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Antiderivatives math homework

  1. Jul 26, 2007 #1

    daniel_i_l

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    1. The problem statement, all variables and given/known data
    Let f:[0,2]->R be defined as:
    if 0 =< x =< 1 then f(x) = 4(x^3)
    if 1 < x =< 2 then x = x^2 + 2
    Prove or disprove:
    There exist c_1 , c_2 in R so that F:[0,2]-R defined as:
    if 0 =< x =< 1 then f(x) = x^4 + c_1
    if 1 < x =< 2 then x = (x^3)/3 + 2x + c_2

    2. Relevant equations



    3. The attempt at a solution

    Now my question is, why don't any c_1,c_2 make F an AD of f?
    Can any shed some light on this?
    Thanks.
     
  2. jcsd
  3. Jul 26, 2007 #2
    definition of antiderivative

    let [tex]f(x)[/tex] be a function which is differentiable in an interval C then there exists a g such that [tex]g(x)=f^{'}(x)[/tex] then f is called the antiderivative of g.

    ur function is not differentiable at [tex]x=1[/tex]
     
  4. Jul 26, 2007 #3

    Dick

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    I suppose, except for any c_1,c_2, F is not differentiable at x=1.
     
  5. Jul 26, 2007 #4

    daniel_i_l

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    Thanks for your answers! I just want to clear one thing up:
    In this case for example, if f wasn't defined at x=1 then would F be an AD of f for all c_1,c_2?
    Thanks.
     
  6. Jul 26, 2007 #5

    Dick

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    Absolutely, if f only needs to be defined on [0,1)U(1,2].
     
  7. Jul 26, 2007 #6

    daniel_i_l

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    Thanks for clearing that up!
     
  8. Jul 26, 2007 #7
    yes that's right it just depends on ur domain of definition
     
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