Derivative problem

Question

Could anyone help me? Let $f:[0,1] \to \mathbb{R}$ be a function of class $C^1$ such that $f(0)=0$ and there exists $a \in ]0,1[$ with $f(a)f’(a)<0$. Show that there exists $b\in ]0,1[$ with $f’(b)=0$

I don't understand: how is different between $f'(a)$ and $f'(b)$. $a$ and $b$ are both elements of $]0,1[$, and $f(a)f'(a)< 0$, so $f'(a)$ or $f'(b)$ can not be equals to $0$.

thanks

You're told there is one value of $x$, namely, $x=a$, such that $f(x)f'(x)\lt0$. That does not make it impossible for there to be another value of $x$, namely, $x=b$, such that $f'(x)=0$. — Gerry Myerson, Aug 05 '13 at 13:15

score 2 · Answer 1 · answered Aug 05 '13 at 13:15

2

WLG (the other case is similar) assume that $f(a)>0$ and $f'(a)<0$ so by the mean value theorem there's $c\in(0,a)$ such that $$f(a)=af'(c)$$ hence $f'(c)>0$ and since $f'$ is continuous so by the intermediate value theorem and since $f'(c)>0$ and $f'(a)<0$ there's $b\in (c,a)$ s.t. $f'(b)=0$.

answered Aug 05 '13 at 13:15

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Yes you're welcome. – Aug 05 '13 at 17:09
Do you know about Ecole Polytechnique. the university of France.? and u have any exam of this school? plz share it with me – Phương Ami Aug 05 '13 at 17:13
In this site http://concours-maths-cpge.fr/fichiers.php you find all the French exams for "les écoles d'ingenieurs" including l'école Polytechnique. – Aug 05 '13 at 17:22
thank u so much. but this site has english or just French language? – Phương Ami Aug 05 '13 at 17:31
It's just for the French language. – Aug 05 '13 at 18:22

score 2 · Answer 2 · answered Aug 05 '13 at 13:16

You can prove the assertion by contradiction. Assume that for all $b\in]0,1[$, $f'(b)$ is different from zero. This implies $--$ because $f$ is assumed to be of class $C^1$ $--$ that the derivative $f'$ is continuous and therefore either positive or negative on $]0,1[$.

If the derivative $f'$ is positive, then, for any $x\in]0,1[$, both $f(x)$ and $f'(x)$ are positive, contradicting the existence of an $a\in]0,1[$ such that $f(a)f'(a)<0$. The derivative being negative leads to a similar contradiction.

Derivative problem

2 Answers2