So I've been trying to prep for an upcoming math competition by going through some problems. This one has me a little stuck,

"Suppose [latex]f[/latex] is a differentiable function function on [0, 2] then there exists a point [latex]c\in [/latex] such that:

[latex]f''©=f(0)-2f(1)+f(2).[/latex]"

I am not sure this statement is true under these conditions, and think twice differentiable is probably required. Assuming that [latex]f[/latex] is twice differentiable I have tried applying the mean value theorem, and have been able to show that there exists [latex]a,b\in [0,2][/latex] and [latex]c\in [f'(a), f'(b)] [/latex] such that:

[latex]f(0)-2f(1)+f(2)=f'(a)+f'(b)=f''©.[/latex]

 

However, I am not seeing that that [latex]f'(a), f'(b)\in [0,2].[/latex]

 

Any ideas on how to continue in this problem?

a would be between 0 and 1, while b would be between 1 and 2 (your intermediate expression should be f'(b)-f'(a)), and c between a and b, so definitely between 0 and 2.

Sorry I was sloppy when I stated what I had done towards a solution:

 

I did get that

[math]f(0)-2f(1)+f(2)=-f'(a)+f'(b)=f''©(b-a).[/math]

where c is in [0, 2]. But the there is still the (b-a) term that needs to be taken care of somehow, and that is where I was stuck. Sorry for the poor description in the OP.

I haven't tried to work it out, but I believe an approach would be as follows: The mean value theorem is essentially a corollary of Rolle's theorem. The is a generalization of Rolle's theorem for higher derivatives, which could be used to answer your question.

 

http://en.wikipedia.org/wiki/Rolle%27s_theorem