General Continuity Proof Question

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Discussion Overview

The discussion revolves around proving the continuity of the function \(3x^{2}-2x+1\) at a specific point, likely \(x=2\). Participants explore the epsilon-delta definition of continuity, particularly focusing on how to handle the expression \(3x+4\) in the context of this proof.

Discussion Character

  • Technical explanation, Conceptual clarification, Homework-related

Main Points Raised

  • One participant expresses confusion about handling the term \(3x+4\) in the continuity proof.
  • Another participant suggests that as \(x\) approaches \(2\), \(3x+4\) approaches \(10\), proposing a relationship involving \(|3x+4 - 10| < 3\delta\).
  • A different participant proposes restricting \(\delta\) to be less than or equal to \(1\) and shows how to derive \(|(x - 2)(3x + 4)| < 13\delta\) to satisfy the epsilon condition.
  • One participant critiques the lack of clarity in the initial question, suggesting that the original poster should specify the type of continuity proof they are attempting to perform.
  • The original poster acknowledges understanding but expresses uncertainty specifically regarding quadratics in continuity proofs.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best approach to the problem, and there are varying levels of understanding and clarity regarding the continuity proof process.

Contextual Notes

There is an implicit assumption that the continuity proof is being conducted at \(x=2\), but this is not explicitly stated. The discussion also reflects varying levels of comfort with the epsilon-delta definition among participants.

kathrynag
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Ok, let's say I had [tex]3x^{2}-2x+1[/tex]
I know we have lx-2l<[tex]\delta[/tex]
Also l(x-2)(3x+4)l<[tex]\epsilon[/tex]
My problem with these types of questions is dealing with the l3x+4l. I just don't really know what to do.
 
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I'm not sure what you are looking for. However x~2 means that 3x+4~10. To be more precise |3x+4 - 10| < 3(delta).
 
We can restrict δ > 0 to be as small as we want, so let's say δ ≤ 1. Then using |x - 2| < δ ≤ 1, show that |3x + 4| < 13, so that |(x - 2)(3x + 4)| < 13δ. In order for |(x - 2)(3x + 4)| to be less than ε, you can just pick any δ ≤ ε / 13; δ = max{1, ε / 13} does it.
 
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It is generally a good idea to actually say what you mean! you say "these types of questions" without saying WHAT types of questions. Probably you mean "prove that 3x2- 2x+ 1 is continuous at x= 3". that's what adriank assumed an he gave a very nice explanation.
 
Ok, I understand that. I just wasn't sure what to do when I got a quadratic because I understood these questions just fine otherwise.
 

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