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I understand the conditions for Rolle's theorem in english : a function $f$ has to be continous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$. But I'm studying in Russian, and they write the conditions like this:

a) function $f$ is continous on $[a,b]$,

b) function $f$ has at all points of the interval $(a,b)$ a finite or definite sign infinite derivative,

c) $f(a)=f(b)$.

I'm having trouble understanding point b), what does 'definite sign infinite derivative' mean? Is there a difference between that, and simply being differentiable on $(a,b)$? Maybe the translation isn't quite right, my russian isn't perfect, but I think that's what it means in English.

All I could think of was, maybe they are refering to when the limit of $f(x)$ when $x \to a$ or $x \to b$ is equal to $\pm\infty$, but then $f$ wouldn't be continous on $[a,b]$. So, I'm lost.

Do you guys have any ideas? Thanks a lot in advance.

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    $\begingroup$ Look at the paragraph titled "Generalization" here $\endgroup$
    – saulspatz
    Nov 10, 2019 at 16:39

3 Answers 3

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$f$ having a finite derivative in $\xi$ means that it has a derivative in $\xi$ in the usual sense: $f'(\xi) = \lim_{x\to \xi} \dfrac{f(x) - f(\xi)}{x-\xi}$ is a real number.

$f$ having a definite sign infinite derivative means that $f'(\xi) = \lim_{x\to \xi} \dfrac{f(x) - f(\xi)}{x-\xi} = \infty$ or $f'(\xi) = \lim_{x\to \xi} \dfrac{f(x) - f(\xi)}{x-\xi} = -\infty$. This is usually denoted as an improper derivative at $\xi$. Geometrically it means that that the graph of $f$ has a vertical tangent at the point $(\xi,f(\xi))$. An example is $f(x) = \sqrt[3]{x}$. You have $f'(0) = \infty$.

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  • $\begingroup$ Nicely done, Wikipedia explanation of generalized version might be more of what she is looking for. $\endgroup$
    – cemsicles
    Nov 10, 2019 at 16:57
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For example $y=x^{1/3},$ a continuous function on all of $\mathbb R,$ is not differentiable at $0$ in the usual sense. But it has a "definite sign infinite derivative" of $+\infty$ at $0.$ I.e., both the left and right derivatives of $x^{1/3}$ at $0$ are $+\infty.$ Contrast this with $y=\sqrt {|x|},$ which has left and right infinite derivatives at $0$ of $-\infty,+\infty$ respectively.

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I believe English definition you are referring to is the standard Rolle's theorem, while Russian is the generalized theorem.

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