- Having a continuous differentiable function $f(x)$, and denote $Z(\cdot)$ number of zeros (assume real line), and $(\cdot)^\prime$ first derivative, I would like to know if following inequality holds: $$Z\left(\left(\frac{f^\prime(x)}{f^2(x)}\right)^\prime\right)\leq Z(f^{\prime\prime}(x)).$$
From Rolle's theorem we know that $Z(g(x))\leq Z(g^\prime(x))+1$, meaning that $$Z\left(\left(\frac{f^\prime(x)}{f^2(x)}\right)^\prime\right)\leq Z\left(\left(\frac{f^\prime(x)}{f^2(x)}\right)^{\prime\prime}\right)+1.$$ But I don't know how to use this fact.
- Another question is the following. Is it true that $$Z\left(\frac{f^\prime(x)}{f^2(x)}\right)\leq Z(f^\prime(x))?$$ My reasoning is that number of zeros are determined only by the nominator ($f^\prime(x)$) and therefore if we divide the function by some other function (e.g. $f^2(x)$) we can in the worst case only remove some of the zeros in nominator, but we can never add new zeros. Is the thinking ok?
Thanks.
Your conditions are not strong enough: consider the real-valued function $f$ defined on $\mathbb{R}$ by: $$f:x\longmapsto\begin{cases}0&\text{if $x=0$}\\\mathrm{e}^{-1/x^2}&\text{if $x\neq0$.}\end{cases}$$ Then for all $x\in\mathbb{R}$: \begin{align*} f'(x)&=\begin{cases}0&\text{if $x=0$}\\\dfrac2{x^3}\mathrm{e}^{-1/x^2}&\text{if $x\neq0$}\end{cases}& f''(x)&=\begin{cases}0&\text{if $x=0$}\\-2\dfrac{3x^2-2}{x^6}\mathrm{e}^{-1/x^2}&\text{if $x\neq0$}\end{cases}\\ \frac{f'(x)}{f(x)^2}&=\begin{cases}\text{DNE}&\text{if $x=0$}\\\dfrac2{x^3}\mathrm{e}^{1/x^2}&\text{if $x\neq0$}\end{cases}& \left(\frac{f'}{f^2}\right)'(x)&=\begin{cases}\text{DNE}&\text{if $x\neq0$}\\-2\dfrac{3x^2+2}{x^6}\mathrm{e}^{1/x^2}&\text{if $x\neq0$}\end{cases} \end{align*} Then $Z\bigl(f''\bigr)=3$ and $Z\left(\left(\dfrac{f'}{f^2}\right)'\right)=0$, so 1 is false.