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Math Help

  1. Jun 6, 2005 #1
    Last edited: Jun 6, 2005
  2. jcsd
  3. Jun 6, 2005 #2

    OlderDan

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    Notice that when x = 2, |x - 2| and |f(x) - 7| are both zero. If you graph those two absolute value functions you will see a "V" within a "V". Suppose [tex]\delta[/tex] is 1. For that value of [tex]\delta[/tex], x will have a maximum and minimum value that satisfies [tex]|x - 2| < \delta[/tex]. If you use those limiting values of x, and look to the graph of |f(x) - 7| you will see that those values correspond to a certain value of |f(x) - 7|. That value is the value of [tex]\epsilon[/tex]. For all values of x between those limiting values, [tex]|f(x) - 7| < \epsilon[/tex]. You can find a relationship between [tex]\delta[/tex] and [tex]\epsilon[/tex] that shows they are proportional.
     
  4. Jun 7, 2005 #3
    [tex]\epsilon[/tex] divided by 3 is what i got... i dont know if i did it right
     
  5. Jun 7, 2005 #4

    OlderDan

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    Looks good to me
     
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