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I would, if possible, to give a rigorous proof of this question

can someone help me?

Let $f: [a, b​​] \to\mathbb {R}$ a differentiable function such that $$ \begin{align*} f(a)&=f(b)\\ f'(a) &= f'_{+}(a)>0,\\ f'(b) &= f'_{-}(b)>0. \end{align*} $$ Prove that there exists $c\in (a,b​​)$ such that $f(c) = 0$ and $f '(c) \le 0. $

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    do you want $f(a)=f(b)=0$?2011-12-12
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    no i want only $f(c)$, and is $f(a)=f(b)$ and no $f(a)=f(b)=0$2011-12-12
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    this is simply not true without any additional assumption. Just look at $\sin(x) + 2$ on $[0, 2\pi]$.2011-12-12
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    You cannot guarantee such a $c$: given any differentiable function $f$, let $m$ be its minimum (exists, since $f$ is continuous on a finite closed interval). Then $g(x)=f(x)+2|m|+1$ satisfies the hypothesis of the claim, but it is always positive so there is no $c$ with $g(c)=0$.2011-12-12
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    derivable not differentiable ... i'm sorry!2011-12-12
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    I'm sorry, but: what's the difference, and how does it affect the proposed examples? (I am assuming that "derivable" is a term of art with which I am not familiar, so perhaps you can enlighten me)2011-12-12
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    Sorry, but since I write ITALY probably, indeed certainly do a bit of difficulty with the language .... for so mean a differentiable function such that admits $ f '(x) $ ....2011-12-12
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    @FrConnection: We say a function $f$ is "differentiable on $[a,b]$" if $f'(c)$ exists for every $c$ in $(a,b)$, and the one-sided derivatives $f'_+(a)$ and $f'_-(b)$ both exist. Is that what you mean? If so, look at the counterexamples given by Thomas and myself.2011-12-12
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    yes it's what I mean2011-12-12
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    If I understand correctly, the problem would be true only if $f(a)=f(b)=0$?2011-12-12

1 Answers 1

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As the comments clearly show, the given statement is false.

To produce a true statement, you either need to assume $f(a)=f(b)=0$ in the hypothesis, or $f(c)=f(a)$ in the conclusion.

The latter case is easily reduced to the former by translation.

I'll take the liberty of stating and proving this:

Theorem: If $f$ is differentiable on $[a,b]$, the appropriate one sided derivatives at the endpoints are positive, and $f(a)=f(b)=0$, then there is a $c\in(a,b)$ with $f(c)=0$ and $f'(c)\le0$.

Proof: Since the derivative from the right at $a$ exists and is positive, $f$ takes a positive value for some $e>a$. Since the derivative from the left exists at $b$ and is positive, $f$ takes a negative value for some $s$ with $e

By the Intermediate Value Theorem, the set $Z=\{x:f(x)=0, e

Let $c$ be the infimum of $Z$. In the case that $Z$ is infinite, by the grace of the continuity of $f$, $f(c)=0$. If $Z$ is finite, we of course have $f(c)=0$.

Now note that $f'(c)$ cannot be positive (otherwise we could find a zero of $f$ greater than e and strictly less than $c$).

Of course, since $c\in Z$, we have $a

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    $c$ is the infimum of $Z$ and $c\in Z$ ?? $c\in(a;b)$, and so $a2011-12-12