f is a differntiable function at [0,1] which maintains 0<=f′(x)<=1 to everyx∈[0,1].

i need to prove that there's a pointx∈[0,1] so

f′(x)=3x/(sqrt(3x2+6))

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- May 2nd 2013, 11:00 AMorirquestion with derivative
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*)=3*x*/(*s**q**r**t*(3*x*2+6)) - May 2nd 2013, 12:26 PMdokrbbRe: question with derivative
Hi there, if I'm not wrong you should do the following:

first - find the derivative of f(x)

second - use the Mean Value Theorem f(c) = (f(1) - f(0))/(1-0),

dokrbb - May 2nd 2013, 08:23 PMorirRe: question with derivative
But i dont have any written funcrion, so how would i find its derivative?

- May 3rd 2013, 05:17 AMdokrbbRe: question with derivative
see here the proof Mean value theorem - Wikipedia, the free encyclopedia

- May 3rd 2013, 06:42 AMhollywoodRe: question with derivative
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 - May 3rd 2013, 09:05 AMjohngRe: question with derivative
Hi,

What you need is the intermediate value theorem for derivatives; you can look up Darboux's theorem on the web for a proof. Here's a proof of your problem:

Attachment 28237 - May 3rd 2013, 05:25 PMhollywoodRe: question with derivative
Yes, johng, you are correct. Thank you. I must have subconsciously changed the conditions from $\displaystyle 0 \le f'(x) \le 1$ to $\displaystyle 0 \le f(x) \le 1$.

- Hollywood - May 3rd 2013, 08:31 PMjohngRe: question with derivative
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.

Attachment 28239