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Finding the particular anti derivative when given f'(x) and TWO f(value)s

  1. Nov 26, 2011 #1
    1. The problem statement, all variables and given/known data

    Here is the question directly from the book:

    f'(x)=x^(-1/3), f(1)=1, f(-1)=-1, find f(x).

    2. Relevant equations



    3. The attempt at a solution

    So there are about ten of these types, and I am very confused. I am confused because if you solve for f(x) with f(a) you get a different value for c than if you solved for f(b).

    So finding f(x) I get f(x)=(3/2)x^(2/3) + c
    If I make f(x)=1 and x=1, I get the value of c to be -1/2
    But if I make f(x)=-1 and x=-1, I get the value of c to be -5/2..

    Thank you so much for your help in advance!
     
  2. jcsd
  3. Nov 26, 2011 #2

    micromass

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    Notice that f(0) is not defined. So you actually have two branches. So your function can be described as

    [tex]f(x)=\left\{\begin{array}{c}(3/2)x^{2/3} + c_1 ~\text{if}~x>0\\ (3/2)x^{2/3} + c_2 ~\text{if}~x<0 \end{array}\right.[/tex]
     
  4. Nov 26, 2011 #3
    So whenever f'(0)=DNE, the function f(x) can be written as a piecewise function?
     
  5. Nov 26, 2011 #4

    micromass

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    Indeed!!

    Do notice that you proved that the anti-derivative of a function is unique up to a constant. But you only proved that for functions that exists everywhere on an interval. If a function does not exist in a point, then the result isn't true anymore!!
     
  6. Nov 27, 2011 #5
    OK, thanks again! You helped tremendously.
     
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