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Slo rolnts) suppose the function €(T) bloodstre-n 6f 4 canine / minutes 056t € alter A unjeclon madels Ibe comcentratlan ot # certuin drug In Writc down...

Question

Slo rolnts) suppose the function €(T) bloodstre-n 6f 4 canine / minutes 056t € alter A unjeclon madels Ibe comcentratlan ot # certuin drug In Writc down formula tor Ck(} Locate any critical values for C{tL Show Your work!Ucethe First Deritautive Test to locate Any local extreme valuas for C(th, or Indicate that they do nof exist Showyourwork Using the number Iine belowxRclutite muXat $ =Rclative min Jt $ =4. (6 points) Consider thc futiction _ fler-3r-24r+30.

Slo rolnts) suppose the function €(T) bloodstre-n 6f 4 canine / minutes 056t € alter A unjeclon madels Ibe comcentratlan ot # certuin drug In Writc down formula tor Ck (} Locate any critical values for C{tL Show Your work! Ucethe First Deritautive Test to locate Any local extreme valuas for C(th, or Indicate that they do nof exist Showyourwork Using the number Iine belowx Rclutite muXat $ = Rclative min Jt $ = 4. (6 points) Consider thc futiction _ fler-3r-24r+30.



Answers

Find all critical points and then use the firstderivative test to determine local maxima and minima. Check your answer by graphing. $$f(x)=3 x^{4}-4 x^{3}+6$$

Let's take a look at function p as shown next to this exercise, you can see that there are three local extreme points. I have point A B. I have point E t and I have point C d. Now which of those air maximums in which of minimums maximums happened at the top of the peaks of the mountains? That means that A, B and C D are both going to be local maximums. They're not necessarily the overall maximum point of the entire function. Just within its particular vicinity, it is the maximum point in its locality. E t is a local minimum. It is the lowest point in its surrounding area, the valley in between the mountains. So function P has to local maximums and one local minimum.

Alright for this problem we are asked to find all critical points of the function F. Of x Y equals E. To the power of negative x squared plus y squared minus four Y. And then indicate whether each point is a max min or a saddle point. So we start by taking the partial derivatives with respect to X. Partial with respect to X. Is going to be negative two X times E. To the power of negative X squared plus y squared minus four Y. Ah By applying chain rule there. Uh Then partial with respect to why is going to be a negative two, Y plus four times E. To the power of negative X squared plus Y squared. Why squared minus four. Why? So we want both of these two equals zero, which tells us that we need X two equals zero and we need uh to Y two equal four, which means we need y two equal to. So our point is going to be 02 next to determine whether it's max min or saddle. We take the second partial derivative, the second partial derivative. This would be applying the product rule here uh on the first partial with respect to X. The second partial with respect to X is going to end up being four times X squared minus two times E. To the power of negative X squared plus y squared minus four Y 2nd partial with respect to why is going to be two times 2, y squared minus eight Y plus seven. Seven. There times E. To the power of negative X squared plus y squared minus four Y. Lastly we need not quite lastly, but next we need to take the mixed derivative which is going to come out to four x times y -2 times E. To the power of E. Or it's the power of negative X squared plus y squared minus four. Y. So d of 02 is going to be equal then to, let's see here will be D of X while right down first going to be four X squared minus two times. So week two times works were minus two times two, Y squared minus eight. Y plus seven times E. To the power of negative two, X squared plus y squared minus four Y minus are mixed partial squared. So that's going to be 16 X squared plus y minus two squared. And eat the power of negative two X squared plus y squared minus four. Why? So the last thing that we need to do is evaluate D of XY at the .02 Which will give us two times negative two times two times four minus eight times two plus seven. Yeah. Times E. To the power of negative two times for minus eight, then, because the X there is going to be zero, uh that last term is just gonna drop off and I'm just gonna calculate this off screen here. All right, we get a final answer of four E. To the power of eight. And the second derivative with respect to X is going to be negative there. So that tells us that since we have negative second derivative with respect to X positive D. That we have a max.

For this problem. We want to find our local extreme points. So we're gonna look at the derivative graph, and we see that the local extreme point will be located at 0.0.368 Um, and if we graph the function, we see that in fact, at 0.368 there is a local minimum.

Alright for this problem we are asked to find all the critical points of the function F of x Y equals x, Y squared minus six, X squared minus three Y squared and then determine whether the function is a minimum maximum or saddle at those points. So we start by taking the first partial derivatives which will have why squared minus 12 X. We want that to equal zero, which means that we have y squared needs to equal 12 X. Or Y needs to equal plus or minus the square root of 12 X. Then the partial derivative with respect to why it's going to give us two xy minus six, Y is equal zero. So we can factor at the Y there. Why times two x minus one needs to equal zero. So this is going to equal zero when either Y equals zero or when two x minus one equals zero. So that tells us that we need to x to equal one or X needs to equal one half. Uh huh. So, what that tells us then is that we have three different possible candidates, candidates. So when X equals zero, Or I'll be a little bit careful here, 1/2. So we'll have a solution when x equals zero because when X equals zero, actually, yeah, when X equals zero, then route 12 actually equal zero, which means Y equals zero. So we'll have both of those. Um Both of those will equal zero at 00 And then we'll have the points 1/2 and positive route six as well as one half And Negative Route six. So, having that, we now need to figure out the nature of the function of those points, which we do by taking the second partial derivatives. So the second partial derivative with respect to X is going to be negative 12. Second partial derivative with respect to Y. It's going to be two X -6. And the mixed partial derivative uh is going to be uh to Y. So D of X Y Is going to be negative 12 Times two x -6 -4 Y Squared. Now, all we need to do is calculate an hour D function there at each one of our points, which I'll calculate offscreen. Alright, so at 00 we should get 72. At one half positive route six, We should get 36 and also at one half negative route six, we should get 36 as well. So we also have that the second partial derivative with respect to X is less than zero. So that means that at each one of these points we are going to have a let me double check here. So I don't say something dumb from theorem. See that means that we are going to have a maximum.


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