(a) Investigate the family of polynomials given by the equation $ f(x) = 2x^3 + cx^2 + 2x $. For what values of $ c $ does the curve have maximum and minimum points?
(b) Show that the minimum and maximum points of every curve in the family lie on the curve $ y = x - x^3 $. Illustrate by graphing this curve and several members of the family.
we want to investigate family. A polynomial is given by the equation of accident to two x cubed plus C x squared plus two X, and we want to find for what values of seed does the curve have a maximum and minimum points. And then we want to show that these minimum the maximum points lie on the curve of why is equal to X minus excuse. All right, so first, let's go ahead and figure out what values of sea will give them maximum. So person, when you do, is take the derivative of every Becks and then find our critical values by saying the derivative equal deserve So taking that derivative there, we're going to get that f prime of X is equal to Well, we'll use powerful to take those derivatives. So we get six at squared plus two c x lost to physical zero, and we could go ahead and just divide all this by two. And I would give us three x squared. What's C. X plus one is a good zero, and now we can go ahead and used the quadratic equation too solved for this and just trying that out is going to be, and you actually go ahead and put lives above it. So three is a see is B and I should put quotation there on that. And one is C with quotations around that. So I'm just gonna go ahead and write out with standard quote. Our equation is so negative B plus or minus east where minus for a c all over two a and then plug it. Everything in we'll get negative C plus or minus the square root of C squared minus Well, a is three c is one, so we'll get 12 all over and then two times a is six. So we get this here for all of our possible solutions. And so the thing we're going to need to make sure is that this radical here is defied. So that's going to happen when let me go. Let me just move this up here when C squared minus 12 is strictly greater than zero. So I can add 12 over I get see greater than or equal to 12 c squared greater than a good 12 and then we can go ahead and take the square root. So I was gonna say CIA strictly greater than the square root of 12. Well, that's one solution. But if C is negative, we can also have that they're a swell And then we could just multiply that negative over and we get see lesson Every go to the negative square root. So these two here will be the values of See that will give. I'll say maximum Berman. So let me just go ahead and write down here. So I see is in the interval of negative infinity two negative 12 Union 12 to Infinity than this here will give us a maximum or a minimum for that class of functions. All right, now, the next thing we want to do is to show that thes maximums. So remember, the maximum men that we have is really this value over here. So we want to show really that X equal to negative C plus or minus this word of C squared minus 12 over six. Uh, you go ahead and make that Maur like negative Chlo. So, Myers 12 we want to show that this gives the same output or our function f of X as well as this line y minus X cubed. So Let's just set these two functions equal to each other and see what we get. So I'm gonna set f of X equal to why so F of X is two x cubed plus see X squared plus two x and why was X minus X cute social? See where these two lines would intersect so I can subtract X and add Execute Gilbert, which would give three x cute plus c X squared plus X is equal to zero. Let's go ahead and factor out an X So we get X times three X squared plus c X plus one is equal to zero. So this here tells us by the zero product property either X is equal to zero War three x squared plus c X plus one is equal to zero. And you might know this, that we've already solved for this because when we looked at the derivative over here and we said equal zero and we divided it by two, we ended up with that exact same what drug? So the solutions, forearm axes and men's will all be the same, so you could go ahead and solve this and then, um, do it that way. But I think this is good place to just go ahead and stop