Mean value theorem variation proof

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Homework Help Overview

The discussion revolves around a proof related to the Mean Value Theorem, specifically concerning a differentiable function defined on the interval [0,1] with given derivative conditions at the endpoints. The original poster presents a statement that claims the existence of a point c in (0,1) where the derivative equals the function value.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the validity of the original statement and the implications of providing a counterexample. There is a discussion about the necessity of additional conditions on the function values at the endpoints.

Discussion Status

The conversation reflects differing interpretations of the statement's validity. Some participants assert that the existence of a counterexample indicates the statement is false, while others clarify the distinction between a statement being unprovable and being false. The dialogue suggests an ongoing examination of the assumptions involved.

Contextual Notes

There is mention of potential missing conditions, such as the values of the function at the endpoints, which may affect the validity of the original claim.

gruba
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Homework Statement


Let f is differentiable function on [0,1] and f^{'}(0)=1,f^{'}(1)=0. Prove that \exists c\in(0,1) : f^{'}(c)=f(c).

Homework Equations


-Mean Value Theorem

The Attempt at a Solution



The given statement is not true. Counter-example is f(x)=\frac{2}{\pi}\sin\frac{\pi}{2}x+10.
Does this mean that the statement can't be proved?
 
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I suspect you should also have ##f(0) = 0## and ##f(1) = 1##.

Otherwise, your counterexample is fine
 
And, yes, if a statement is not true, it cannot be proved!
 
gruba said:

Homework Statement


Let f is differentiable function on [0,1] and f^{'}(0)=1,f^{'}(1)=0. Prove that \exists c\in(0,1) : f^{'}(c)=f(c).

Homework Equations


-Mean Value Theorem

The Attempt at a Solution



The given statement is not true. Counter-example is f(x)=\frac{2}{\pi}\sin\frac{\pi}{2}x+10.
Does this mean that the statement can't be proved?

There is a difference between "can't be proved" and "is false". If a statement has a counterexample, it is false; that is stronger than the claim that it 'cannot be proved' (but, of course, it cannot be proved as well).
 

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