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Derivative of trig & continuity

  1. Oct 8, 2009 #1
    1. Is there a value of b that will make...
    x+b, x<0
    g(x) = <
    cosx, x=>0
    continuous at x = 0?
    differentiable at x = 0?
    give reasons.


    2. I'm not sure what are related equations for this. Limits?

    3. So I try to find how to make it continuous at x = 0
    limx->0(cosx) = cos0 = 1
    so, limx->0(x+b)=1
    b=1
    so if b=1, the graph is continuous at x=0.
    (did i do that right?)
    now, i am not sure about making the graph differentiable at x=0 though.
    because if b is 1, it is differentiable(1) and cosx(1), so does that mean the graph is always differentiable?


    So, yes, my English isn't that great so maybe I misread something and made a simple question seem over complicated. So please let me know that if it's the case. Thanks before-hand though :)
     
  2. jcsd
  3. Oct 8, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, the first part is correct. The limit "from the left" is b while the limit from the right is 1. In order to be continuous, those two limits must be the same: b= 1.

    If a function is differentiable at a point, the derivative is not necessarily continuous at that point but it must satisfy the "intermediate value property". And essentially that means that taking the limit of the derivatives from right and left must be the same: The derivative of cos(x) is sin(x), which goes to 0 as x goes to 0, and the derivative of x+ b is 1. No choice of b can make those the same.
     
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