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 $\displaystyle k>0$, 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 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...$\displaystyle f(q)>f(p)$

2. 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 $\displaystyle \frac{f(q)-f(p)}{q-p}$
Explanation:
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.

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!