Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose that a function $f(x)$ is differentiable $\forall x \in [a,b]$. Prove that $f'(x)$ takes on every value between $f'(a)$ and $f'(b)$.

If the above question is a misprint and wants to say "prove that $f(x)$ takes on every value between $f(a)$ and $f(b)$", then I have no problem using the intermediate value theorem here.

If, on the other hand, it is not a misprint, then it seems to me that I can't use the Intermediate value theorem, as I can't see how I am authorised to assume that $f'(x)$ is continuous on $[a,b]$.

Or perhaps there is another way to look at the problem?

share|improve this question
Google "Darboux's Theorem". –  David Mitra Jan 5 '13 at 13:19
I don't knowf it's relevant, but somewhere in M. Spivak's book "Calculus," there is something called the "Sunrise lemma", or maybe "Rising Sun Lemma", which tells you that a derivative has better properties than you might expect. –  Geoff Robinson Jan 5 '13 at 13:22

3 Answers 3

This is not a misprint. You can indeed prove that $f'$ takes every value between $f'(a)$ and $f'(b)$. You cannot, however, assume that $f'$ is continuous. A standard example is $f(x) = x^2 \sin(1/x)$ when $x \ne 0$, and $0$ otherwise. This function is differentiable at $0$ but the derivative isn't continuous at it.

To prove the problem you have, consider the function $g(x) = f(x) - \lambda x$ for any $\lambda \in (f'(a), f'(b))$. What do you know about $g'(a)$, $g'(b)$? What do you conclude about $g$ in the interval $[a, b]$?

share|improve this answer

This is a result known as Darboux's Theorem. A proof can be found in the Wiki article here.

share|improve this answer

If it is a misprint then this follows from the Intermediate Value Theorem

But I am pretty certain this is not a misprint. This fact is called Darboux's Theorem and doesn't need $f^{\prime}$ to be continuous. That is why the Intermediate Value Theorem can't be used. Here is a proof sketch of this theorem:

Suppose $f^{\prime}(a)<r<f^{\prime}(b)$ and define the function $g:[a,b]\to \mathbb{R}$ as \begin{equation}g(t)=f(t)-rt\end{equation}

By the Extreme Value Theorem on $[a,b]$, \begin{equation}\exists \xi\in [a,b]:\forall x\in [a,b]\ g(\xi)\le g(x)\end{equation}

If we show $\xi\in (a,b)$ we are done (Fermat Theorem).

Obviously $g$ is differentiable and $g^{\prime}(a)=f^{\prime}(a)-r<0$ while $g^{\prime}(b)=f^{\prime}(b)-r>0$ .

Observe $\exists \delta_1,\delta_2>0$ so that $\forall x\in (a,a+\delta_1) \ g(x)<g(a)$ and $\forall x\in (b-\delta_2,b) \ g(x)<g(b)$. Thus, \begin{equation}\exists x_1,x_2\in (a,b): g(x_1)<g(a)\text{ and } g(x_2)<g(b)\end{equation} This yields that $\xi\in (a,b)$ and we are done

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.