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(a) Investigate the family of polynomials given b…

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Problem 39 Hard Difficulty

(a) Investigate the family of polynomials given by the equation $ f(x) = cx^4 - 2x^2 + 1 $. For what values of $ c $ does the curve have minimum points?
(b) Show that the minimum and maximum points of every curve in the family lie in the parabola $ y = 1 - x^2 $. Illustrate by graphing this parabola and several members of the family.

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Calculus 1 / AB

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 6

Graphing with Calculus and Calculators

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Video Transcript

we want to investigate the Valley of Polynomial. Lt's given by the equation of ecstasy to see times extra four minus two x squared plus one. And we want to come in for what I use it. See, Does the curve have a minimum point or minimal points? And then we want to show that the minimum and maximum points of our function will lie on the problem. Why isn't one minus X squid? So first thing we're going to need to do is to take the derivative, he goes, We will need to know that to find our minimum values, not so get f Prime of X is equal to sell. To take the derivative of that poor know me over is going to be powerful breach of the terms someone get or c X cubed minus for X. Now find a critical points. You need to set this equal to zero, actually, so we need to set. That is absurd. And I'll solve for that off on the side here so I could go ahead and factor an X ena Thor out. So it's gonna be four x see X squared minus one is zero. So that's gonna tell us either X is equal to zero or see X squared minus one is 0 to 0. And so we can just go ahead and move things around to solve for X here, so it's acceptable to plus or minus the square root of one of her seat. Now, since we're dividing by sea, that tells us that C has to be bigger than our equal to you. Zero there has to not be zero. And since we're taking the square root, then is it needs to be strictly grave. So at least right now, we know that C has to be strictly great with them. Zero Um, that is, at least if X is equal to zero, does not give us a minimum, but at least for a second solution for this other quadratic scene used to be sure. So let's go ahead and use the second derivative test to determine if these three values that you have here are minimum or maximum. So second derivative will give us clothes the X squared minus four, and again, we're going to need to set this here equal to Europe or actually outside into that. Let's just go ahead and plug in X zero and excessively to foster my s one. What's the minus? Descriptive. One of so a double crime of zero. No problem. In zero we get negative for Well, that's strictly less than zero. Which implies the function is Khan cave down at that point. And the second derivative test tells us this is a local max. So now this actually does kind of give us the fact that well, possibly if we are going to have a minimum, it has to be see strictly greater than zero. Since we've already determined that X is equal is there will always be local Max. All right, now let's go ahead on and look at the other value. Now this here will give the same value with the plus or the minus. Since we only have X squared and square gets rid of the plus and the minus essentially, song is gonna go and put this end and solve it once. So during that 12 times, see, and then if we square plus or minus one of this, we're route of see Well, that's just going to be won over C minus or well, those things come out just to get one way of 12 minus four, which is eight, which is strictly greater than zero. And so this year implies con que up at both of those values for X or we get a local So we only get a local men when C is strictly grave and zero and that local men will occur at X is equal to plus or minus one of the square. See what? So now we know where our minimums are going to be if they exist for that family. Now we want to show that the Max Men of that is also on the curve. Why is it one minus X squid? So, in other words, what this is really wanting us to show is that when we set, FX is equal to one minus X squared that it intersects used to curves at X is equal to zero and execute to cluster. Myers won over. See? All right, so let's go ahead and plug and protects. So that's going to be C X to fourth Maya's to X, where, +11 mile square. Well, we could go ahead and subtract one over and add swear and doing that, We'll get a C minus, X squared. Busy zero Alice, factor out squeak. So we get X squared time. See X squared minus one is zero. And now this is going to say either X, where is equal to zero or C X word minus one is zero. Well, X squared is zero only when x is equal to zero. And you might notice that this year Excellent use use cream. So you might notice that this here is exactly what we saw for oh, here, when we're looking for are critical bodies. So this also gives the solution Exes it too. What's the minus one? Oversee square root. So we get that both of power Critical points will also have the same output for our class of function or a family function as well a cz one minus x squared. So all the maximums will lie on that

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