What can be derived from the fact that $\lim_{x\to 0}\frac{f(x-3)f(x+3)}{x^3}$ is convergence?

Question

Problem is : There is a cubic function $f(x)$ , with its positive coefficient. And,

1. $\lim_{x\to 0} {\frac{f(x-3)f(x+3)}{x^3}}$ is convergence.

2. there is only one natural number $k = k_1$ which makes function $\left| f(x)-f(k) \right|$ not differentiable on exactly three points.

3. $f(k_1) = -15$

$f(8) = ? $

I can see that $f(x)$ has to have local max/min and $f(k_1)$ is placed between those values. since otherwise it would be contradictory to condition 2. And to deal with the uniqueness of "natural number" $k$, I had to figure out more about the function $f(x)$.

But all I think from condition 1 is $f(x-3)f(x+3) = x^3Q(x)$, and I can't proceed further. Maybe a little hint would be really helpful?

Answer 1

The point $x=0$ is a zero of order (at least) $3$ of $f(x+3)f(x-3)$. What does that say about the numbers $-3$ and $3$ as roots of $f$?

EDIT
Or to continue from your notation that $x^3$ divides $f(x+3)f(x-3)$, how can this division go? Either

1. $x^3$ divides $f(x+3)$
2. $x^2$ divides $f(x+3)$ and $x$ divides $f(x-3)$
3. $x$ divides $f(x+3)$ and $x^2$ divides $f(x-3)$
4. $x^3$ divides $f(x-3)$

Try all the cases. Remember $f$ is cubic. And "with its positive coefficient", do you mean its leading coefficient is posivite?

Further hint: $x^k$ divides $f(x+a)$ is equivalent to $(x-a)^k$ divides $f(x)$.