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Proof Regarding a Twice Differentiable Function

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)

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 . ...

Solution Summary

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

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