be f(x) Continuous function in [0,2] , and Differentiable in (0,2).

0<f(1)<f(0)<f(2).

Prove that the Differentiable become zero in any point in the open Interval (0,2).

thanks

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- Nov 28th 2010, 01:43 AMZOOZProving the Differential of the Function
be f(x) Continuous function in [0,2] , and Differentiable in (0,2).

0<f(1)<f(0)<f(2).

Prove that the Differentiable become zero in any point in the open Interval (0,2).

thanks - Nov 28th 2010, 02:11 AMFernandoRevilla
Is it included in the hypothesis $\displaystyle f'$ continuous in $\displaystyle (0,2)$ ? . In such a case, you only need to apply Lagrange's theorem to the intervals $\displaystyle [0,1]$ and $\displaystyle [1,2]$ .

Regards.

Fernando Revilla - Nov 28th 2010, 02:12 AMHallsofIvy
I think what you want is a proof that the derivative is 0 at

**some**point in (0, 2). "Any point" would mean**every**point and that is not true.

Use the "mean value theorem" on the intervals (0, 1) and (1, 2). There exist a number, a, between 0 and 1, such that (f(1)- f(0))/(1- 0)= f(1)- f(0)= f'(a). Since f(0)> f(1), f'(a) is negative. There exist a number, b, between 1 and 2, such that (f(2)- f(1))/(2- 1)= f(2)- f(1)= f'(b). Since f(1)< f(2), f'(b) is positive.

Now, do you have the theorem that says that, though the derivative of a given function is not necessarily continuous, it still must satisfy the "intermediate value property"? If so that immediately tells you that there is some point, c, between a and b and so between 0 and 2, such that f'(c)= 0.

If you do not have that theorem, you should prove it! Define g(x) on [0, 2] by g(x)= (f(x)- f(0))/x as long as x is not 0 and g(0)= f'(0). You can show that g is continous on [0, 2] and so the intermediate value property applies to g: g(x) takes on all values between g(0)= f'(0) and g(2)= (f(2)- f(0))/2. Define h(x) on [0, 2] by h(x)= (f(2)- f(x))/(2- x) as long as x is not 2 and h(2)= f'(2). Then h(x) takes on all values between h(0)= (f(2)- f(0))/2 and h(2)= f'(2). But by the mean value theorem, f'(x) takes on all of those values itself.

This does NOT belong in "pre-calculus"! - Nov 28th 2010, 02:19 AMFernandoRevilla
Excellent comment

**HallsofIvy**. For that reason I asked about continuity of $\displaystyle f'$ .

Regards.

Fernando Revilla - Dec 2nd 2010, 05:21 AMZOOZ
thanks to both of you.