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Sketch the graph of a continuous function f satisfying 1. f' (x) > 0 if |zl < 2 and f' (x) < 0 if |z] > 2. 2_ f' (-2) = 0. 3 Jin If'...

Question

Sketch the graph of a continuous function f satisfying 1. f' (x) > 0 if |zl < 2 and f' (x) < 0 if |z] > 2. 2_ f' (-2) = 0. 3 Jin If' (c)| =0 4 f" (c) > 0 if x # 2.

Sketch the graph of a continuous function f satisfying 1. f' (x) > 0 if |zl < 2 and f' (x) < 0 if |z] > 2. 2_ f' (-2) = 0. 3 Jin If' (c)| =0 4 f" (c) > 0 if x # 2.



Answers

Sketch the graph of a continuous function $f$ that satisfies the given conditions. $$\begin{aligned} &f(0)=1 . f(2)=-1 ; f^{\prime}(0)=f^{\prime}(2)=0, f^{\prime}(x)>0 \quad \text { for }\\ &|x-1|>1, f^{\prime}(x)<0 \quad \text { for } \quad|x-1|<1 ; f^{\prime \prime}(x)<0 \quad \text { for }\\ &x<1 . f^{\prime \prime}(x)>0 \text { for } x>1 \end{aligned}$$

Mhm. Yeah. Yeah, Just

Mhm. And this problem they That's just to find a function that satisfies all of these conditions. I didn't actually find an exact function for this, but it should be if you get a polynomial of high enough order, you should be satisfied everything. So they only they only asked us to sketch it, so they didn't ask us to actually find the function. So we want FF two minus two to me. Uh Six. So here's minus two, and here's six. So I kind of went off here a little bit. Um and then f of one equals two. So here's one, here's two After three equals 4. Okay, and then F prime at one and three are zero. So we need we need a horizontal tangents at this point at this point, so that's satisfied. And now um in in the range in this range between one and three, we need to have uh Between between one and 3 we need to have positive slope. So again we have to have this somehow that this comes and then this outside of that range, we need negative slope. So we have to have something like that. Um Then we we also know that, let's see here, the curvature needs to be negative Outside of Whenever X is greater than two. Yeah, And then when X is between -2 and one. Okay, all right, so we need negative curvature is here, we need positive curvature between zero and 1012 here, this and then positive curvature Whenever X is thus the -2. So we have this, so I have an inflection point here, an inflection point here and an inflection point here. Um This they basically said that the slope needs to be positive here, so this can't be, you know, we don't have something that looks like this. Um It has to have positive slope there with Yeah, I guess so. Again, if you get a high enough order polynomial, you could find a function that fits all this stuff. But I didn't go through that exercise.

Yeah. And this time we're given a bunch of conditions um We're also told I guess I didn't write this down that death is an even function. Um That was pretty obvious from the other conditions that they gave us. So the first derivative is zero at zero. The function value. The function that evaluates to one at zero plus or minus two, it's one half. Um and because I already said that one and then that plus or -2, the second derivative is zero. So if we start off with an even function We really have three conditions to satisfy um You know F double primary two is zero. F prime of zero is zero and if At two is 1 half. Mhm. Mhm. Um Well No and FF- 00. This one kind of comes for free because we haven't even function um that at least has doesn't go to doesn't go to infinity and zero. So this one kind of comes for free because if we have an even function then we know that the derivative is an odd function so that when it comes for free. So I said it probably looks something like this, right? Um I thought okay maybe we'll probably have an ass until here. I suppose it could go off to infinity here. There's nothing including that. But I thought okay, it looks like maybe a rational polynomial. Um That would do something like this. So I said okay we need we have three unknowns. So I said okay let's make something like this palm. We can get something that looks roughly like this with this kind of form. And then we need to shift it up or down to make it satisfied the, you know, these requirements. So I said, okay, let's look at the over X squared plus A plus C. And so we take, you know, a first derivative of the second derivative. And we said, you know, make sure this is satisfied, this is satisfied and this is satisfied. That gives us three equations and A B and C for the for three unknowns. And solving those isn't too difficult. And we get a S. 12 B is 24 C is minus one. So what we get is 24. Mhm. All over X squared plus 12 minus one. Or we can write that this way is 12 minus X. Grad all over X squared plus 12. And so we can see here that we're in X0, you didn't get one. Um you can check that the second derivative at zero, we have inflection points here and here and they evaluate to one half and then this happens to us until 2 -1. Um But that didn't they didn't tell us anything about how this goes off. You know how this goes to infinity. So there there is a there is a function that satisfies our requirements

All right, this is chapter two section for problem number 18. And for this one, uh, we're just required to sketch a function. Um, and for this function, we want d ah, first derivative of the function to always be positive for the second derivative toe. Always be negative when X is less than two. And to always be positive when X is more than two. Um, so for this first derivative, that's not anything. Doesn't put any huge restrictions on a graph. It just always has to be increasing. So it could look like that line. It could look like that. It could swoop downwards like that. There are a lot of options for this one. Now, for the second derivative, it just means it has to be negative. So going downwards, less than two positive. So curving upwards. Greater than two. Um, except in this case, we can't have these segments that I just crossed out because during those segments, um, if the first derivative is negative and first we want the first rivers. You positive? So, actually, if I redraw those, we're here. We can't have that. We can't have that. So if we put our two more right there. Okay, We have this positive section of the curve right here. We can drag that over and put it to the right of to. And this satisfies our thing so far, because the first derivative is positive. It's increasing. And the second driven to this positive because it's curving up and X is greater than two. Um, and we can take this other part over here and put this on the left of to right, because this is a negative second derivative, because its curving down, which is good because it's less than two, and it is still increasing as well. It never has a negative slope. Um, so this is a function that fits the criteria were looking for, um, So there you have it again. It's always positive and increasing. Um, the second derivative. His negative blow to and positive above two. Here you go.


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