Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
add comment

2 Answers

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 $h(a)=h(b)$...).

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

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 $[a,b]$.

share|improve this answer
@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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.