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Calculus and Geometry with a Cubic Function and Straight Line

  1. Nov 4, 2011 #1
    1. The problem statement, all variables and given/known data

    I am asking about part iv).

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


    2. Relevant equations

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

    3. The attempt at a solution

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

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

    At x = b... [itex] b^3 - b - m(b - a) = 0 [/itex]

    At x = c... [itex] c^3 - c - m(c - a) = 0 [/itex]

    So they're both equal to 0...


    [itex] b^3 - b - m(b - a) = c^3 - c - m(c - a) [/itex]

    so [itex] b^3 - b - mb = c^3 - c - mc [/itex]

    so by part i) [itex] b^3 - b - b(3b^2 - 1) = c^3 - c - c(3b^2 - 1) [/itex]

    so [itex] b^3 - b - 3b^3 + b = c^3 - c - 3b^2c + c[/itex]

    so [itex] -2b^3 = c^3 - 3b^2c [/itex]

    so [itex] b^2(3c - 2b) = c^3 [/itex]

    but this doesn't look useful...

    Thank you.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 4, 2011 #2

    hotvette

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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).
     
  4. Nov 4, 2011 #3

    Yes..

    Really? I think I did use it...

    I got [itex] b^3 - b - m(b - a) = 0 [/itex] by putting [itex] x = b [/itex] in the given relation...
     
  5. Nov 4, 2011 #4

    hotvette

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    You might try doing the algebra first before substituting.
     
    Last edited: Nov 4, 2011
  6. Nov 4, 2011 #5
    I did try that and got up to [itex] b^3 - b - mb = c^3 - c - mc[/itex] which is what I wrote ...

    Maybe you mean something else?
     
  7. Nov 4, 2011 #6

    hotvette

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    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.
     
  8. Nov 5, 2011 #7
    Unfortunately, I don't think that's the best and proper way in math to do this? You'd have to work to show [itex] c = -2b [/itex] and not assume it's true already...
     
  9. Nov 6, 2011 #8

    hotvette

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    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.
     
    Last edited: Nov 6, 2011
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