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Math Help - Find Delta's Algebraically

  1. #1
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    Find Delta's Algebraically

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

    I set it up like this:

    \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?

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

    Combine terms:

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

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

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

    I set it up like this:

    \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?


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

    Combine terms:

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

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

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

    Stuck.
    We need to show that

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

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

    So,

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

    Assuming m>0 we obtain

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