I have something that doesn't work out for me

  1. I'm going to use this theorem of differential analysis: if g(x) is differentiable between a and b, then there is a c for which [tex]\frac{g(b)-g(a)}{b-a}[/tex]=g'(c)

    Let f be differentiable twice, and let f(0)=f'(0)=0 and let f''(x)[tex]\geq[/tex]1 for all x>0

    choose g(x) as f'(x), a = 0, b = x
    then there is a c so that:
    [tex]\frac{g(b)-g(a)}{b-a}[/tex]=[tex]\frac{f'(x)}{x}[/tex]=f''(c)[tex]\geq[/tex]1 (f''(x) was already bigger then 1)

    so f'(x) [tex]\geq[/tex] x (for all x > 0)

    now I am going to use the theorem again, saying g(x)=f(x), a = 0, b = x

    then there is a c so that:

    [tex]\frac{g(b)-g(a)}{b-a}[/tex]=[tex]\frac{f(x)}{x}[/tex]=f'(c)[tex]\geq[/tex]x

    so f(x) [tex]\geq[/tex] x^2 (for all x>0)

    but this isn't true. The function 1/2 x^2 also applies for all those things I wanted the function to be, but is smaller than x^2. Anyone knows what I've been doing wrong? This was an old exam exercise, and it's been troubling me since before the holidays.

    Ps. this is the first time I did everything with tech on this site, and it's horrible unclear to me if I entered it allright. Hope it worked. If it doesn't, I hope I'll have enough time to edit a bit before someone reads it ^_^
     
    Last edited: Sep 9, 2007
  2. jcsd
  3. HallsofIvy

    HallsofIvy 40,943
    Staff Emeritus
    Science Advisor

    Here's your error. You proved above that [itex]f'(x)\ge x[/itex]. That does NOT give [itex]f'(c)\ge x[/itex], only [itex]f'(c)\ge c[/itex].

     
  4. ah of course. that's great. thanks a lot.
     
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