. These are not in points. So these are this is not the farthest to the left of the farthest to the right. Then you would use a bracket, bracket in involve endpoints parentheses. Anything else? So that's the interval in which is increasing. Let's see what this person did. A negative 6 to 0 and 4 to 10. However I would strongly advise them to use parentheses um for 10. Now let's figure out the intervals in which it is decreasing so decreasing. Okay so negative two negative four negative six. Negative eight negative 10 negative 12. So that's the first interval. So so that's the first endpoint. So negative 12 comma. Then where does it stop groups? Where does it stop decreasing? Stops decreasing it negative six. So we have to say negative six but we have to put a parenthesis. So brackets at endpoints parentheses at any other point. So Union and then oops again zero comma four zero up zero comma four parenthesis. Again. Union. And then we have to use um 2468 10, 10. Okay, so 10 to 16. So 10 comma 16 and bracket. So that is our answer. Let's see what this person did. So they did 12 to native six. Pulling negative 60 to 4. The other 4 10 to 16. However, what they did not do is they did not indicate the brackets. So that was the mistake here. So because it's at the interval notation, and when you're talking about interval notation, you have to use brackets.

Right. So in this problem we have the picture of a part of a parabola and it is defined by f of X equals four minus x squared. And then we're fitting a rectangle inside the parabola and it's going to have boundaries of the X axis and the Y axis. And the location where the one vertex of the rectangle meets that upside down parabola is called X. Y. The point down here is X zero. The point here is 00 And the point right here is going to be zero. Why? And the first thing we're going to do is we're going to write the perimeter of the rectangle as a function of X. We're then going to calculate the absolute minimum and maximum of that new function and then we're going to determine the dimensions of the rectangle with the largest perimeter. We're then gonna start all over and we're going to express the area as a function of X. Find the absolute minimum and maximum of that new function and then find the dimensions of the rectangle that yield the largest area. So let's think about what the definition of perimeter is. Definition of perimeter is to go all the way around the rectangle. So if I was to draw the rectangle and I knew from here to here we went X. That means our um base or length of this rectangle is X. And I know we went from here to here to find the height of that rectangle. So we're going to call it why but why is actually defined right here? Why is the same thing as for minus X squared? So therefore for part A we could say the perimeter of this rectangle is X plus four minus X squared plus X plus four minus X squared. Because we're going across up across and down to find our perimeter. So the perimeter, when I combine all my late terms would be negative two X squared plus two X plus eight. So that's part a part B is asking you to determine the absolute minimum and absolute maximum of that function on the closed interval from zero 22 And that's going to lead us to the extreme value theorem. And the extreme value theorem says that a continuous function must obtain absolute minimum and absolute maximum values on a closed interval. Yeah, it also says these extreme values. Mhm. Yeah. Are obtained either on a relative extra MMA or on the endpoints of the interval. So we have the end points of the interval. Those are here zero and two. So what we need to do is we need to find any relative extra MMA that's in that interval as well. So in order to find relative extra MMA we are going to take the derivative of our function. And the excrement occurs either at where the derivative is zero or fails to exist. So we are going to do the derivative of our perimeter function. So we're going to start with P of X equals negative two X squared plus two X plus eight. So the derivative of that would be negative four X plus two. So in order to find the relative extra MMA we are going to set the derivative equal to zero. So we get four X equals two or X equals one half. So now I'm going to set up a chart utilizing my three values of X. My minimum of the interval, my maximum of the interval and the relative extra money I just but shaved. So my perimeter formula again was negative two X squared plus two X plus eight. So I'm going to substitute zero in and I'm gonna get negative two times zero squared plus two times zero plus eight or eight, negative two times a half squared plus two times one half plus eight. And that's going to give me 9.5. And now I'm going to put the two in so I'll get negative two times two squared plus two times two plus eight and that's going to yield a four. So therefore our absolute maximum of Pf X will be 9.5 because this is the highest number and are absolute minimum of P. Of X is for and let's do part C. Now part C is asking you to find the dimensions of the rectangle that yield that largest perimeter. So if you think about what our rectangle looked like, we said this was X. And this was four minus X squared. We just found that X being a half gets us that largest perimeter. So that means mhm. That this is one half and this would be four minus one half squared, or four minus a quarter, leaving us three and three quarters. So our dimensions of the rectangle, yielding that largest perimeter will be one half by three and three quarters. Yeah. And then sure enough, if I do that, if I say one half plus three and three quarters plus one half plus three and three quarters for this side, I will get that area. Now we're ready to start part two of this problem. So we're gonna start with that same image again. So we started with the X. Y axis, we had that part of a parabola again and we had our rectangle fitting inside. And we said this point here was X zero. This point here was X. Y. This point here was 00 at this point here was zero. Why? And I'm going to pull that rectangle out of the picture. So if I were to measure from here to here that my base would be X. And from here up my height would be why? And because I have the function defined as four minus X squared, then I could say that why is the same thing as four minus X squared? So for part A you need to determine a function A. Of X. That represents the area of this rectangle in the area of a rectangle is found by doing length times. With so therefore we will have X times four minus X squared. Or if we distribute that X in we could say four X minus x. Cute! Now we're ready to talk about part B, part B is again applying that extreme value theorem, we need to determine the absolute maximum an absolute minimum and again just to reiterate what the extreme value theorem says is that on a closed interval any continuous function will always have absolute minimum and absolute maximum values. And they will occur either at the end points of that interval. And remember our endpoints of our interval are from 0 to 2 or at any I'm sorry at any relative extra MMA So to find our relative extreme A we are going to have to take our function and take its derivative. So our function is four x minus X cubed. So we're going to find the derivative of that which would end up being four minus three X squared. And we will set that equal to zero. So we could say three x squared equals four X squared equals four thirds X equals plus or minus two over radical three. And when we rationalize that we get two RADICAL 3/3. Now, because of our defined interval from 0 to 2, we are going to end up only using the positive version of this because the negative version doesn't fall in that interval. So again, we're going to utilize a chart. Oops. There we go. And we are going to put our X values the endpoints zero and two along with our relative Ekstrom A of two RADICAL 3/3. Our area formula is for x minus X cubed. So we're going to say four times zero minus zero cubed. And we're gonna get a zero. Well then do four times two radical 3/3 minus two RADICAL 3/3 cubed. And I'm going to just move this to down a little bit that I can show you the work or the solution to this. So we're going to end up with eight radical 3/3 minus eight. Radical 27/27. And Radical 27 will simplify and clean up a little bit. So we can then turn this fraction into 24 Radical 3/27 which again that can clean up a little bit more. If we simplify the 24 27 we will get eight RADICAL 3/9. And when I clean that up, I end up with 16 RADICAL 3/9. And now I'm going to use the other end point to and we will get four times two minus two cubed, which will end up being zero as well. So therefore our absolute maximum of a X. Is 16 Radical 3/9 and our absolute minimum of the function f X would be zero. And finally we're going to do part C and part C. In this problem is asking you two find the dimensions of that rectangle that has the largest area. So again, let's draw our rectangle. We said that this was X and this was for minus X squared and we just found the largest area occurred when X was too radical 3/3. So that means here is too radical 3/3 and here would be for minus two RADICAL 3/3 squared, which would be four minus four times 3/9, which would be for minus four thirds, which would end up being two and two thirds. So the dimensions of the rectangle with the largest area is going to be two radical 3/3. Bye two and two thirds. And that's going to yield the largest area of the 19 radical three over, sorry, the 16 radical 3/9. So, I hope this solution process was helpful.


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