# Having trouble understanding condition for Rolle's Theorem (russian translation)

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.

• Look at the paragraph titled "Generalization" here Nov 10, 2019 at 16:39

$$f$$ 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 $$f'(0) = \infty$$.
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.