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Finding Delta

  1. Aug 24, 2007 #1
    Find an open interval about a on which the inequality | f(x) - L | < E holds. Then give a value for D > 0 such that for all x satisfying 0 < | x - a | < D the inequality | f(x) - L | < E holds.

    f(x) = x^2, L = 4, a = -2, E = .5

    To solve I setup the inequality

    Step 1: -.5 < x^2 - 4 < .5

    Step 2: 3.5 < x^2 < 4.5

    This is where I get stuck, need help with the algebra.

    I got sqrt 3.5 < x < sqrt 4.5 & thought my interval for E would equal (sqrt 3.5, sqrt 4.5), but the book gives (-sqrt 4.5, -sqrt 3.5). I know if you take the sqrt, you have to take the positive & the negative of that number, but how do you know which to choose in this case & by doing this, does the inequality signs always reverse (like < to >)? Going to skip the rest of the problem, because I didn't have a problem solving the rest.
     
  2. jcsd
  3. Aug 24, 2007 #2

    Dick

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    Your answer is correct for a=+2, but the question says a=-2. Draw a graph of f(x)=x^2 and look around the points x=2 and x=-2. When you are taking a sqrt and you have choice of +/- you have to think about the problem.
     
  4. Aug 25, 2007 #3

    HallsofIvy

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    Your problem is not to find bounds on x2 or x but on x- a which, here, is x+2.
    From -.5< x2-4< .5, factor to get -.5< (x-2)(x+2)< .5 and then think about dividing by x-2. With x "close" to -2, say between -1 and -3, what is the largest x-2 can be? What is the smallest?
     
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