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Fundamental therom problem

  1. Sep 1, 2004 #1
    My professer told us to think about this problem. I have the answer in my solutions manual but I want to know how I would go through the thinking process to solve it...(and others like it) The question says

    "Find a function f and a number a such that

    [tex]
    6+\int_{a}^{x}\frac{f(t)}{t^2}dt=2\sqrt{x}
    [/tex]
    for all x greater than zero"
     
    Last edited by a moderator: Sep 1, 2004
  2. jcsd
  3. Sep 1, 2004 #2

    Tide

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    Differentiate the equation!
     
  4. Sep 1, 2004 #3
    There was no point in that reply.. :grumpy:
    Anyone out there that can really help?
     
    Last edited: Sep 1, 2004
  5. Sep 2, 2004 #4

    Tide

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    Begging your pardon but I told you exactly what you need to do!
     
  6. Sep 2, 2004 #5
    WTF? Tide's post was immensely helpful and practically gives you the entire solution.
     
  7. Sep 2, 2004 #6

    Gokul43201

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    I second that. Tide's given you a correct way to solve the problem.
     
  8. Sep 2, 2004 #7

    HallsofIvy

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    Since you did not grasp what Tide said, here's another way of looking at it:
    Your equation is equivalent to
    [tex]\int_{a}^{x}\frac{f(t)}{t^2}dt=2\sqrt{x}- 6[/tex]


    Do you notice that the right hand side is a constant?
     
  9. Sep 2, 2004 #8
    The right hand side is not a constant...
     
  10. Sep 2, 2004 #9

    HallsofIvy

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    OMIGOD! I stared at that repeatedly and saw [itex]2\sqrt{2}[/itex].
     
  11. Sep 2, 2004 #10

    HallsofIvy

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    Alem2000: as Tide said, differentiate both sides:
    The derivative of [tex]\int_{a}^{x}\frac{f(t)}{t^2}dt[/tex] is [tex]\frac{f(x)}{x^2}[/tex] (that's the "fundamental theorem your title referred to) and the derivative of [tex]2\sqrt{2}= 2(x^{1/2})[/tex] is [tex]x^{-1/2}[/tex].

    Set them equal and solve for x.
     
  12. Sep 2, 2004 #11
    :rofl: :rofl: :rofl: OOOOOO! I think I made that way more complicated then it was. Thanks alot Tide...sorry about the frustration :wink:
     
  13. Sep 2, 2004 #12

    HallsofIvy

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    I did it again! I wrote [tex]2\sqrt{2}[/tex] when I meant [tex]2\sqrt{x}[/tex]!
     
  14. Sep 2, 2004 #13
    And now you are throwing factorial signs about, tisk tisk ;)
     
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