Given $f^3(x)+xf(x)+x^3=0$ prove that $-x^2<f(x)<0$ and more

Question

Given the continuous function: $f: (0,+\infty)\to \mathbb R$ for which it applies: $$f^3(x)+xf(x)+x^3=0, \forall x\in(0,+\infty)$$ I) Prove that $-x^2<f(x)<0$

II) Prove that the equation: $6-xf(x)=x^2+5x$ has got at least one root at $(1,2)$

III) Find the limit: $\lim\limits_{x \to 0^+}{(f(x)\sin{1\over x})}$

IV) Find the limits: $\lim\limits_{x \to 0^+}{f(x) \over x}$ and $\lim\limits_{x \to 0^+}{f(x) \over x^2}$

Personal work:

I) For $x>0$ we have: $$f^3(x)+xf(x)+x^3=0 \iff {f^3(x) \over x}+f(x)+x^2=0$$

II) $$6-xf(x)=x^2+5x \iff 6-x^2-5x=xf(x) \iff {6-x^2-5x \over x}=f(x)$$

Bolzano doesn't seem to work as $f(1)=0$ and $f(2)=-4$. Also, there doesn't seem to be any obvious root(s) in $(1,2)$

Answer 1

This addresses the original version of the question.

No, as $f(x)=\dfrac{6-x^2-5x}{x} = -\dfrac{(x+6)(x-1)}{x}$, $x\neq0$, so $f$ has two distinct real roots $-6$ and $1$ outside the open interval $(1,2)$.

Answer 2

We have $f'(x)=- \frac{6}{x^2}-1 <0$ for all $x \ne 0 $. Since $f(1)=0$ we get

$f(x) <f(1)=0 $ for all $x \in (1,2)$. Hence $f$ has no root in $(1,2)$.

Answer 3

II) The eqn $6-xf(x)=x^2+5x \iff 6-xf(x)-x^2-5x=0.$

We consider the continuous function $H(x)=6-xf(x)-x^2-5x.$ And $H(1)=-f(1)>0, H(2)=-[2^2+f(2)]<0$. Thus, ... exits $x_0\in (1,2): H(x_0)=0.$