f is a differntiable function at [0,1] which maintains 0<=f′(x)<=1 to every x∈[0,1].
i need to prove that there's a point x∈[0,1] so
f′(x)=3x/(sqrt(3x2+6))
see here the proof Mean value theorem - Wikipedia, the free encyclopedia
The integral of your function $\displaystyle g(x)=\frac{3x}{\sqrt{3x2+6}}$ from 0 to 1 is approximately 0.55, so it would seem that you could set 0 < f(0) < 0.45 and f'(x) slightly greater than g(x) to get a function which stays between 0 and 1, but the derivative is never equal to g(x).
- Hollywood
Hi again,
I took my own advice and surfed the web for the intermediate value theorem for derivatives. The proof in wikapedia, Fermat's theorem (stationary points) - Wikipedia, the free encyclopedia, has an error or at least an incomplete argument. PlanetMath has the same proof. So here's a complete proof of the theorem.