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Let $f(x),g(x)$ be differential functions, and $f'(x)g(x)\neq f(x)g'(x)$ for all $x\in\mathbb R$. Prove that between two roots of $f(x)$ there is a root of $g(x)$.

I guess this has to do with Rolle's theorem. I saw that when $f'(x)=0$, $g(x)\neq0$ and when $f(x)=0$, $g'(x)\neq0$, but I didn't manage to prove the conjecture. Thanks for any help!

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up vote 7 down vote accepted

Suppose that $a,b$ are roots of $f$ with $a<b$ and $g(x)\neq0$ for $x\in (a,b)$.
Consider the function $h(x)=\dfrac{f(x)}{g(x)}$ in $[a,b]$ to derive a contradiction
(is well defined, differentiable on [a,b] with $h(a)=h(b)$...).

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I think this answer hits the nail in the very head. +1 – DonAntonio Jan 12 '13 at 13:28

Suppose $f$ has two consecutive roots $a$ and $b$, and for a contradiction, suppose $g$ has no roots on $[a,b]$. Divide both sides of the inequality by $f(x)$ and $g(x)$ to get

$$F(x) := \frac{f'(x)}{f(x)} \neq \frac{g'(x)}{g(x)} =: G(x). \qquad x \in (a,b)$$

Then $F(x)$ is continuous on $(a, b)$ and

$$\lim_{x \to a^+} F(x) = +\infty \\ \lim_{x \to b^-} F(x) = -\infty$$

Since $G$ does not diverge near $a$ or $b$, this implies that

$$\lim_{x \to a^+} [F(x) - G(x)] = +\infty \\ \lim_{x \to b^-} [F(x) - G(x)] = -\infty$$

So the function $F - G$ is continuous on $(a,b)$, and goes from $+\infty$ to $-\infty$. By the mean value theorem, $F - G$ has a root on $(a,b)$, say $c$. But this implies that $F(c) = G(c)$, which contradicts our inequality. So $g$ must have a root on $[a,b]$.

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@Pambos: I've edited the answer to clarify. Basically, $F$ is continuous and goes from $+\infty$ to $-\infty$, and $G$ does not intersect $F$. That's impossible if $G$ is continuous and finite. – TMM Jan 12 '13 at 12:51

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