Can't decide if my proof is right or not ?inverse functions

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Discussion Overview

The discussion revolves around the validity of a proof concerning the derivative of an inverse function, specifically addressing the conditions under which the derivative exists for a one-to-one continuous function that is differentiable. The participants explore the nuances of the proof derived from Spivak's calculus, focusing on the definitions and limits involved in the argument.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant describes their approach to proving that if a function \( f \) is one-to-one, continuous, and differentiable, then the derivative of its inverse exists, starting from the definition of \( f^{-1}(b) \).
  • Another participant questions the clarity of the proof, specifically asking for the definition of \( b \) and the interval involved, suggesting that more precision is needed.
  • A third participant agrees that the proof seems intuitively correct but points out that the notation used, such as \( \lim_{f(x) \rightarrow a}{g(x)} \), is not formally defined and may require clarification or rephrasing for a formal proof.
  • Some participants express confusion regarding the interchange of variables in limits and whether the function remains unchanged during this process.

Areas of Agreement / Disagreement

There is no consensus on the correctness of the proof. While some participants find the approach intuitively appealing, others raise concerns about the formal definitions and clarity of the notation used.

Contextual Notes

Participants note the need for precision in defining variables and intervals, as well as the potential ambiguity in the notation used for limits, which may affect the formal validity of the proof.

Andrax
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so before i start this is extracted from spivak's calculus proof
now here is the problem the teacher who taught us limits and dervatives for the first time told us not to use f(x)-f(a) instead of f(a+h)-f(a)
so spivak wanted us to prove that if f a one-one function and is continious on an interval and differentiable on f^{-1}(b) then f'(b) exists and (f^-1)'(b)=\frac{1}{f'(f^-1(b))}(f^-1 = inverse of f )
i did the opposite of what spivak did , spivak started from the definition of f'(b) while i started wtih the definition of f^{-1}(b), so is my proof right?
let f(a) = b then f^-1(b)=a so
f'(f^{-1}(b) = f'(a) = \lim_{x \to a}\frac{f(x)-f(a)}{x-a}=\lim_{x \to f^-1(b)}\frac{f(x)-b}{x-f^-1(b)}
now when x approches a=f^-1(b) in the domain of f , f(x)approaches f(a)=b in the domain on f-1 and we can set f(x)=x later on
so =\lim_{f(x) \to b}\frac{f(x)-b}{x-f^-1(b)}
replacing f(x) by the new x in the domain of f^-1 and b by f^-1(b) and x by f^-1(x)= a(I hope I'm right in this phase, please correct me if I'm wrong)
=\lim_{x \to b}\frac{x-b}{f^-1(x)-f^-1(b)} = \frac{1}{(f^-1)'(b)}
i've always been confused when changing the x to f(x) : do we also change the function or it remains intact?Limx->af(x) <=>Limf(x)->f(a) [f(f(x))) or f(x)? ] if i made a fault it would be this, thanks for reading.
 
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What is b? And you haven't named the interval. I think you need to be more precise.
 
verty said:
What is b? And you haven't named the interval. I think you need to be more precise.

It doesn't really matter with this proof but okay let b belongs to [n, m] where f is defined...
 
I think your proof is intuitively right. However, an expression such as

\lim_{f(x)\rightarrow a}{g(x)}

is not formally defined. At least, I don't think Spivak defines it or works with it. So to get a formal proof, you either have to define precisely what the above means, or you need to write your proof to avoid the above notation.
 
micromass said:
I think your proof is intuitively right. However, an expression such as

\lim_{f(x)\rightarrow a}{g(x)}

is not formally defined. At least, I don't think Spivak defines it or works with it. So to get a formal proof, you either have to define precisely what the above means, or you need to write your proof to avoid the above notation.
Well yeah I'll try that I thought it's obvious since the pair x, f(x) is in f so f(x), x in f-1
 

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