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

WORK:

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

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

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

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

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

$\displaystyle =(q-p)(q^2+qp+p^2)-k^2(q-p)$

$\displaystyle =(q-p)(q^2+qp+p^2-k^2)$ ...this isequation 1

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

Fromequation 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...$\displaystyle f(q)>f(p)$