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If $f(x)$ and $g(x)$ are differentiable in an interval $I$ and: $$f(x)g'(x) - f'(x)g(x)\not=0 \qquad\forall x \in I$$ then there is always a root of $g$ between two roots of $f$.

How do I prove this?

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analysisgb
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    No, but the condition is equivalent to $(f/g)'\neq 0$ (on intervals where $g$ has no zero). – Jochen Dec 12 '19 at 13:22
  • Ah, yep there is a minus. That was sneaky. Same trick as language app. Should make you doubt yourself shouldn't it? :) – mathreadler Dec 12 '19 at 13:23

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Consider $y=\frac{f(x)}{g(x)}$ and suppose there to be no root of $g(x)$ between two roots of $f(x)$.

Then, on the interval between the two roots, $y$ has no turning points and no discontinuities and is therefore monotonic. Then it cannot have two roots unless it is identically zero, in which case $f(x)g'(x) - f'(x)g(x)=0$ anyway.

  • How do i mathematically prove this? And why do i consider $f(x)/g(x)$? What does that tell me? Thanks for your answer! – analysisgb Dec 12 '19 at 13:24
  • You consider $f/g$ because the condition you are given involves the numerator of the quotient rule. –  Dec 12 '19 at 13:25
  • But then i would have to consider $g(x)/f(x)$ , right? – analysisgb Dec 12 '19 at 13:31
  • No. Considering f/g is better because you are assuming g has no roots and so you are dealing with a nice function with no discontinuities. –  Dec 12 '19 at 13:32
  • I understand this problem better now. By using the quotient rule i rule out the possibility that $g(x) = 0$ because otherwise i would be dividing by 0. – analysisgb Dec 12 '19 at 13:48
  • Good. Strictly it's the other way round though - because we are assuming g is not equal to 0 we can use the quotient rule. –  Dec 12 '19 at 13:50
  • Ok you are right.. that makes sense. So $h(x) = f(x)/g(x)$ and $h(a)=h(b)=0$ and $h(x)$ is therefore monotonic between these two roots (a and b), but wouldn't it be possible that there is a local extrema/or a turning point? I am not quite sure how to continue from this point on.. – analysisgb Dec 12 '19 at 13:56
  • All local extrema occur at points where the derivative is zero or undefined. So you only need consider x=a and x=b. –  Dec 12 '19 at 14:03
  • Remember that the numerator of the derivative of $h(x)$ is given to be non-zero. –  Dec 12 '19 at 14:04
  • Ahhh so if i assume that there is no root of $g(x)$ in [a,b] i would only be able to get my two roots a,b if the numerator of the derivative is equal to zero. But that's a contradiction, because we are given the inequality: $f(x)g'(x)-f'(x)g(x) \not = 0$ so that means, that there must be a root of g(x) to satisfy that inequality. Am i right? (: – analysisgb Dec 12 '19 at 14:15
  • Not quite how I would put it but I'm sure you've got the idea now. Good. –  Dec 12 '19 at 14:17
  • I would be very interested in your correction of my previous comment (detailed explanation of your thought process if you have time)? Thanks for your time and effort! – analysisgb Dec 12 '19 at 14:19
  • OK-will comment tomorrow (travelling today). –  Dec 13 '19 at 00:47
  • It's not so much a correction as a way of expressing the steps. You wrote g has no root (1) implies numerator is equal to zero (2). But numerator is not zero (3) and therefore g has a root (4). Which is fine but if you were splitting it into equal sized deductions then it would be : (1) implies f/g has no TP and no discontinuities (1a) therefore f/g is monotonic (1b) therefore f/g is zero on the interval. Only then would it be "Therefore numerator is zero (2) but numerator is not zero (3) and therefore g has a root (4)." –  Dec 13 '19 at 15:53