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Hard Anaylsis Problem

  1. Mar 8, 2010 #1
    Is there a function f , differentiable for all real x, such that | f (x) |< 2 and f (x)f ′ (x) ≥ sin(x)?

    I noticed that [f(x)*f(x)]' = 2f(x)*f'(x) = [f(x)]^2

    So I tried multiplying that inequality by 2.
    2f (x)f ′ (x) ≥ 2sin(x)

    Then I tried integrating both sides.

    [f(x)]^2 ≥ -2cos(x).
    If I integrate both sides from 0 to Pi
    We will get [f(pi)]^2-[f(0)]^2 ≥ -2[cos(pi) - cos(0)]
    This gives us [f(pi)]^2-[f(0)]^2 ≥ 4.

    This gives us [f(pi)]^2-[f(0)]^2 ≥ 4. I am not sure what this tells me. Do I need to find a function with this property? I also need | f (x) |< 2
  2. jcsd
  3. Mar 8, 2010 #2


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    Homework Helper

    Why did you start a new thread instead of replying to the old one? You've got f(pi)^2>=4+f(0)^2. Is that compatible with |f(x)|<2?
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