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Math Help - Epsilon (Did I do this correctly)

  1. #1
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    Epsilon (Did I do this correctly)

    I have a problem that says

    f(x)= x^2 L=4 Xo= -2 e= 0.5

    I have to find the value of delta.


    Do I start by setting it up as

    -.5 < x^2 -4 < .5??

    sq root 3.5 < x < sq root 4.5

    Then I take the larger absolute value of the 2....

    sq root 4.5 - 2 = .12


    Did I do this problem correctly?
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  2. #2
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    You are correct in performing reversible operations all the way up to

    3.5 < x^2 < 4.5.

    Here, we must remember that x could be positive or negative, and therefore that

    \sqrt{x^2}=|x|, not x.

    Hence, we have

    \sqrt{3.5} < |x| < \sqrt{4.5}.
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  3. #3
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    Quote Originally Posted by KarlosK View Post
    I have a problem that says

    f(x)= x^2 L=4 Xo= -2 e= 0.5

    I have to find the value of delta.


    Do I start by setting it up as

    -.5 < x^2 -4 < .5??

    sq root 3.5 < x < sq root 4.5

    Then I take the larger absolute value of the 2....

    sq root 4.5 - 2 = .12


    Did I do this problem correctly?
    Take δ =1/2 => |x-(-2)| =|x+2|<1/2 => |x+2|<1 => |x|-2<1 => |x|<-1 =>|x|+2<1 => |x-2|<1 => |x+2||x-2|<1/2 => |x^2-4|<\frac{1}{2}
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  4. #4
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    In the future, please state the problem completely. People who reponded guessed that you mean "Find \delta> 0 such that is |x- 2|< \delta then |x^2- 4|< \epsilon" but they shouldn't have to guess!

    And, of course, there exist an infinite number of answers to that question!
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    In the future, please state the problem completely. People who reponded guessed that you mean "Find \delta> 0 such that is |x- 2|< \delta then |x^2- 4|< \epsilon" but they shouldn't have to guess!

    And, of course, there exist an infinite number of answers to that question!
    I copied the problem exactly how it was worded in the book. I come on here to get math help, not to be talked down to. That would be like me saying please spell check your answer before posting it afterall I assume you meant responded by "reponded"
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  6. #6
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    Quote Originally Posted by KarlosK View Post
    f(x)= x^2 L=4 Xo= -2 e= 0.5
    Start with bounding |x+2| by 1. Why do that? Because (x+2) is the other factor in x^2-4.
    \left| {x + 2} \right| < 1 \Rightarrow  - 3 < x <  - 1 \Rightarrow  - 5 < x - 2 <  - 3 \Rightarrow \left| {x - 2} \right| < 5.
    Find a delta that works \delta  = \frac{{0.5}}{5} = 0.1
    Note that if |x+2|<\delta<1 we get what we want.
    \left| {x^2  - 4} \right| = \left| {x - 2} \right|\left| {x + 2} \right| < (5)(0.1) = 0.5.
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