Solutions of two differentiable functions

Question

I'm trying to prove that for two functions, $f(x)$ and $g(x)$, both diffrerentiable at every real $x$,

$f(x)$ having at least two solutions to the equation $f(x) = 0$

And if $f'(x)g(x) - f(x)g'(x) \neq 0$,

Then there is a solution to $g(x) = 0$ between the two solution of $(f)x$.

I tried to use the quotient rule for point $c$ (from Rolle theorem, for two solution of $f(x) = 0$ such that $f(a)=f(b) = 0$), but I got stuck.

I also tried to use the quotient rule for $a$ and $b$ mentioned before, but I still can't proceed.

Answer 1

Let $h(x)=\frac{f(x)}{g(x)}$, if $f(a)=f(b)=0$ then $h(a)=h(b)=0$, so that there has to be some max point or min point $c\in(a,b)$, so, $h'(c)=0\rightarrow f'(c)g(c)-f(c)g'(c)=0$ .

Given that there isn't, we have to say that $g(c)=0$, so $h(c)=\frac{f(c)}{g(c)}$ doesn't exist, which is the only reason why the derivative has no zero point.

Answer 2

Let's call $h(x)=\dfrac{f(x)}{g(x)}$.

$f$ and $g$ are differentiable $\forall x$,

AND let's assume that $\forall x\in]a;b[, g(x)\neq 0$,

Then $h$ is as well.

Then you have, since $f(a)=f(b)=0$ ($a

This means $\exists c \in ]a;b[, h'(c)=0$

But this means $f'(c)g(c)-f(c)g'(c)=0$

This is in contradiction with the statement $f'(x)g(x)-f(x)g'(x)\neq 0$.

Our assumption is then false, and $\exists c\in]a;b[$ such that $g(c)=0$