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Math Help - IVT and Jump discontinuity

  1. #1
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    IVT and Jump discontinuity

    As I understand it IVT states that there will always be one value between an interval that satisfies it. Which is pretty easy to understand. If it doesn't work it means you have some kind of discontinuity. However, this problem in this book is asking me draw a graph of a function f(x) on [0,4] with the given property.

    Jump discontinuity at x = 2, yet does satisfy the conclusion of the IVT on [0,4]. I've tried to wrap my head around this but It has me stumped. I assume it's a trick question but before I lay it to rest I have to ask.

    Thanks.
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  2. #2
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    I just had an idea is it a Tan function? Because technically it can break at x =2 but still satisfies every M for every c between [0,4] But is that considered jump discontinuity or an infinite
    Last edited by Newskin01; February 11th 2011 at 04:48 PM. Reason: add more thoughts :)
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