# Problem about Mean Value Theorem

Let $f: [0,1] \rightarrow \mathbb{R}$ be differentiable, with $f(0) = 0$, $f(1) = 1$ and $f'(0) = 1/2$. Can we conclude that $\exists a \in (0,1) \ni f'(a) = 2/ \pi$?

I don't know if this is true. I have to apply the Mean Value Theorem, because $f$ verifies the conditions, but I don't know the answer.

• What does the ]0, 1[ notation mean? – Mark Pattison Aug 8 '15 at 17:41
• @MarkPattison: My guess is that it means the open interval $(0,1)$. The books in our local schools, for example, use this ugly notation. My freshman calculus students have a bit of unlearning to do. A fresh batch awaits me in three weeks time. Sigh. – Jyrki Lahtonen Aug 8 '15 at 18:41
• @JyrkiLahtonen The notation $]a;b[$ is a lot better than $(a;b)$ and more consistent with geometry. – Paracosmiste Aug 8 '15 at 19:07
• @whatever: I'm familiar with its pros and cons (having been forced to use it up to high school). The writers of advanced textbooks disagree with you, and it's my duty to bring my students in touch with the real world :-) – Jyrki Lahtonen Aug 8 '15 at 19:12
• @JyrkiLahtonen I know $(a,b)$ is used in most (english) textbooks but that doesn't change the fact that it's inconsistent with geometry and ambiguous. Moreover, Bourbaki used $[a;b]$. – Paracosmiste Aug 8 '15 at 19:17

First, the MVT says that there's a point $c$ in the interval with $f'(c) = 1 > \frac{2}{\pi}$.
Second, you know that $f'(0) = \frac{1}{2} < \frac{2}{\pi}$.
So you'd like to say "well, between $0$ and $c$, the derivative must take on the value $\frac{2}{\pi}$, by continuity."