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Neeeeed helppppp for a question,, (not hw)

  1. Mar 16, 2007 #1
    neeeeed helppppp for a question,,urgent!!(not hw)

    Hi, I have been working on this question for some time,, but still couldn't solve it,,does anyone have an idea? thanks for helping!

    Prove that if f is a twice differentiable function with f(0)=0 and f(1)=1 and f '(0)=f '(1)=0, then / f ' ' (x) / >=4 for some x in [0,1].
    In more picturesque terms: A particle which travels a unit distance in a unit time, and starts and ends with velocity 0, has at some time an acceleration>=4

    Hint:Prove that either f ' ' (x)>4 for some x in [0, 1/2] or f ' '(x)<-4 for some x in [1/2, 1].

    and show that in fact we must have/ f ''(x)/ >4 for some x in [0,1]
  2. jcsd
  3. Mar 16, 2007 #2


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    Perhaps you could show some of your working?

    And how urgent can it be if it's not "hw"?
  4. Mar 16, 2007 #3
    it will probably be a test question on the test tmr....

    using contradition? or not..i dont have enough time to think about it now, just want to see some clues.
    Last edited: Mar 16, 2007
  5. Mar 16, 2007 #4
    i m not sure if my answer is completed,, can anyone check it for me? thx!
    suppose f''(x)<4 for all x in [0, 1/2],apply mean value theorem,if0<=x<=1/2
    (f'(x)-f'(0))/(x-0)<4 ==> f'(x)<4x,one more time mean value theorem,if0<=x<=1/2
    (f(x)-f(0))/(x-0)<4x ==>f(x)<4x^2 f(1/2)<1
    suppose f''(x)>-4 for all x in [1/2,1]
    (f'(1)-f'(x))/(1-x)>-4 ==> f'(x)<4(1-x)
    (f(1)-f(x))/(1-x)<4(1-x)=>f(x)>1-4(1-x)^2 f(1/2)>1 contradition
    therefore, either f''(x)>=4 for some x in [0, 1/2],orf''(x)<=-4 for some x in [1/2,1]
  6. Mar 16, 2007 #5

    matt grime

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    That can't possibly be right.

    Consider an f satisfying f''=0 identically on [0,1/2]. Obviously it fails the first half of your condition. And it is easy to complete it to a smooth function satisfying the boundary conditions with f'' =>0.
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