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Show that f(s.u) =4+y' has infinitely many critical points (all along 0) and that although it is elear that each one of them is global Mini_ tlu Sxold derivati...

Question

Show that f(s.u) =4+y' has infinitely many critical points (all along 0) and that although it is elear that each one of them is global Mini_ tlu Sxold derivative test is inconclusive for cach one.

Show that f(s.u) =4+y' has infinitely many critical points (all along 0) and that although it is elear that each one of them is global Mini_ tlu Sxold derivative test is inconclusive for cach one.



Answers

Show that $f(x, y)=x^{2}$ has infinitely many critical points (as a function of two variables and that the Second Derivative Test fails for all of them. What is the minimum value of $f ?$ Does $f(x, y)$ have any local maxima?

Two problem ese showed that affects wise. They're goingto x squared us all. I square minus or ex wife has to has an infinite number off critical points. Andi at thes e go to zero at each one, then showed that half has a local minimum at each critical point. First we compute pasha laugh tracks. Why, Max, this is two ax minus four. Max, why is zero to eight? Why minus four Laks, If we light also with them you got two zero half X is equal to you are with was employment to life? Why? Why come be any number? House appoints practice at the critical point of affects. Why? Then we compute at black slacks. This is a one to two facts lie. It's secret Connect four After Why why would you write So the sequel to to come straight minus negative four square. So this is equal to zero before reflects why we know this is equal to your ex minus to life a squared plus two. This is always going to them too on one x equal to to lie. The function has a local minimal. This proof that it has a local minimum at each critical point

Yeah. Here we are given a function F of x Y equals x squared plus four, Y squared minus four X Y plus two. We are asked to show some things about it. We want to show that it has an infinite number of critical points and the indiscriminate zero at each of them. Then we want to show that at each critical point despite being inconclusive and the second derivative test, we have local and absolute minimum. And so first we have to find the critical points. So let's go ahead and set are X and Y partials equal to zero. The x partial will be two X minus four Y. And the y partial will be eight Y minus four X. And so we see that we can kind of solve this system of equations here. Um This is equivalent to saying X equals two. Why? And if we were to plug this into the second term we would have eight y minus for times two. Why equals zero or eight? Y minus eight. Y equals zero. Which of course gives us zero equals zero. And so this shows that there are an infinite number of critical points that lie along this lying X equals two. Why that is if you plug in any point of the form a comma or X equals two wise. So it would be to a comma a. Then this is going to give you a critical point. And so we know that this is the line along which they follow because that's the substitution that we make in order to um achieve this, zero equals zero. And so now let's go ahead and find our second derivatives. So we can do the second derivative test on these infinite critical points. So F x x is just too F Y Y is eight and the mixed partial F X Y is four or negative for rather. And so plugging in are critical point to a comma a to get the F X. X. Y. Y. Mixed partial the discriminative it which is given by fxx, F Y Y minus F X Y squared. And then our conclusions. Okay, so here fxx it's of course to F F F Y Y is eight, F x Y. Is negative four. And so the discriminative is 16 minus four squared which is zero. And so as the question told us, uh this is inconclusive. No. And so we cannot conclude anything about these points based solely on the second group they've test alone. However, um if we look back at the original function, there is some clever manipulation we can do so we will just rewrite it here for clarity. Ethics, Y equals X squared plus four, Y squared minus four X Y plus two. Yeah. And so if we imagine as kind of separating mhm this part off from the rest of the um function the plus two. In an effort to kind of factor this um we can actually factor this into x minus two Y squared and this gives the X squared, this gives a minus for explain, it gives the plus four Y squared. So we are good and then we retain this plus two on the outside here. And so because this squared term here is always going to be greater than or equal to zero because you can never have something square to be a negative number unless it's imaginary, which we're not dealing with here. What this means is that the function, the value of the function can be at most or at or at least to the function never is less than two because we always have this constant plus two being added to something that's greater than or equal to zero. And where do we achieve this minimum value of two? Will we achieve it When x minus two, Y equals zero. Or in other words when we are lying along the line, X equals two. Why? Which is just the line that are critical points are lying along. And so what this shows is that the critical points achieve the minimum value of function? Yes, I think mm F x Y equals two and Mhm. Um So we can conclude that they are of course both local. Yeah. And absolute minimum. And yeah. So despite the second derivative test being inconclusive by manipulating the original expression of it, uh To note this fact here, we were able to show that they are local and absolute minimum for the function.

For a given person. We were take. Casey will do needle by putting case equals zero Exercise a little X squared plus y school must take part surgery radio with respect to X in this case, variable wise question for FX physical do wicks Now take are surely would you respect white? In this case, variable X is constant for a bicycle. Go away now to find the political points Take affects people to zero and f y is equal to zero So what do your taxes? Your ex is going to zero gives access it, you know And if by is too white for Isaac zero views quiet before zero so over. Critical point is 00 When daisy do deedle Now when ke is not equal to zero would think partial delivery to with respect Rex in this case variable wise constant. So, Daddy, radio X squared gives two weeks David, you off cakes by gives anyway and very rich you squeeze zero because variable wise constant So if X is equal to X plus key away now take partial daily work you with respect Why in this case variable X is constant. So if Isaac will do Okay, X plus two away. Now, to find a political points take F X is equal to zero and therefore is equal to zero. Was we take we express care Bicycle boo zero gives why Isaac would do negative to expect. Okay, now take f y is equal to zero. So okay, exposed white people. Do you now put the really off white in this equation? By simply five days, we get X is equal to zero and Casey called plus or minus two. Now put the really off X and kay Why? So you're wise equal to negative Wickes by cake We have returned. Why's it called negative baggage? Put X is equal to zero and a reminder to wise you know So again we have 00 is the only critical points when candy is not equal to zero. In any case, 00 is or do you think Thank you


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