If $f,g : (a,b) → \mathbb{R}$ are differentiables and for all $x \in (a,b)$, $f(x)g'(x) \neq f'(x)g(x)$, in between two zeros of $f$ is a zero of $g$.

Question

If $f,g : (a,b) → \mathbb{R}$ are differentiables and for all $x \in (a,b)$, $f(x)g'(x) \neq f'(x)g(x)$, in between two zeros of $f$ is a zero of $g$.

I am trying to prove this but am quite stuck.

Does this proposition hold at the first place? If it does, how can we prove this? (We call $x$ a zero of $f$ if $f(x)=0$).

Answer 1

Let $u,v\in(a,b)$ be zeroes of $f$ and assume that $g$ has no zeroes in $(u,v)$. From $f(u)=f(v)=0$ and the given inequality, we obtain that $g(u)$, $g(v)$ are non-zero, hence also positive. This allows us to consider $h(x)=\frac{f(x)}{g(x)}$ on $[u,v]$. We have $h(u)=h(v)=0$. By Rolle $h'(x)=0$ for some $x\in(u,v)$. But $h'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}\ne0$ for all $x\in (u,v)$.

Answer 2

Let $W = f' g - g' f$. Then $W \neq 0$ by the condition, and if $x_1,x_2$ are successive zeros of $f$, then $$ W(x_1) = f'(x_1) g(x_1), \qquad W(x_2) = f'(x_2) g(x_2), $$ and these both have the same sign. But $f'(x_1)f'(x_2)<0$ because otherwise the mean value theorem and the intermediate value theorem would imply that $f$ has another zero between $x_1$ and $x_2$. Hence we also have $g(x_1)g(x_2)<0$, so the intermediate value theorem implies that $g$ has a zero between $x_1$ and $x_2$.