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Jeffery W.

If $f(x)=x+\sqrt{2-x}$ and $g(u)=u+\sqrt{2-u},$ is it true that $f=g ?$

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Fangjun Z.

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Simon E.

Felicia S.

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all right, so we want to find the local Maxima local minima, and then any saddle points of this function. F x y is X squared plus y squared plus y squared X plus four. And I recall that are candidates for local extreme or saddle points, for that matter are the points for the Grady int is equal to zero a k a. We want to find the critical points first and so recalled it. What I really mean by saying the grading of ethics equal to zero is it? The Grady int vector is zero. So that happens exactly when the partial with respect to x zero and the partial with respect, why is zero? So what's the partial with respect to X? Well, that's going to be two x plus. Why squared? And I'll set that equal to zero and then the partial with respect to why is to why, plus to why X and I said that equal to zero. So this is something that's gonna play a central role. I mean, here and also in the next topic, solving systems of equations like this. So we want to be really careful and kind of do some factoring and look case by case. Um, So what we have here? If I factor out of two, why have to y and then one plus x zero? So here, either y zero or X is negative one. So if why is zero? Well, then, if I plug in zero for why I have two x zero. Well, that means that x zero. And then what about if X is equal to negative one? Well, I plug a negative one for X and I'll add over two. And so I'll have y squared is equal to two. So then I have why is equal to plus or minus square it of tittle? Okay, so my critical points are going to be 00 and then minus one and then plus or minus root, too, for life. Okay, so now what I want to do next is actually find this test function deep. And so in order to do that, I need to find the remaining partial drift. This So the second partial driven this. So I need f X x. Well, that's going to be the derivative of this. So that's just going to be too, and then f why Why is going to be to plus two X and then f x y will be the same thing as FBI X, which is to y And so if I evaluate adds se 00 and remember that this d function we write it up here. We're down here so d is going to be f x x times f Why? Why? Minus f x y swear. So I'm just going to take this function and evaluated at zero. So that's going to be too times two minus zero. So it's gonna be four, which is greater than zero. So that tells me I either have a local max or a local men. And now, because F x x of 00 is two, which is greater than zero, I can conclude that f of 00 which is four is a local men by my second derivative test. So I d 00 is positive. My first derivative Sorry. My second derivative of X, evaluated in 00 is positive. So this point is where a local man occurs that local men is actually for okay and then now let's do the same thing for negative one plus or minus Root, too. Now I'm doing this because we'll see that actually, the same thing will happen. So if I take f x x f y y fxx is too f y y is two plus two times negative one, which is zero. And then what do I do next? Well, okay, so it's gonna be zero minus if I plug in, plus or minus route to and for. Why here? I'm gonna have a number squared. So this is going to be well in both cases, so it's gonna be four times too. So it's gonna be eight. So negative eight. Sorry. So, again, that's coming from if I plug in route to or minus root, too Here, that's going to be eight when I square it. But I'm subtracting it. So this was zero and I subtract, get negative eight, which is less than zero, which means I have saddle points so negative one plus or minus. Root too, are saddle points. Okay, and that's it. So I have one local minima here that for I had to saddle points at minus one have plus or minus root too

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