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Math Help - manipulative thingie

  1. #1
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    manipulative thingie

    What can I do to this to make it independent of x. i.e. manipulate it somehow to cancel out the g(x)?


    \mid\frac{\varepsilon_1}{g(x)}\mid + \mid\frac{\varepsilon_2L}{g(x)M}\mid


    btw \varepsilon_1 = \mid f(x) - L \mid and \varepsilon_2 = \mid g(x) - M \mid
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  2. #2
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    Quote Originally Posted by Caity View Post
    What can I do to this to make it independent of x. i.e. manipulate it somehow to cancel out the g(x)?


    \mid\frac{\varepsilon_1}{g(x)}\mid + \mid\frac{\varepsilon_2L}{g(x)M}\mid


    btw \varepsilon_1 = \mid f(x) - L \mid and \varepsilon_2 = \mid g(x) - M \mid
    As it stands you can't

    CB
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  3. #3
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    This is the original problem that I need to simplify to solve for \varepsilon_1 and \varepsilon_2 for f(x) - M and g(x) - N respectively.
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  4. #4
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    Quote Originally Posted by Caity View Post



    This is the original problem that I need to simplify to solve for \varepsilon_1 and \varepsilon_2 for |f(x) - M| and |g(x) - N| respectively.
    Take note of those absolute value signs that should be there.

    To deal with this problem, you'll certainly need to know that N is not zero. Given that that is the case, suppose that you choose \varepsilon_2 to be less than |\tfrac12N|. By the triangle inequality, that will imply that g(x) lies between \tfrac12N and \tfrac32N. That should enable you to make some progress.
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  5. #5
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    Actually, I'm trying to prove that \frac{f(x)}{g(x)} = \frac{M}{N}

    I apologize for any confusions... and of course N does not = 0
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