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Thread: Find Delta's Algebraically

  1. #1
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    Find Delta's Algebraically

    $\displaystyle f(x) = mx+b$ when $\displaystyle m>0$, $\displaystyle L = (m/2)+b$, X sub 0 = 1/2, $\displaystyle \epsilon = c>0$

    I set it up like this:

    $\displaystyle \arrowvert mx+b+\frac{m}{2}+b\arrowvert< c>0$

    Okay, so it looks like my restrictions are that it has to be greater than 0 for M and C. But I don't know how to deal with that in an inequality. Does that mean that I simply put C on both sides without one being a negative and a positive?

    $\displaystyle c<\arrowvert mx+b+\frac{m}{2}+b\arrowvert< c$

    Combine terms:

    $\displaystyle c<\arrowvert mx+\frac{m}{2}+2b\arrowvert< c$

    What do I do here? I want X by itself? So.

    $\displaystyle \frac{c}{m}<\arrowvert x+\frac{m}{2}+2b\arrowvert< \frac{c}{m}$

    Stuck.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by A Beautiful Mind View Post
    $\displaystyle f(x) = mx+b$ when $\displaystyle m>0$, $\displaystyle L = (m/2)+b$, X sub 0 = 1/2, $\displaystyle \epsilon = c>0$

    I set it up like this:

    $\displaystyle \arrowvert mx+b+\frac{m}{2}+b\arrowvert< c>0$

    Okay, so it looks like my restrictions are that it has to be greater than 0 for M and C. But I don't know how to deal with that in an inequality. Does that mean that I simply put C on both sides without one being a negative and a positive?


    $\displaystyle c<\arrowvert mx+b+\frac{m}{2}+b\arrowvert< c$

    Combine terms:

    $\displaystyle c<\arrowvert mx+\frac{m}{2}+2b\arrowvert< c$

    What do I do here? I want X by itself? So.

    $\displaystyle \frac{c}{m}<\arrowvert x+\frac{m}{2}+2b\arrowvert< \frac{c}{m}$

    Stuck.
    We need to show that

    $\displaystyle |f(x)-L|<\epsilon$ whenever $\displaystyle |x-a|<\delta$

    And we have $\displaystyle a=\frac1{2}$ and $\displaystyle L=\frac{m}{2}+b$

    So,

    $\displaystyle |(mx+b)-(\frac{m}{2}+b)|=|m(x-\frac{1}{2})|<\epsilon$

    Assuming $\displaystyle m>0$ we obtain

    $\displaystyle |(x-\frac{1}{2})|<\frac{\epsilon}{m}=\delta$
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