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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?

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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

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 $[a, b]$?

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

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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

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