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Given fcxsw) = 2x2+2y2- #X-5 (A)fial the criticAl pointc 0f fva) mh hu Xy-phne by folvigg VftxwJ- & (B ` ue the Lagr € Ange multiplier methol to frnd +he Onftralned: Critical Pointf of f(x)) on 1e crrc / C ; X+3 2 =/6 bo 9($) Solving Vfixi)= ^ Va (pd) x2+92 =16(c) fma t WIAR cd mm Valuef of fio3 On Kha ciacle X 1+97=16 amdl stnte na Pohtc where here va Wer occur (D) fial the hax and min Va luej of f(yy) On to difc x2+y7 < 16 aA State n Jomt Where mee Va luer 6(( Ur_NOTE PoL blem 0 ku ov

Given fcxsw) = 2x2+2y2- #X-5 (A)fial the criticAl pointc 0f fva) mh hu Xy-phne by folvigg VftxwJ- & (B ` ue the Lagr € Ange multiplier methol to frnd +he Onftralned: Critical Pointf of f(x)) on 1e crrc / C ; X+3 2 =/6 bo 9($) Solving Vfixi)= ^ Va (pd) x2+92 =16 (c) fma t WIAR cd mm Valuef of fio3 On Kha ciacle X 1+97=16 amdl stnte na Pohtc where here va Wer occur (D) fial the hax and min Va luej of f(yy) On to difc x2+y7 < 16 aA State n Jomt Where mee Va luer 6(( Ur_ NOTE PoL blem 0 ku overlaps wfti exectire 22" 'f </4.8 On P1017



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For $f(x)=x^{3}-k x,$ where $k \in \mathbf{R},$ find the values of $k$ such that $f$ has a. no critical numbers b. one critical number
c. two critical numbers

For this problem, we want to find the critical points of a few functions determine whether or not whether they are max men or neither and determine whether or not the tangent line will be parallel to the horizontal axis at that critical point. So our first function is negative six X cubed plus 18 X squared plus three. And that is labeled as for this problem one moment. That is labeled as H. Of X. So taking H prime of acts, we will get just 36 x minus 18 X squared having that. Uh we have a common factor of uh in fact it's a common factor of 18 X between the two terms. So we can actually factor out a negative 18 X and write this as negative 18 x times x minus two. Which tells us that we'll have a critical point at X equals zero and a critical point at X equals two. Now, since we have a negative leading coefficient on an odd polynomial, that means that will essentially have double negatives when x is less than zero. So we'll be coming in positive to the left of X equals zero. Then turning too negative in between zero and two. Then turning to positive again. Now that means that we'll have, oh excuse me. You should be careful here. We have on our derivative. We have a negative leading coefficient on an even uh even um polynomial, even degree polynomial. And so that actually means that we'll be coming in negative turning to positive then turning too negative again. So what that means going from negative to positive means that we have amen and going from positive to negative means that we have a max for the second part of the problem barbie we have GFT equals T to the power five first eq It means that G prime of T. Is going to equal five T to the power of four plus three. T squared can factor out the t squared from the two. We have t squared times five T squared plus three. This will only ever equal zero when T equals zero. So we have a critical point where T equals zero. Here, we have everything positive in an even degree polynomial, which tells us that we won't have a max or amen we'll be going from positive derivative to positive derivative, but we still have a critical point there at T equals zero. And because of the because of the fact that it is corresponding to a derivative equal zero critical point, that does mean that we will have a tangent line parallel to the horizontal axis there. For part C. We have the function why equals x minus five to the power of 1/5. Which then in turn means that y prime is going to equal by applying quotient rule will find that Y prime is going to be 1/5 times x minus five. The power of 4/5. Now this has no point where it will equal zero, but we do have a point where the derivative will be undefined when X equals five. So here it will, the tangent line will not be parallel to the horizontal axis, it will be perpendicular than for part D. We have the function F of X equals x squared minus one to the power of 1/3. That means then that f prime of X. Again, we're applying the in this case will be applying the chain world, we'll have to X over three times X squared minus one. The power of to over three. Now this will equal zero when X equals zero. So we have a critical point there. Oh, we'll also have that coming in, we'll have negative values. So we'll be switching from having negative derivatives to positive derivatives, which means that we'll have at X equals zero. That will be a minimum, and it will be parallel to the horizontal axis and we'll also have a critical point at X equals plus or minus one. Now those will be a critical points where we are perpendicular to the horizontal axis. The tangent line will just be a straight line there. So since they are perpendicular to the horizontal axis, those are neither maximum nor minimum.

Okay now we've got the function part A F of X equals X cubed minus six X squared plus X plus two. Okay. And we've got to find the critical values. So we gotta take the derivative of this three X squared minus 12 X plus one. Set that equal to zero. Um And I'm going to try to factor this three X something something. Uh huh. 12 It doesn't seem to be working out. So let's just use the quadratic formula X. Equals negative B plus or minus the square root of B squared, which is 144 minus four times a times C. Which is just three over two times three. 144 minus 12 is 100 32. Right, so that's gonna be 12 plus or minus the square root of 132 over six. Right now 132 divided by four is 33. So I can simplify that to 12 plus or minus to root 33/6. And now we can simplify this uh 26 plus or minus square root of 33 over three. Let's just make sure that did this right? Three x squared minus 12 x plus one. Okay so 12 plus or minus the square root 12 squared minus four times three over two times 34 times three is 1244 minus 12. Okay so we've got six plus or minus the square to 33/3. All right, so now we need to check f of what's the interval for this thing, negative 1 to 5. Okay well six minus the square root of 33. That would be in the interval six plus the square root of 33. There's about six plus six is 12. 12 divided by three is four. There are both in the interval. So we're gonna need to do f of negative one f of six minus the square to 33/3 f of six plus. Why is it not writing So square root of 33/3? It does not like to write right now. And the last one was five. Okay I'm just gonna go ahead and graph this and get these numbers off of the graph. So X q minus six X square. Let's X plus two. Okay. So we've got a critical point at. Yeah. Okay. Okay. So yeah I found the two critical points. The one critical point gives a value of two and the other critical point would give a value of negative 26 point 042 Trying to get that in there. Um Those are the extreme so we don't really need to check negative one or five. Now. For part B. What if we were doing the absolute value? Well the minimum value would occur uh whenever the function equals zero so the minimum would be zero. Um We would have critical points where the function equals zero. So that would be X squared minus six X squared plus x x squared minus six X squared plus X plus 12 equals zero. Yeah I can grab that and X cubed and find where it equals zero. So the added critical points would be what's her interval negative 1 to 5 negative 0.484 Um 0.717 Which I think is the square to 2/2 and five point 766 But that's not in the interval. So we would have to add these critical points. So half of these points zero now would be the minimum value and then the maximum value is just gonna be the absolute value of this or where was the other one? Uh This but obviously 26.42 is much larger to the max will occur at six plus the square root of 33/3. And it's going to be 26.42 because everything is just going to be changed to positive since it's the absolute value. Mhm.

Okay, so let's start shaking. Are critical points or finding are critical points. So we re taking preservative. So we have four times six. That's 24 except for three minus 16 times three. That's 48 x squared 45 times to that's negative 90 x and in close to 54. So we're gonna set this equal to Darryl and factor are following Ono meal. Okay, so we see that we can practice into we contract out of six and then we have, um two X minus one a two x plus three and Leslie, it's minus one or expert mystery. Okay, so we have three critical points. Um, that's when it's most amount, actually. So, Well, estar en points as well. So stuttering. What? Make it a five. And then we have, um X is equal to negative 3/2 X is equal to 1/2 exit with drink and X equals five. And now let's plug in these X points into our function after backs to see our puppets. Okay, so we see that we get the following outputs for our ex terms. So we see that our absolutes Max, is when X is equal to negative five and our absolute minimum. That's when X is equal to negative 3/2

In disclosing. Well, even the function f thanks. I think a Jew exclaim, Let us tram! Thanks less than 36 you find it pretty And the critical numbers. When did you find the AARP? Graham? The next now and by the power was rendered Teoh. Thanks. I was drowned here and we said that epidemic all Jews zero. And therefore again the next week you called your trout evenly. But you never to six. And we found a critical numbers. Yeah, And now we used a secondary. If the test to in clarifying every sinner, it is a maximum or minimum. So the F number Bremner Thanks. We could you chew and then funds and embarrassed that I remember brown on the critical numbers here you quoted to and it will be created and zero. And by the second the roof. The test doesn't implies that it will f X has no a relative minimum at x A good you six


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