# Can Rolle's Theorem be true for the critical point where derivative doesnt exist?

there is the problem that I met 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.

• $c = 0$, while not a critical point, is an extreme point, and a global minimum. – Arthur Dec 1 '14 at 9:49
• 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 – aukxn Dec 1 '14 at 10:06
• 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. – Siminore Dec 1 '14 at 10:47
• 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 ? – aukxn Dec 1 '14 at 12:30
• Let me be clear: the function $x \mapsto x^{2/3}$ is not, not, and not differentiable at zero. – 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.

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