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Thread: What is being asked?

  1. #1
    May 2009

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

    $\displaystyle f(x)=x^3-k^2x$
    $\displaystyle f(q)=q^3-k^2q$
    $\displaystyle f(p)=p^3-k^2p$

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

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

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

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

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

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

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

    Don't hesitate to ask for further clarification if necessary.
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  3. #3
    May 2009
    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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