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busy_bee · May 27, 2009, 08:14 AM

I have the solution of the problem below, but I don't understand it. I don't understand exactly what I am being asked to do in this question. In the solution, you start of by showing that f(q) - f(p) is divisible by q - p. After that, I really don't understand what happens. Any help would be greatly appreciated. Thanks in advance.

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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 $f(q)>f(p)$ .

Unknown008 · May 27, 2009, 10:56 AM

Well, I no more can view the link... Here's your question:

Let $f(x)=x^3-k^2x$

For $p \not= q$ , divide $f(q)-f(p)$ by $q-p$

If $0\leq{p}<q<\frac{k}{\sqrt3}$ , then show that $f(q)<f(p)$

Then, if $p>q$ , then show that $f(q)>f(p)$

Correct me I I missed something, for I cannot recall every thing that was in the question.

Firstly, I did the first part, that was easy and I got:

$\frac{f(q)-f(p)}{q-p}=q^2-p^2-k^2$

Thereafter, I don't know how to do it exaclty, but you have to show that $f(q)<f(p)$ if $0\leq{p}<q<\frac{k}{\sqrt3}$ .

I did some things but not sure they are good. You'll have to wait for people better than me in this field.

EDIT: oh wait, I missed that :o

$\frac{f(q)-f(p)}{q-p}=q^2+qp+p^2-k^2$

busy_bee · May 27, 2009, 01:00 PM

Originally Posted by Unknown008

Well, I no more can view the link... Here's your question:

Correct me I I missed something, for I cannot recall every thing that was in the question.

Firstly, I did the first part, that was easy and I got:

$\frac{f(q)-f(p)}{q-p}=q^2-p^2-k^2$

Thereafter, I don't know how to do it exaclty, but you have to show that $f(q)<f(p)$ if $0\leq{p}<q<\frac{k}{\sqrt3}$ .

I did some things but not sure they are good. You'll have to wait for people better than me in this field.

Thank you very much for trying it! This part is fine, in terms of my understanding it. However, I can't put two and two together of why we are doing what we are doing in the second part (I will write this out for referance, and to save people hassle of doing it themselves - I did not know how to use the coding here when I originally posted).

The first part should be
$\frac{f(q)-f(p)}{q-p}=q^2+qp+p^2-k^2$

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... $f(q)>f(p)$

Whoa! That took a long time! So, if anybody can please answer this - why are we doing what we are doing at the two bullet points?