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Thread: Precise Definition of a: f(x)=x^2, L=49, c=-7, epsilon=0.4

  1. #1

    Precise Definition of a: f(x)=x^2, L=49, c=-7, epsilon=0.4

    For the given value of f(x) and values of L, c, and epsilon>0, find the largest open interval about c on which the inequality absolute value of f(x)-L<epsilon holds.
    f(x)=x^2 L=49 c=-7 epsilon=0.4




    here's the work I did (sorry this is my first post and I couldn't figure out how to do special characters or absolute value symbols)
    x^2 -49<.4
    48.6<x^2<49.4
    6.9713<x<7.0285
    So I thought my interval would be (6.9713,7.0285) but the answer was supposed to be (-7.0285,6.9713)
    Where did I go wrong?

  2. #2
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    Quote Originally Posted by Anastasia3rd View Post
    For the given value of f(x) and values of L, c, and epsilon>0, find the largest open interval about c on which the inequality absolute value of f(x)-L<epsilon holds.
    f(x)=x^2 L=49 c=-7 epsilon=0.4




    here's the work I did (sorry this is my first post and I couldn't figure out how to do special characters or absolute value symbols)
    x^2 -49<.4
    48.6<x^2<49.4
    6.9713<x<7.0285
    So I thought my interval would be (6.9713,7.0285) but the answer was supposed to be (-7.0285,6.9713)
    Where did I go wrong?
    Perhaps spend some more time on the Absolute Values and the Squares.

    Is x < 0, potentially?
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  3. #3
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    Quote Originally Posted by Anastasia3rd View Post
    For the given value of f(x) and values of L, c, and epsilon>0, find the largest open interval about c on which the inequality absolute value of f(x)-L<epsilon holds.
    f(x)=x^2 L=49 c=-7 epsilon=0.4




    here's the work I did (sorry this is my first post and I couldn't figure out how to do special characters or absolute value symbols)
    x^2 -49<.4
    48.6<x^2<49.4
    6.9713<x<7.0285
    So I thought my interval would be (6.9713,7.0285) but the answer was supposed to be (-7.0285,6.9713)
    Where did I go wrong?
    No wonder you are confused. Either you misread the answer or the book has a typographical error.

    [tex]|x^2 - (-\ 7)^2| < 0.4 \implies |x^2 - 49| = 0.4 \implies - \ 0.4 < x^2 - 49 < 0.4 \implies[/tex]

    [tex]48.6 < x^2 < 49.4 \implies 6.9713 < x < 7.0285 \text { or } -\ 7.0285 < x < - \ 6.9713.[/tex]

    But we are looking for a neighborhood close to MINUS 7. So the first neighborhood is irrelevant.

    Thus we get [tex]-\ 7.0285 < x < - \ 6.9713[/tex] as the answer.

    It is quite clear that [tex]-\ 7.0285 < x < 6.9713[/tex] is an error.

    x = 0 lies in that interval and

    [tex]|0^2 - 49| = 49 \not < 0.4.[/tex]

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