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Assume that $f$ is a differentiable function on $[0,\infty]$ and $f(0) = 0$. Furthermore, assume that for all $x \in (0,\infty)$ there is a $c \in (0,x)$ such that $ 0 < f'(c) = \frac{f(x) - f(0)}{x} = \frac{f(x)}{x} $ Must there exist a point $p \in (0,\infty)$ such that for all $x \in (0,p)$ we have $f'(x) > 0$?

This is not a book or homework problem, this is a a question of my own. So far I have tried a contradiction, assume that for all $p \in (0, \infty)$ there is at least one $x \in(0,p)$ such that $f'(x) \leq 0$. Consider such a point $x$, now, by Darboux's Theorem there should be a point $x_0 \in (0,x)$ such that $f'(x_0) = 0$. Let $p = x_0$, there must be another point $x$ such that $f'(x) \leq 0$ , and hence, another point $x_1$ such that $f'(x_1) = 0$ (it is of course trivial that $x_1 < x_0$). As this process keeps repeating, there should be a infinite sequence of points $x_0,x_1,x_2,...$ converging to $0$ (the sequence is bounded below by $0$ by definition so the Monotone Convergence Theorem should guarantee convergence to $0$ right?) such that $f'=0$ at any point in this sequence. And here I'm stuck. Any convergent sequence in $R$ is Cauchy could that help at all? Or is this the entirely wrong approach? This is my first semester of Advanced Calculus, so please try not to laugh too hard if this question is stupid. Thanks!

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    inf$\{x_n: n \in N\} = 0$ and the sequence is strictly decreasing, so the limit of the sequence should be $0$ correct?2012-11-22

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A counterexample is $f(x)=x^2(2+\sin(1/x))$, when $x\neq 0$, and $f(0)=0$. Below is an image from Mathematica to help you believe that it works, but you can calculate the derivative to check that it works.

enter image description here

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    Thank you so much. The question was bothering me for a long time!2012-11-22