Exercise with lagrange and derivatives

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Homework Help Overview

The discussion revolves around a problem involving Lagrange's theorem in the context of continuous and differentiable functions. The original poster is tasked with demonstrating the existence of a point \( t \) in the interval \( (a,b) \) that satisfies a specific equation relating the function values and derivatives.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply Lagrange's theorem but struggles with the appearance of an additional variable in their reasoning. Another participant suggests defining a new function \( g \) to simplify the problem, leading to a question about the existence of \( t \) that satisfies a derived condition. There is also a concern about adhering strictly to the requirements of using Lagrange's theorem.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to the problem. Some guidance has been offered regarding the use of a new function to facilitate the solution, but there is no explicit consensus on the best path forward. The original poster expresses frustration with the iterative nature of their attempts.

Contextual Notes

Participants are navigating the constraints of the problem, specifically the requirement to utilize Lagrange's theorem while also addressing the complexities introduced by the function's behavior. There is an acknowledgment of the potential for circular reasoning in their attempts.

Felafel
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Homework Statement



Being a>0 and f:[a,b]--->R continuos and differentiable in (a,b), show that there exists a t ##\in## (a,b) such that:
## \frac{bf(a)-af(b)}{b-a}=f(t)-tf'(f)##

The Attempt at a Solution


For lagrange's theorem, we have:
## \frac{f(a)-f(b)}{b-a}= -f'(t) ##

thought i could find f(t) from the equation ##f(x) = f(t) + f'(t)(x-t)+R_1(x-t)## ignoring R.
but then a "x" would appear and I don't know how to deal with it.
 
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To simplify the equation, consider g:[a,b]->R with
$$g(x)=f(x)-f(a)-(x-a)\frac{f(b)-f(a)}{b-a}$$
This is chosen to get g(a)=g(b)=0 and ##g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a}##

Is there some t ∈ (a,b) such that ##0=g(t)-tg'(t)##?
Working backwards, you should be able to get the original equation.
 
sounds good, but I'm explicitly asked to use Lagrange's theorem
 
I think you can use that theorem to solve the reduced problem ;).
 
another hint? i don't get it, i always end up in a loop :(
 

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