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Derivative of a function

  1. Mar 5, 2014 #1
    Problem statement

    ImageUploadedByPhysics Forums1394018429.382539.jpg
    My question is for number 27.
    Revelant equation
    None

    Attempt at a solution

    I'm not sure where to start.
    ImageUploadedByPhysics Forums1394018545.381550.jpg ImageUploadedByPhysics Forums1394018553.886597.jpg

    This is my teachers answer. I understand how the slope is 1 for x greater than -1 and that it is -2 at x greater than -1 and that there is a point at (0,-1) but I don't understand how they connect to form that final pic. I think I'm missing something ,can someone help me?
     
  2. jcsd
  3. Mar 5, 2014 #2
    Never mind I think I understand it now. I think I was just confused by the previous picture leading up to the answer.
     
  4. Mar 5, 2014 #3

    Maybe someone could elaborate on this for me? Thanks
     
  5. Mar 5, 2014 #4

    SammyS

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    What does it mean for a function to be continuous ?
     
  6. Mar 6, 2014 #5
    That there are no breaks or holes
     
  7. Mar 6, 2014 #6

    HallsofIvy

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    If f'(x)= 1, for x< -1, then f(x)= x+ C1 for some constant C1, for x< -1, so the graph is a straight line with slope 1.

    If f'(x)= -2, for x> -1, then f(x)= -2x+ C2 for some constant C2, for x> -1, so the graph is a straight line with slope -2.

    Since f is continuous, the two lines must meet at x= -1. That means that -1+ C1= -2(-1)+ C2.

    That, together with f(0)= C2= -1 is sufficient to determine both C2 and C1.
     
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