# How to prove that derivatives have the Intermediate Value Property

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?

• 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. Aug 1, 2011 at 1:04
• @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, 2011 at 1:14
• @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, 2011 at 1:43
• @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". Aug 1, 2011 at 1:58
• Possible duplicate of Darboux Theorem
– mlc
May 3, 2017 at 6:35

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...

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

• 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, 2011 at 1:39

A proof without words:

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.

• This is the same proof given by Apostol in his Mathematical Analysis Second Edition 1974. May 20, 2018 at 2:43
• Wow @ParamanandSingh I find it amazing that the paper got published in AMM despite that! Feb 26, 2023 at 13:20
• @Shinrin-Yoku: editors of reputed journals can also make mistakes once in a while. Let's give them that much room. Feb 26, 2023 at 16:59
• 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 $X$. May 3, 2023 at 17:34

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 $$g(a) = f'(a) and $$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 $$h(x) := \frac{f(b) - f(x)}{b-x}$$

Suppose $$a and $$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 $$g(a), then since $$g'(b)<0$$ there's some $$x_1 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 $$g'(x_3)=0$$.