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A Cubic Function and a Straight Line with One-variable Calculus

  1. Nov 5, 2011 #1
    Hi. I posted this in the Homework question but after 152 views with no right answer, my question looks analytical and rigorous enough to be posted here.

    Thank you....

    ----

    I am asking about part iv).

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

    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 6, 2011 #2
    FOR PART 4:
    You just expand RHS of give eq. in and equate coefficients of x^2 on both sides and you'll get it.
     
  4. Nov 6, 2011 #3
    Hi omkar13. Thanks so much for your answer.

    Now I feel like an idiot...how did you see to expand the RHS of the given eq.?

    Somehow I didn't see it....I thought there was something harder going on....
     
  5. Nov 7, 2011 #4

    hotvette

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