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University of North Texas

# 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 basic shape of the curve changes.$f(x) = x^2 + 6x + c/x$ (Trident of Newton)

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##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

##### Michael J.

Idaho State University

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### Video Transcript

we want to describe The graph of f of X is equal to X squared plus six X plus C over X as seeing Berries. And we want to then Graff several members of the family to illustrate the trends that we discover. And in particular, we want to determine how the maximum minimum points and points and inflection move as C changes. We should also identify any kind of transitional values of C average. The basic shape of the curve changes. All right, So first, maybe what we might do is to try toe, get some inspiration for what are X intercept should be. So let's go ahead and make this into a rational function or at least one fracture, I should say, So we can revive, says X cubed plus six X squared plus C all over X. Now I don't know a good way to solve for them, you know you're here, so I'm just not going to attempt it. Um, so let's move on to the first derivative and maybe see if we can get some inspiration for that. So they take the derivative of each of those terms with meetings powerful. So we get to X plus six minus C over X square Because, remember, see divided by exes really see times X to the negative First power. All right, so let's do the same thing. Combine this into one fraction. We get to X cube plus six X squared minus c all over x weird. Now, just like before, I have no idea how to solve that Cubic in the numerator, so I'm just gonna ignore it. For now, let's move on to the second derivative and hopefully get some inspiration from this. So take the door of the skin would get to plus two C over X cube. Using the power rule and combining this into one fractured will end up with two x cubed plus two C all over. Excuse Now the numerator. This is something I know how salt pork. So maybe grow up to be able to find some of the collection points. So that's going to give us two x cubed plus two c is zero. Um, subtract the tool over and divide by two would get X cubed is eager to negative. See taking the cube root of you decide to get X is equal to the negative cube root of C, so at least we'll know how are inflection Point will change, and we know it will be an inflection point. Um, since we have a cubic in the numerator we know at one point is gonna be causing the one point is going to be negative just due to its so we at least know that will be our point of inflection. So we really didn't get much from this. But you might know this if, well, let's see equal to zero. This fractional part goes away and we're just left with the quadratic X cubed plus six sex. So why don't we go ahead and look at what happens when C is equal to zero and maybe we can get some inspiration from that? Well, that tells us after Becks isn't going to equal to X squared plus six X Now, if we set this here equal to zero, well, we could go ahead and factor it, and it's gonna get those X X plus six is equal to zero. So either X is equal to zero. R X is equal to negative six. So something we might just assume my happens is we're going to have these two ex intercepts and they're just going to get shifted, left or right, depending on our values first. All right. So now the first derivative when c 00 willpower rule game to get to expose six almost set this equal to zero. So subtract the six over. We get two exes, You to 96 to 5. You decide by two you would get excessive to me. And since the graph of this here is going to look something along the lines of this, we will know that this will be a minimum, since around exited with a negative three of the function is decreasing to elect a bit of an increasing. So we might say that our minimum for this family should be around exiting into negative three and then our second derivative that gives us two. And, well, that's strictly greater than zero. So this implies is always con que up. So something slightly different from what we have for the more general case. So it looks like we'll have exit except around excessive zero. Next issue with negative six. A minimum somewhere around X equals negative three and his longest X does not equal to zero. Um, we will have our inflection point at X is equal to negative que group of sea. So now let's go ahead and look at these graphs. So at C 00 we end up with this transitional of value, as you can see, because when cuz good, negative one. And cuz Goto one, we end up with a distinct differences in grafs and you might notice that at least for cuz it's a negative one sees a negative 10. We do end up with our two ex intercepts being pretty close toe negative six and zero. And you might also notice at about ecstasy but a negative three, which is about right here we have pretty much where minimum should be and also on the other side for our minimums. Exes go too negative. Three. Looks like we have those minimums there. Um, the only difference is four are positive values for C. We also have these two maximums and another minimum that ends up coming. So we didn't quite get everything from making that replacement for season desirable. At least it gave us a little bit more information also for inflection points so we can see about, um, negative keeper of seat. You should have our inflection points. So over at C zero. Negative one. That should be at one. And that looks like our inflection point there at negative 10. Well, that should be something about two ends. Someone of the earth. Not there. But somewhere around here, it looks like we have where our point of inflection is first season with negative 10. Well, it should be around negative, too. So we get our plan into action there and seizing the one that should be a negative one. And it looks like we do have a corn inflection.

University of North Texas

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Derivatives

Differentiation

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##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

##### Michael J.

Idaho State University

Lectures

Join Bootcamp