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Really difficult calculus problem

  1. Mar 7, 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).


    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

    I honestly have no idea if such a function even exists. Does anyone know what to do?

    f i do what Dick said, and 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. I am not sure what this tells me. Do I need to find a function with this property? I also need | f (x) |< 2
     
    Last edited: Mar 8, 2010
  2. jcsd
  3. Mar 7, 2010 #2

    Dick

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    You are being a little sloppy. Do a definite integral from 0 to a on both sides of 2*f(x)*f'(x)>=2*sin(x). Put a=pi. What do you conclude?
     
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