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I know that if $f:I\to R$ is differentiable and $f'$ ascending (or descending), then $f'$ is continuous ( $I$ is an interval and $R$ the real number). Is the above statement true whenever $f'$ is one to one?

Aliakbar
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Yes, thanks to the middle value property of the derivative - namely that $f^\prime \restriction_{[x,y]}$ attends every value between $f^\prime(x)$ and $f^\prime(y)$. The property itself is an easy exercise.

Alexander Shamov
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  • How do you deduce the continuity of $ f' $ from the fact that it has the Intermediate Value Property? There are discontinuous functions that possess this property. – Haskell Curry Jan 08 '13 at 06:20
  • I did not downvote you, by the way. I just wanted to say that I don’t see the implication right away. – Haskell Curry Jan 08 '13 at 06:22
  • @HaskellCurry: Sure, but together with injectivity this implies that $f^\prime$ is monotone. – Alexander Shamov Jan 08 '13 at 06:27
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    Monotonicity implies continuity, that’s for sure as it is guaranteed by the Intermediate Value Property. However, how injectivity implies monotonicity is something that I fail to see. I hope you don’t mind my being a little fussy over the facts. :) – Haskell Curry Jan 08 '13 at 06:27
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    @Haskell: Here are some proofs that injective functions with IVP are continuous: http://math.stackexchange.com/questions/7172/injective-functions-with-intermediate-value-property-are-continuous-better-proo. (Although in some of them it is taken for granted that it is straightforward to show that an injective function with IVP is monotone.) – Jonas Meyer Jan 08 '13 at 06:28
  • Alright! I see it now. +1 for your answer. :) – Haskell Curry Jan 08 '13 at 06:30