DJBruce Posted January 16, 2012 Share Posted January 16, 2012 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? Link to comment Share on other sites More sharing options...

mathematic Posted January 16, 2012 Share Posted January 16, 2012 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. Link to comment Share on other sites More sharing options...

DJBruce Posted January 17, 2012 Author Share Posted January 17, 2012 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. Link to comment Share on other sites More sharing options...

mathematic Posted January 17, 2012 Share Posted January 17, 2012 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 Link to comment Share on other sites More sharing options...

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