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Calculus question

  1. Dec 13, 2004 #1

    Could someone help me with this question please ?

    A set of curves which all pass the origin, have equations :

    y=f3(x)...... where f ' n(x) = fn-1(x) and f1(x) = x^2

    1.) find the expression for fn(x)

    2.) find f2(x) and f3(x)

    I don't know where to begin, especially the first question .


  2. jcsd
  3. Dec 13, 2004 #2


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    How is the derivative of a function related to the function itself? Through the operation of the integral. So

    [tex]f_n'(x) =\frac{df_n}{dx} = f_{n-1}(x) \Leftrightarrow ...[/tex]
    Last edited: Dec 13, 2004
  4. Dec 13, 2004 #3
    Could somebody else explain further please ?
  5. Dec 13, 2004 #4
    Actually quasars tip is totally ok. I would have given the same answer... Just try it...f1 is x² and the derivative of f2 equals f1 = x²...So in order to find f2, just integrate x² with respect to x. You get x³/3...can you move on from here...

  6. Dec 13, 2004 #5


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    Assuming you have genuinely been trying to solving the thing for the past 45 minutes, I'll complete the reasoning...

    (Assuming f is continuous,)

    [tex]\Leftrightarrow df_n = f_{n-1}(x)dx \Leftrightarrow \int_0^x df_n = \int_0^x f_{n-1}(x)dx \Leftrightarrow f_n(x) - f_n(0) = \int_0^x f_{n-1}(x)dx \Leftrightarrow f_n(x) - 0 = \int_0^x f_{n-1}(x)dx \Leftrightarrow f_n(x) = \int_0^x f_{n-1}(x)dx[/tex]

    Because if they pass the origin, when x = 0, f = 0.
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