# Roots of a function between two roots of another function

I've been asked the following problem:

Let $f$ and $g$ be two functions differatiable in the interval $I$. Proove that if $$f(x)g'(x) - f'(x)g(x) \ne 0 \quad \forall x \in I$$ there is one root of $g$ between every two roots of $f$

Now, the problem is, this question was asked in German, and the word for 'one' and the word for 'a' are both the same in German. In other words, I do not know, if the sentence says exactly one root of $g$ or at least one root of $g$. I've already spent a few hours on this problem and managed to show that there is at least one root of $g$ between two roots of $f$:

$$f(x)g'(x) - f'(x)g(x) = c, \quad c \ne 0$$ $$\Leftrightarrow g(x) = \frac{f(x)g'(x) - c}{f'(x)}$$ $$x_1, x2 \in I, \quad f(x_1) = 0, \quad f(x_2) = 0$$ $$g(x_1) = \frac{f(x_1)g'(x_1) - c}{f'(x_1)} = -\frac{c}{f'(x_1)}$$ $$g(x_2) = \frac{f(x_2)g'(x_2) - c}{f'(x_2)} = -\frac{c}{f'(x_2)}$$ Next, we can say If $g(x_1)$ and $g(x_2)$ have different signs, there exists a $\xi$ for which $g(\xi) = 0$ (IVT)

$g(x_1)$ and $g(x_2)$ have different signs if and only if $f'(x_1)$ and $f'(x_2)$ have different signs. Using Rolle's Theorem we can say:

There is an $\eta \in (x_1, x_2) : f'(\eta) = 0$

As $x_1 < \eta < x_2$ we can say, that either $f'(x_1)$ is negative and $f'(x_2)$ is positive or vice versa. Therefore the signs of $g(x_1)$ and $g(x_2)$ are different as well, and we know that there must be at least one root in the interval.

However, I just do not manage to create a proof that shows that there can't be more than 1 root in the interval. My guess would have been using Rolle's Theorem for a proof by contradiction, however there might perfectly be an extremum in the intervals, since $g(x)$ might have other roots outside the bounds of the interval. (Or did I get something wrong here)

So my question is:

Is the initial statement true, and if it is, how can I proove it?

• I believe the German statement asked for "a root". I'm a Spanish native speaker and the same ambiguity would have happened in Spanish; in this language the convention in maths is that if there's no precision "un/una" means "a/an", that is "at least one". – Alejandro Nasif Salum Dec 14 '17 at 22:54
• Also, the uniqueness comes from the fact that the statement is symmetric in $f$ and $g$, so under those conditions you proved that there is "at least one" root of $g$ between two given roots of $f$, but you've also proved that there is "at least one" root of $f$ between two given roots of $g$. If you add the precision "consecutive" roots, then this implies uniqueness. On the other hand, between non-consecutive roots of, say, $f$ there can be more than one root of $g$; precisely one more than the number of roots $f$ you have left in between. – Alejandro Nasif Salum Dec 14 '17 at 22:58
• This does not address your main concern, but the proof itself might be simplified by observing that under the assumption that $g$ has no roots, $f'(x)g(x)-f(x)g'(x)$ is just $g^2(x)$ times the derivative of $\frac{f(x)}{g(x)}$, and Rolle says that this must be zero somewhere in-between. – Hagen von Eitzen Dec 14 '17 at 23:01
• Just to be sure: Does this mean i can say: Assume $g(x)$ has no roots. Let $h(x) = \frac{f(x)}{g(x)}$. Then $h(x_1) = h(x_2) = 0$ because $f(x_1) = f(x_2) = 0$. Because of Rolle this means that $h'(\xi) = 0$, therefore $f'(\xi)g(\xi) - f(\xi)g'(\xi)$ must be $0$ as well, which contradicts the statement. Because of that, $g(x)$ must have roots. – zockDoc Dec 14 '17 at 23:21

The uniqueness comes from the fact that the statement is symmetric in $f$ and $g$, so under those conditions you proved that there is "at least one" root of $g$ between two given roots of $f$, but you've also proved that there is "at least one" root of $f$ between two given roots of $g$. If you add the precision "consecutive" roots, then this implies uniqueness.
On the other hand, between non-consecutive roots of, say, $f$ there can be more than one root of $g$; precisely one more than the number of roots $f$ you have left in between.
Anyway, I would say your proof works fine only under the hypothesis of consecutive roots. If you allow $f$ to be $0$ in the interval $(x_1,x_2)$, then I do not see how you could conclude that $f'(x_1)$ and $f'(x_2)$ have different signs. Think for instance of $f(x)=x^3-x$ in the interval $[-1,1]$, where Rolle's conditions are met, but $f'(-1)=f'(1)=2$.