Is the discontinuity of f(f(f(x))) at x=0 and x=1 non-removable?

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Discussion Overview

The discussion revolves around the nature of discontinuities in the function f(f(f(x))) where f(x) = 1/(1-x), specifically at the points x=0 and x=1. Participants explore whether these discontinuities are removable or non-removable, engaging in theoretical reasoning and algebraic manipulation.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant asserts that f(f(f(x))) has discontinuities at x=0 and x=1, questioning the classification of these as removable based on a reference book.
  • Another participant suggests that if any cancellation occurred during the calculation of f(f(f(x))), it may have removed a removable singularity.
  • A different viewpoint posits that neither x=0 nor x=1 are actual discontinuities since they are excluded from the function's domain, emphasizing the importance of domain considerations in function manipulation.
  • A later reply questions whether this means the discontinuities are non-removable, indicating uncertainty about the classification.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the discontinuities, with no consensus reached on whether they are removable or non-removable.

Contextual Notes

Participants highlight the importance of domain considerations and the effects of algebraic manipulation on the classification of discontinuities, but the discussion does not resolve the underlying mathematical questions.

phymatter
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if f(x)=1/(1-x) , then the points of discontinuity of f(f(f(x))) are x=0 and x=1 ,
b'cos f(f(x)) is (x-1)/x and f(f(f(x))) is x , but my book (I.A. Maron S.V.C) says that both x=0 and x=1 are removable discontinuities , but i don't think so! pl.confirm me!
 
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When you calculated f(f(f(x))) if you canceled anything, then you were removing a removable singularity
 
Neither are actually discontinuities since they are removed from the domain of the function. But that’s largely semantic. The idea they are getting at I think is best understood with an example.

If f(x) = (x-1)(x+1)/(x-1) and g(x) = x+1 with standard treatment of the domain are they same? The answer is no, because g(1) = 2 and f(1) is undefined.

You have to be careful when you do algebraic manipulations to functions of how you alter the domain. so when you got f(f(x)) = (x-1)/x, x=0 is removed from the domain at that point.
 
So this discontinuity would be non-removable?
 

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