there is the problem that I met

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At 0 the derivative of f(x) doesn't exist so 0 is the critical number but the conclusion of Rolle's theorem is the f'(c) (here c=0) must be 0. Are there any explaination in this case


A critical point is a point where the derivative exists and equals zero. So $c=0$ is not a critical point of $f$. The point here is that $f$ fails to satisfy all of the assumptions of Rolle's theorem, and indeed the conclusion fails, too.

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    $\begingroup$ $c = 0$, while not a critical point, is an extreme point, and a global minimum. $\endgroup$ – Arthur Dec 1 '14 at 9:49
  • $\begingroup$ sorry, I must say critical number to be exact as the book I use so I still don't have any explaination for this case $\endgroup$ – aukxn Dec 1 '14 at 10:06
  • $\begingroup$ Please, read the assumptions of Rolle's theorem. Is your function $f$ differentiable at any point of the open interval $(-1,1)$? I do not think so. Hence there is no contradiction, since Rolle's theorem is not applicable to your function. $\endgroup$ – Siminore Dec 1 '14 at 10:47
  • $\begingroup$ I try using the definition of derivative at a point to find the derivative at c=0. The result is f'(c)=0 but if I use general derivative then there is no derivative at x=0. So if that's the case then there is point c that derivative of f will be 0 so the question is wrong because actually there is point c that f'(c) = 0 ? $\endgroup$ – aukxn Dec 1 '14 at 12:30
  • $\begingroup$ Let me be clear: the function $x \mapsto x^{2/3}$ is not, not, and not differentiable at zero. $\endgroup$ – Siminore Dec 1 '14 at 14:53

If f is differentiable, it also be continuous and have critical point. According to Fermat theorem, if f is differentiable at this point it's derivative must be 0. So your example does not meet the prerequisite.

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  • $\begingroup$ This is incorrect. The function need only be differentiable. $\endgroup$ – JP McCarthy Dec 1 '14 at 9:59
  • $\begingroup$ You are correct. It just uses Fermat theorem, not IVT. $\endgroup$ – user193702 Dec 1 '14 at 10:08
  • $\begingroup$ This is still not precise enough. The function isn't differentiable on $(-1,1)$ --- that is the problem. $\endgroup$ – JP McCarthy Dec 1 '14 at 11:01

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