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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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    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 is nonempty.

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, and we are done.

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