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I'm reading a book which gives this theorem without proof:

If a and b are any two points in an interval on which ƒ is differentiable, then ƒ' takes on every value between ƒ'(a) and ƒ'(b).

As far as I can say, the theorem means that the fact ƒ' is the derivative of another function ƒ on [a, b] implies that ƒ' is continuous on [a, b].

Is my understanding correct? Is there a name of this theorem that I can use to find a proof of it?

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Continuity implies Intermediate Value Property. The converse implication does not necessarily hold. –  André Nicolas Aug 1 '11 at 0:50
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One standard example is $f(x)=x^2\sin(1/x)$ if $x \ne 0$, $f(0)=0$. It is everywhere differentiable, its derivative has IVP, of course, but is not continuous. I think the general result is due to Darboux. –  André Nicolas Aug 1 '11 at 1:04
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@André: you're thinking is right. See the entry on Wikipedia on the Darboux theorem. There are much worse examples and almost as easy to describe: see Conway's base 13 function. This thread here is closely related. –  t.b. Aug 1 '11 at 1:14
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@ablmf: this is sometimes called Darboux-continuous; @Jack: nothing wrong with the title (except capitalization); @André: I apologize for that horrible typo at the beginning of my last comment. ` –  t.b. Aug 1 '11 at 1:43
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@ablmf There's a special name for the class of functions $f$ such that $f'$ exists and $f'$ is continuous, namely $C^1$. I sometimes use this as a "mnemonic" to remind myself that continuity of derivative is indeed a stronger property than just "$f$ is differentiable". –  Srivatsan Aug 1 '11 at 1:58

2 Answers 2

up vote 6 down vote accepted

The result is commonly known as Darboux’s theorem, and the Wikipedia article includes a proof.

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Et pour ceux qui maîtrisent le français: Gaston Darboux, Mémoire sur les fonctions discontinues, Annales scientifiques de l'École Normale Supérieure, Sér. 2, 4 (1875), p. 57-112 (IX. Définition d'une classe singulière de fonctions, p.109). –  t.b. Aug 1 '11 at 1:39

This is actually a nice exercise. (In fact, if I recall correctly, it was given as a problem on the very first math exam I took in college. Unfortunately all I was able to say was that it was true if $f'$ was assumed to be continuous, for which I received zero credit.)

Let me set it up a little bit and leave the rest to the interested readers: it is easy to reduce the general case to the following: suppose that $f'(a) > 0$ and $f'(b) < 0$. Then there exists $c \in (a,b)$ with $f'(c) = 0$.

Here's the idea: an interior point with $f'(c) = 0$ is a stationary point of the curve (and conversely!). In particular the derivative will be zero at any interior maximum or minimum of the curve. Recall that since $f$ is differentiable, it is continuous and therefore assumes both a maximum and minimum value on $[a,b]$. So we're set unless both the maximum and minimum are attained at the endpoints. Perhaps the sign conditions of $f'$ at the endpoints have something to do with this...

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