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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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    $\begingroup$ 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. $\endgroup$ Aug 1, 2011 at 1:04
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    $\begingroup$ @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. $\endgroup$
    – t.b.
    Aug 1, 2011 at 1:14
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    $\begingroup$ @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. ` $\endgroup$
    – t.b.
    Aug 1, 2011 at 1:43
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    $\begingroup$ @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". $\endgroup$
    – Srivatsan
    Aug 1, 2011 at 1:58
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    $\begingroup$ Possible duplicate of Darboux Theorem $\endgroup$
    – mlc
    May 3, 2017 at 6:35

5 Answers 5

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

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    $\begingroup$ 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). $\endgroup$
    – t.b.
    Aug 1, 2011 at 1:39
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A proof without words:

enter image description here

The slope of the secant varies continuously from $f'(a)$ to $f'(b)$, so takes on every value in $[f'(a), f'(b)]$. By the mean value theorem, so does $f'$.

For the details, you can read the original proof by Lars Olsen that this animation is based on. Remarkably, this proof only seems to have been discovered in 2004.

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    $\begingroup$ This is the same proof given by Apostol in his Mathematical Analysis Second Edition 1974. $\endgroup$
    – Paramanand Singh
    May 20, 2018 at 2:43
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    $\begingroup$ Wow @ParamanandSingh I find it amazing that the paper got published in AMM despite that! $\endgroup$ Feb 26, 2023 at 13:20
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    $\begingroup$ @Shinrin-Yoku: editors of reputed journals can also make mistakes once in a while. Let's give them that much room. $\endgroup$
    – Paramanand Singh
    Feb 26, 2023 at 16:59
  • $\begingroup$ There are a few problems with this picture: $f(a)=f(b)$, and $f'(a)=f'(b)$, and the variable points are labelled $A$ and $B$ instead of $X$. $\endgroup$
    – mr_e_man
    May 3, 2023 at 17:34
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Here's what I came up with off the top of my head: Let $c$ be between $f'(a)$ and $f'(b)$, and want to show there exists $x_0 \in (a,b)$ such that $f'(x_0) =c$.

Case 1: $c \leq \frac{f(b)-f(a)}{b-a}$.

Then set $$ g(x) := \frac{f(x) - f(a)}{x-a} $$ Then $g$ is clearly continuous for $x>a$, and $g(a) = f'(a)$, so $g$ is continuous at $a$ as well. Since $g(a) = f'(a) <c$ and $g(b) \geq c$, by the intermediate value theorem there exists a point $x_1 \in (a,b]$ such that $c=g(x_1)$. Then by the mean value theorem, there exists $x_0 \in (a,x_1]$ such that $f'(x_0) = c$.

Case 2: If $c > \frac{f(b)-f(a)}{b-a}$, then try a similar argument with $$ h(x) := \frac{f(b) - f(x)}{b-x} $$

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Suppose $a<b$ and $f'(a)\neq f'(b)$. Given any $k$ between $f'(a)$ and $f'(b)$, let $g(x)=\pm(f(x)-kx)$, so that $g'(x)=\pm(f'(x)-k)$, where the sign is chosen such that $g'(a)>0>g'(b)$.

If $g(a)=g(b)$, then by Rolle's theorem there's some point $x\in(a,b)$ where $g'(x)=0$, that is, $f'(x)=k$.

(Here's my proof of Rolle's theorem using the IVT.)

If $g(a)>g(b)$, then since $g'(a)>0$ there's some $x_1>a$ where $g(x_1)>g(a)>g(b)$. Then the IVT gives a point $x_2\in(x_1,b)$ where $g(x_2)=g(a)$. Then Rolle gives a point $x_3\in(a,x_2)$ where $g'(x_3)=0$.

If $g(a)<g(b)$, then since $g'(b)<0$ there's some $x_1<b$ where $g(x_1)>g(b)>g(a)$. Then the IVT gives a point $x_2\in(a,x_1)$ where $g(x_2)=g(b)$. Then Rolle gives a point $x_3\in(x_2,b)$ where $g'(x_3)=0$.

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