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Proof Regarding a Twice Differentiable Function

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Context: We are learning Rolle, Lagrange, Fermat, Taylor Theorems in our Real Analysis class. We just finished continuity and are now studying differentiation.

Question:
Let f: [a,b] --> R, a < b, twice differentiable with the second derivative continuous such that f(a)=f(b)=0.

Denote M = sup |f "(x)| where x is in [a,b]

and g:[a,b] --> R, g(x)=(1/2)(x-a)(b-x)

i) Prove that for all x in [a,b], there exists

Cx in (a,b) such that f(x)= - f "(Cx)g(x).

ii) Prove that if there exists x0 in (a,b) such that

|f(x0)| = Mg(x0), then f = Mg or f=-Mg.

Notes:
Cx is a constant dependent on x ,

x0 is a particular x in (a,b)

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Solution Summary

The solution is comprised of an attachment of two pages of carefully formatted and worked calculations to give the required proof.

Solution Preview

In your problem, you mentioned Cx. It should be a number which is determined by x. It is not C times x.
Please see the attached file for the detailed proof.

Proof:
1. Since , is a closed interval, then should have maximum and minimum value in . Suppose , . I want to show that is between and for any . Then by the Mean Value Theorem, there is a , such that . This implies . ...

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