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i am just out of ideas here... that is the question, its from a test in my college

let $ f(x), g(x)$ be differentiable functions. for each $x$ it is true that $ f'(x)g(x) \neq g'(x)f(x)$. We also know that $ f(a)=f(b)=0$

prove that exists a $c, a<c<b$ so that $g(c)=0$

Ive tried to build some functions to use the Intermediate value theorem or Lagrange theorem. but no success...


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Do you mean for every $x$, $f^\prime(x)g(x)\not= g^\prime(x)f(x)$, or there exists $x$ with the property? – Patrick Mar 4 '12 at 18:44
up vote 2 down vote accepted

suppose that $g$ has no zero on $(a,b)$ then $f/g$ is well defined, and his derivative have a constant sign and then $(f/g)$ strictly increase or decrease ( because $(f/g)' = \frac{f'g-g'f}{g^2}$) but $f(a)/g(a) = f(b)/g(b) = 0$ that's a contradiction.

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hah Ive actually tried that, but you explained it perfectly - now i got it... thank you!! – YNWA Mar 4 '12 at 18:42
To nitpick: you need to prove that $f/g$ is defined on $a$ and $b$ too. – Aryabhata Mar 4 '12 at 18:43
It's a pleasure :). $g(a)$ and $g(b)$ couldn't be $0$ by the first relation, because $f(a)=f(b)=0$. – Adrien Boulanger Mar 4 '12 at 18:51


Can $\displaystyle g(a)$, $g(b)$ be $0$?

Now assume $\displaystyle g(x) \ne 0$ for any $\displaystyle x$, and consider the function $\displaystyle h(x) = \frac{f(x)}{g(x)}$.

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thank you! very helpfull :) – YNWA Mar 4 '12 at 18:42
@YNWA: You are welcome! – Aryabhata Mar 4 '12 at 18:49

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