How Can You Prove a Symmetric Derivative Property for a Continuous Function?

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Homework Help Overview

The discussion revolves around proving a property related to symmetric derivatives for a continuous function defined on the interval [0,1]. The function is continuous on [0,1] and differentiable on (0,1), with the condition that the function values at the endpoints are equal.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the Mean Value Theorem (MVT) as a potential starting point for the proof. There is a recognition that the problem may involve more complexity than initially anticipated. One participant suggests defining a new function, g(x), to aid in the exploration of the problem.

Discussion Status

The discussion is ongoing, with participants sharing initial thoughts and approaches. There is an acknowledgment of the complexity of the problem, and a new function has been introduced to facilitate further exploration.

Contextual Notes

Participants are working under the constraints of the problem's conditions, specifically the continuity and differentiability of the function, as well as the equality of function values at the endpoints.

johnson12
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Let f be continuous on [0,1] and differentiable on (0,1) such that f(0)=f(1), prove that there exist a 0 < c < 1 such that f[tex]\acute{}[/tex](1-c) = -f[tex]\acute{}[/tex](c).

thanks for any suggestions.
 
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Just from the givens it made me think of MVT...
 
That was my first impression but it seems to be a little trickier than that.
 
johnson12 said:
That was my first impression but it seems to be a little trickier than that.

It's a little more complicated, but not much. Define g(x)=f(x)+f(1-x). g(0)=g(1), right?
 

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