1. ## 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}, then $f(q),
whereas, if $\frac{k}{\sqrt3}\leq{p}, 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}, $q-p>0$
$q^2+qp+p^2-k^2<(\frac{k}{\sqrt3})^2+(\frac{k}{\sqrt3})(\frac{ k}{\sqrt3})+(\frac{k}{\sqrt3})^2-k^2$
$q^2+qp+p^2-k^2<\frac{k^2}{3}+\frac{k^2}{3}+\frac{k^2}{3}-k^2$
$q^2+qp+p^2-k^2<0$
which implies... $f(q)-f(p)=(+)(-)<0$
which implies... $f(q)

• If $\frac{k}{\sqrt3}\leq{p}, 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... $f(q)>f(p)$

2. 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}
$

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.

3. 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!