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Math Help - curve sketching

  1. #1
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    curve sketching

    let f(x)= ((3x-1)|x-1|) / (x-1) for x !=0
    (a) sketch the graph of f(x)
    (b)evaluate lim x->1+ f(x)
    (c) evaluate lim x->1- f(x)
    (d)evaluate lim x->1 f(x)
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by bobby77 View Post
    let f(x)= ((3x-1)|x-1|) / (x-1) for x !=0
    (a) sketch the graph of f(x)
    This behaves linearly in x except when x=1, where there is a jump discontinuity. This is because |x-1|/(x-1) is a constant equal to +1
    when x>1, and a constant equal to -1 when x<1.


    For x>1 we have:

    f(x) = 3x-1

    and for x<1, we have:

    f(x) = -3x+1,

    with a jump discontiuity where f changes grom -2 to the left of x=1
    to ~2 to the right of x=1.

    Now I have a sketch of this but the upload function appears to be
    brocken

    (b)evaluate lim x->1+ f(x)
    If the diagram had uploaded it would be clear that the limit as x goes to
    1 from above is the limit as x goes to 1 of 3x-1 which is equal to 2.

    (c) evaluate lim x->1- f(x)
    If the diagram had uploaded it would also be clear that the limit as x goes to
    1 from below is the limit as x goes to 1 of -3x+1 which is equal to -2.

    (d)evaluate lim x->1 f(x)
    As the two previous limits are not equal there is no such limit.

    RonL
    Last edited by CaptainBlack; October 10th 2006 at 06:29 AM.
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  3. #3
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    Quote Originally Posted by bobby77 View Post
    let f(x)= ((3x-1)|x-1|) / (x-1) for x !=0
    (a) sketch the graph of f(x)
    ...
    Hi,

    I'll try to upload the graph. It's working!

    Then here is the graph of your function:
    Attached Thumbnails Attached Thumbnails curve sketching-fkt_sprung.gif  
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