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Math Help - What is being asked?

  1. #1
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    What is being asked?

    Hi Everybody! I have a problem with a math question. I typed out the said question (in bold at the beginning) and the solution underneath. I don't understand what is being done at the two bullet points - why we are doing what we are doing. So, if somebody would be so kind as to explain, I would be very grateful - it's been bugging me for two days (I asked for help on some other forum, but sadly, I only got one reply...explaining how to do the first part.)

    For a fixed k>0, let
    f(x)=x^3-k^2x, x is an element of R
    For  p is not equal to q, divided f(q)-f(p) by q-p.
    Prove that is 0\leq{p}<q<\frac{k}{\sqrt3}, then f(q)<f(p),
    whereas, if \frac{k}{\sqrt3}\leq{p}<q, then f(q)>f(p).


    WORK:
    f(x)=x^3-k^2x
    f(q)=q^3-k^2q
    f(p)=p^3-k^2p

    f(q)-f(p)=(q^3-k^2q)-(p^3-k^2p)
    =(q^3-p^3)-k^2(q-p)
    =(q-p)(q^2+qp+p^2)-k^2(q-p)
    =(q-p)(q^2+qp+p^2-k^2) ...this is equation 1
    \frac{f(q)-f(p)}{q-p}=q^2+qp+p^2-k^2






    From equation 1
    • If 0\leq{p}<q<\frac{k}{\sqrt3}, q-p>0
      q^2+qp+p^2-k^2<(\frac{k}{\sqrt3})^2+(\frac{k}{\sqrt3})(\frac{  k}{\sqrt3})+(\frac{k}{\sqrt3})^2-k^2
      q^2+qp+p^2-k^2<\frac{k^2}{3}+\frac{k^2}{3}+\frac{k^2}{3}-k^2
      q^2+qp+p^2-k^2<0
      which implies... f(q)-f(p)=(+)(-)<0
      which implies... f(q)<f(p)

    • If \frac{k}{\sqrt3}\leq{p}<q, then q-p>0 and
      q^2+qp+p^2-k^2>(\frac{k}{\sqrt3})^2+(\frac{k}{\sqrt3})(\frac{  k}{\sqrt3})+(\frac{k}{\sqrt3})^2-k^2
      q^2+qp+p^2-k^2>0
      which implies... f(q)-f(p)=(+)(-)>0
      which implies... f(q)>f(p)
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  2. #2
    Senior Member I-Think's Avatar
    Joined
    Apr 2009
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    Good day day to you busy bee. Let's hope I'm a good teacher.

    So you understand the process by which q^2+qp+p^2 was obtained from \frac{f(q)-f(p)}{q-p}<br />
    Explanation:
    For any given inequality a-b>0 or a-b<0, we can determine which is greater than which, in the former a>b, in the latter b>a

    In the case where q^2+qp+p^2<0, this means that this expression is negative. By observing the expression \frac{f(q)-f(p)}{q-p}. the only way for it to be negative is if p>q,(not true due to question constraints) or if f(p)>f(q).

    For q^2+qp+p^2>0, the expression is positive, hence f(q)>f(p)


    By letting each of q and p be equal to  \frac{k}{\sqrt{3}}, we establish an expression which is greater than/lesser than q^2+qp+p^2.
    \frac{k^2}{3}+\frac{k^2}{3}+\frac{k^2}{3}-k^2 boils down to 0,so now we can determine whether f(p)>f(q) or f(q)>f(p).



    Don't hesitate to ask for further clarification if necessary.
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  3. #3
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    You were a great teacher - so thank you! I understand now.

    It makes sense in my head now whereas before it felt like mumble-jumble.

    Thank you a million times!
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