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"Suppose that the function $f: \mathbb{R} \rightarrow \mathbb{R} $ is differentiable and that ${x_n}$ is a strictly decreasing bounded sequence with $f(x_n) \leq f(x_n+1)$ for all $n$ in $\mathbb{N}$. Prove that there is a number $x_o$ at which $f'(x_o) \geq 0$."

This question really seems related to the Monotone Convergence Theorem, what with the monotone bounded sequence being explicitly described in the problem. However, I don't understand where I should start. I just got two proofs just now (a triumphant moment, let me tell you) but the sequence seems out of place. I don't see how I can link up this sequence with the newer derivatives definitions, unless I should use the epsilon-delta definition of continuity along with the limit definition of the derivative in order to have some common ground, but even though I can kind of visualize their being related, I am having trouble after recognizing that the sequence converges to the supremum of my defined set.

I know that I can define some set $S = \left\{{a_n | n \in \mathbb{N}}\right\}$ bounded above and by Completeness, $S$ has a supremum. Then I can define $a$ = sup(S) so now I know that ${a_n} \rightarrow a$. At this point I feel I'm just playing with the definitions and now approaching any kind of investigation of inverse derivatives (the chapter) or even derivatives at all! Maybe I'm missing some kind of crucial theorem or something.

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Do you mean $f(x_n) \le f(x_{n+1})$? This seems like a really weird condition to impose. Either way I don't see at all how this is possible ; please confirm your statement you're trying to prove is well-written. –  Patrick Da Silva Oct 21 '12 at 13:01
    
The question is identical to its phrasing in the book. I will attach a photograph for confirmation. 2 sec –  Arthur Collé Oct 21 '12 at 13:34
    
Voila - oi49.tinypic.com/2usy04z.jpg –  Arthur Collé Oct 21 '12 at 13:36

1 Answer 1

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We find inifinitely many such points, namely one between $x_n$ and $x_{n+1}$ for each $n\in\mathbb N$. By the mean value theorem, we have $$f(x_{n+1})-f(x_n) =f'(\xi)(x_{n+1}-x_n)$$ for some $\xi$ between $x_n$ and $x_{n+1}$. The left hand side is $\ge 0$ and the factor $x_{n+1}-x_n$ is $<0$, thus $f'(\xi)\le 0$.

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You need the function to be continuously differentiable to do this. It is not supposed in the question. And OP asks for positive derivative not negative... hence my comment! –  Patrick Da Silva Oct 21 '12 at 13:28

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