# question with derivative

• May 2nd 2013, 12:00 PM
orir
question 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)=3x/(sqrt(3x2+6))
• May 2nd 2013, 01:26 PM
dokrbb
Re: 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, 09:23 PM
orir
Re: question with derivative
But i dont have any written funcrion, so how would i find its derivative?
• May 3rd 2013, 06:17 AM
dokrbb
Re: question with derivative
• May 3rd 2013, 07:42 AM
hollywood
Re: question with derivative
The integral of your function $g(x)=\frac{3x}{\sqrt{3x2+6}}$ 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, 10:05 AM
johng
Re: 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, 06:25 PM
hollywood
Re: question with derivative
Yes, johng, you are correct. Thank you. I must have subconsciously changed the conditions from $0 \le f'(x) \le 1$ to $0 \le f(x) \le 1$.

- Hollywood
• May 3rd 2013, 09:31 PM
johng
Re: 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