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Describe how the graph of $f$ varies as $c$ varie…

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Problem 62 Hard Difficulty

Describe how the graph of $f$ varies as $c$ varies. Graph
several members of the family to illustrate the trends that you
discover. In particular, you should investigate how maximum and minimum points and inflection points move when $c$ changes. You should also identify any transitional values of $c$ at which the basic shape of the curve changes.
$$f(x)=x^{3}+c x$$

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Calculus 1 / AB

Essential Calculus Early Transcendentals

Chapter 4

APPLICATIONS OF DIFFERENTIATION

Section 4

Curve Sketching

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Video Transcript

we need to investigate what happens when we're given the equation. Effects is equal to X cubed minus C X and what happens to our equation, most notably our critical points and our point of inflection when we change the value of C. So to find that I will talk about its first derivative and its second derivative. So it's first derivative weaken. Just take the derivative of our equation. That is three X squared minus c. So we can see that the first derivative does depend on the value, See, But our second derivative will be six X So our second derivative, our point of inflection is just X equals zero. So our second derivative does not depend on our value of C, but if we're talking about, are normally equation and our first derivative than it does depend on it. So let's say that our extreme, our critical points. So you said this equal to zero. Who would then we would solve four. See, So we would have you would put move, see to the other side so that a B C's equal to three x squared we was all facts sorry and then X we can get one or two values the square root of sea over three, plus or minus the square to see over three, and it would determine you can use this as and as a endpoints of interval and test out if dysfunction if see if this point. But this X value is actually a maximum or a minimum. So see does have some wait on the on the extreme, so the extreme appoints would be the square root of sea over three. And then whatever value it's plug back into the original function, and that is for our first candidate. And then we have the negative square root see over three and then f of negative square root C over three. We could also talk about how see changes the X intercepts. So if we see here, we can just sit this equal to zero. We can talk about the X intercepts so we can factor out in X, so it's going to be X times X squared minus C. So that means our intercepts would be at 00 It would be at the square root of C zero and then the minus square root of C because you said both X equals zero and x four minus sees equal to zero. That is assuming that EC, that see does not equal to zero for these two points. So let's actually try to giraffe some functions. Three said that the value of C does not change the second derivative or any of the points of inflection. We'll just, uh, move it aside for now. So let's discuss some grafts. So let's see, when C is greater than zero, then autograph should have. These few points should have X intercepts. Thus these points of, uh, these critical points. So let's say our graph. Let's say C is equal to three. Then we have, well, X cubed minus three x and I highly recommend you actually use a Hey, what's it called Graph in a graph in calculator are a website to graft. Issa's well, so our intercepts will be will be 00 So square root of three. Let's say that's right here. Um, the extreme will be at 3/3 is just one, and then we plug in one for here for here. Can we just get we actually get so it should be at minus one. We plug that back in, we should get to and then for the other point. It should be one and then minus two. This is a cubic function, so it should cubic function with a positive coefficient. So it should look something like this. Who should have be increasing to this point, decreasing and increasing again. So it should be, um, yes. So let's take another example. Let's do so. Let's see what happens when C is equal to six. So if physical to six, then our equation is X cubed minus six x. Our X intercepts are going to be 00 the square root of six. So the square of 60 and then minus square root 60 The critical points will be 6/3, just square root of two in minus square root of two. They have to plug this back into the equation. And that is around 5.65 minus 5.65 This is 5.65 So we could, uh, graph this. So I listed the points from this equation. I'll try to graph the same The, um, Mint sees equal to six. So we have a point that is to the left of this point. A point over here, so these points are square. It is six zero and minus square root of 60 And then we have the origin. So yeah, and then we have our We have our critical points that will just be to the left of the critical points off. Other equation. So we have. We're you have minus square root of two and then 5.6. So I'll say up here. So this is minus root too. 5.65 Then the same thing for this point. Over here, that is route to minus 5.65 And then, uh, yeah, in my end, up scratching over some of them. But it's going to increase until minus Root, too. Then decrease. Go to the origin and then increase. It crosses the X axis. So as we can see that if as increasing see, uh, when it's greater than zero that we are increasing how high, how high the, um, critical point is and how far the X intercepts are. I always want to point out a or weird aspect, and it is that one C is less than zero. So we can say that our function is Let's say C is equal to negative three. That means our equation will be x third X to the third power plus three X Well, you might have found the ah problem. We have already and it's that a lot of these solutions deal with the square root to take the square root of something and we have a negative number and we're not going to do with imaginary numbers. So what can we dio? But we do have our this point or here. So it does cross the X axis. And if we solve for X here for acts, When he said it to zero here we will only get X is equal to zero. But what about the first derivative? Well, what is our first derivative? It says that we have our how were except our critical points here. But they all involve square roots. So why Why does this happen? Well, if we take our derivative of our equation, you see that well, but they're at this equation. If we try to graph this equation, it actually does not touch the X axis. So and so that means there is no value for X that will make this zero. That means there are no critical points for this equation. So if critique how there are no. If your C is less than zero, then there will be no critical points for your equation. We could do the increasing decreasing. But like I say here that this will always be positive. So it will always be increasing. So it would look something like this. This is one sees equal to negative three. So as a wrap up when sees greater than zero are, as we constantly keep, increase increasing, see, when it's positive, is that we will see a higher a higher magnitude of our extreme A guys we saw here compare to to 5.65 our X intercepts will be more spaced out and when we are, then we have very that when we have negative numbers for C, then we will have no extreme and we can and it will look like this. And in fact, if we do another number like negative six, then it will. This graph will actually mawr look like a linear function, although it's not, but it will very it will resemble it very greatly.

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