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  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)?


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


    btw $\displaystyle \varepsilon_1 = \mid f(x) - L \mid$ and $\displaystyle \varepsilon_2 = \mid g(x) - M \mid$
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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)?


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


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

    CB
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    This is the original problem that I need to simplify to solve for $\displaystyle \varepsilon_1$ and $\displaystyle \varepsilon_2$ 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 $\displaystyle \varepsilon_1$ and $\displaystyle \varepsilon_2$ 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 $\displaystyle \varepsilon_2$ to be less than $\displaystyle |\tfrac12N|$. By the triangle inequality, that will imply that g(x) lies between $\displaystyle \tfrac12N$ and $\displaystyle \tfrac32N$. 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 $\displaystyle \frac{f(x)}{g(x)} = \frac{M}{N}$

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