Calculus and Geometry with a Cubic Function and Straight Line

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Homework Help Overview

The discussion revolves around a calculus problem involving a cubic function and a straight line, specifically focusing on part iv) of the homework. Participants are analyzing the relationship between the cubic function and the line, exploring algebraic manipulations and implications of given relations.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss algebraic attempts to relate the cubic function and the line, questioning how to demonstrate specific relationships such as c = -2b. There is also a focus on whether the original poster has adequately utilized given relations from earlier parts of the problem.

Discussion Status

The discussion is ongoing, with participants providing suggestions for algebraic approaches and questioning assumptions. Some participants have offered guidance on evaluating expressions and rearranging terms, while others emphasize the importance of not assuming relationships without proof.

Contextual Notes

There is mention of earlier parts of the homework that may contain relevant equations, and participants are navigating the constraints of proving certain relationships without making assumptions.

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Homework Statement



I am asking about part iv).

[PLAIN]http://img715.imageshack.us/img715/7977/113ivb.jpg


Homework Equations



I guess they would be the ones in the earlier parts...

The Attempt at a Solution



In the given fact, I think x^3 - x - m(x - a) distance from the cubic function to the line. So for every point in (b, c), this would be negative.

I'm really not sure how to show c = -2b so I just tried to play with some algebra...

At x = b... b^3 - b - m(b - a) = 0

At x = c... c^3 - c - m(c - a) = 0

So they're both equal to 0...


b^3 - b - m(b - a) = c^3 - c - m(c - a)

so b^3 - b - mb = c^3 - c - mc

so by part i) b^3 - b - b(3b^2 - 1) = c^3 - c - c(3b^2 - 1)

so b^3 - b - 3b^3 + b = c^3 - c - 3b^2c + c

so -2b^3 = c^3 - 3b^2c

so b^2(3c - 2b) = c^3

but this doesn't look useful...

Thank you.
 
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Have you answered i, ii, and iii? Re iv, it looks like you didn't use the relation given in the problem (the one you don't have to prove).
 
=hotvette;3598514]Have you answered i, ii, and iii? ).


Yes..

Re iv, it looks like you didn't use the relation given in the problem (the one you don't have to prove).

Really? I think I did use it...

I got b^3 - b - m(b - a) = 0 by putting x = b in the given relation...
 
You might try doing the algebra first before substituting.
 
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hotvette said:
You might try doing the algebra first before substituting.

I did try that and got up to b^3 - b - mb = c^3 - c - mc which is what I wrote ...

Maybe you mean something else?
 
Yep, I see what you mean. I spoke before trying it (gotta stop doing that, it burns me too often). One thing you can do (not sure it would be accepted by your prof though) is to substitute c=-2b into your last expression and see if both sides are indeed equal.
 
Unfortunately, I don't think that's the best and proper way in math to do this? You'd have to work to show c = -2b and not assume it's true already...
 
Got it. Evaluate the given equation at x=a. After rearranging, you'll have an expression for c that is a ratio of polynomials in a & b. Substitute the expression for a and simplify.
 
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