# If the mean value theorem always gives a $c \in (0,\infty)$ such that $f'(c) > 0$ is there an interval starting from $0$ such that $f' >0$

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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If a sequence is decreasing and bounded from below by zero, it converges. but it does not have to converge to zero –  Amr Nov 22 '12 at 5:32
inf$\{x_n: n \in N\} = 0$ and the sequence is strictly decreasing, so the limit of the sequence should be $0$ correct? –  Eric Nov 22 '12 at 5:37

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.
Is there always a point such the the derivative is greater than $0$? –  Eric Nov 22 '12 at 5:40
Yes there is, although I just realized I did not quite include your condition that $f(x)>0$ (I only have $f(x)\geq 0$), so I should adjust it. Edit: Done. (The old example was $x^2(1+\sin(1/x))$, and the $1$ was changed to $2$ so that $f$ is always strictly positive when $x\neq 0$.) –  Jonas Meyer Nov 22 '12 at 5:41
Awesome :D How ever I am wonder what happens if the function is uniformly continuous on $R$? –  Eric Nov 22 '12 at 5:49
@Eric: That causes no problems whatsoever. Only what happens near $0$ is important, and every continuous function on $[0,1]$ is uniformly continuous. Every differentiable function on $[0,1]$ can be extended to a uniformly continuous differentiable function on $[0,\infty)$, for example by extending the tangent line at $(1,f(1))$ to the right as the definition of $f$ on $(1,\infty)$. –  Jonas Meyer Nov 22 '12 at 5:51