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Limits and Continuity

  1. Aug 30, 2006 #1
    I'm having trouble understanding the problem:

    Find the largest open interval, centered at x=3, such that for each x in the interval the value of the function f(x) = 4x - 5 is within 0.01 unit of the number f(3)=7

    The solutions manuel goes on to say that the abs[f(x)-f(3)] = abs [(4x - 5) - 7] = 4 abs [x-3] < 0.1 if and only if abs[x-3] < (0.1)/4 = 0.0025

    I get the first part of the answer where you basically subtract f(x) from f(3), but I'm having trouble understanding the rest of the probelm where the four moves outside the absolute bracket. Also, where does the center at x=3 have to do with anything?
  2. jcsd
  3. Aug 30, 2006 #2
    If I understand the question correctly, all you need to do is find the endpoints of the interval. This can be done by solving:

    7 + 0.01 = 4x - 5
    7 - 0.01 = 4x - 5
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