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Find values for this piecewise function so it can be differentiable

  1. Mar 13, 2013 #1
    1. The problem statement, all variables and given/known data

    Fx= ax^2 + bx + c -infiniti<x<0
    = D x=0
    =x^2 sin(1/x) - 2 0<x<infiniti

    a) FInd all vlaues of a, b, c and d that make the function f differntiable on the domain -∞<x<∞

    b) Using the values founbd in part a, determine lim x-> 0- f'(x), lim x->0+ f'(x) and f'(0). Is f'(x) continuous at x=0. Explain.


    3. The attempt at a solution

    I need help with part b).... a=all real numbers b= all real numbers c= -2 d = -2
     
  2. jcsd
  3. Mar 13, 2013 #2

    SammyS

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    You need help with part a) also.
    To be differentiable, the function must also be continuous. You've accomplished that with your choices of c and d .

    To be differentiable at x=0, f'(0) must exist.

    What is [itex]\displaystyle \ \frac{d}{dx}(ax^2 + bx + c)\ ?[/itex]

    What is [itex]\displaystyle \ \frac{d}{dx}(x^2 \sin(1/x) - 2)\ ?[/itex]

    b is not arbitrary. It must be some specific value.
     
  4. Mar 13, 2013 #3
    f'(x) = 2ax + b
    f'(x) = 2x cos(1/x)(-1/x^2)
     
  5. Mar 13, 2013 #4
    f'(x) = 2ax + b
    f'(x) = (-2/x) cos(1/x).... I guess f'(0) doesnt exist for this?
     
  6. Mar 14, 2013 #5
    lim = 2a(0) + b = f'(0) = lim (-2/x) cos(1/x)
    x-> 0- ..................................................... x->0+

    So B = 0....
     
  7. Mar 14, 2013 #6
    How do you find the
    lim for the f'(x) of the sin function.
    x-> 0+

    Or do you say the derivative is not constant at x=0 b/c there isnt a answer at that point ?
     
  8. Mar 14, 2013 #7

    haruspex

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    That's not how the product rule works.
     
  9. Mar 14, 2013 #8

    SammyS

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    Your f'(x) is wrong for x > 0.

    You need to use the product rule. ( In other words, your derivative is incorrect for x2sin(1/x). )
     
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