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Homework Help: Differentiability and extreme points question

  1. Sep 6, 2011 #1
    f is continues in [0,1] and differentiable in (0,1)
    f(0)=0 and for [tex]x\in(0,1)[/tex] |f'(x)|<=|f(x)| and 0<a<1
    (i)the set {|f(x)| : 0<=x<=a} has maximum
    (ii)for every x\in(0,a] this innequality holds [TEX]\frac{f(x)}{x}\leq max{|f(x)|:0<=x<=a}[/TEX]
    (iii)f(x)=0 for [TEX]x\in[0,a][/TEX]
    (iii)f(x)=0 for [TEX]x\in[0,1][/TEX]
    in each of the following subquestion we can use the previosly proves subquestion.
  2. jcsd
  3. Sep 6, 2011 #2
    So what did you try?? If you tell us what you tried, then we'll know where to help?
  4. Sep 6, 2011 #3
    i dont want solution only starting guidence.
    for 1 i know the a continues function has a maximum
  5. Sep 6, 2011 #4
    If you want guidance, then you'll have to show us what you've tried. You won't get a solution here...
  6. Sep 6, 2011 #5
    there is a difference between solution
    and starting guidence
  7. Sep 6, 2011 #6
    For 1, you indeed use that a continuous function has a maximum. What continuous function do they think they're using here??
  8. Sep 6, 2011 #7
    i only know about the function what was given
  9. Sep 6, 2011 #8
    What function do you need to show continuity of??
  10. Sep 6, 2011 #9
    f is continues in [0,1] and differentiable in (0,1)
    f(0)=0 and for [tex]x\in(0,1)[/tex] |f'(x)|<=|f(x)| and 0<a<1
  11. Sep 6, 2011 #10
    Yes, that is what we are given.

    But we need to prove in (1) that

    [tex]\{|f(x)|~\vert~x\in [0,a]\}[/tex]

    has a maximum. To show this, we must use that every continuous function on a closed interval has a maximum.
    So, what will we take as our continuous function?
  12. Sep 6, 2011 #11
    we cant take a spesific function
    its a proof for all function

    a proof for a spesific is not sufficient
  13. Sep 6, 2011 #12
    You don't understand what micromass is trying to imply. He's not suggesting taking a specific function like f(x) = sinx. He means you need to choose the correct function in your proof, which has to do with f(x) (hint hint), in order to prove that {|f(x)| | 0<= x <= a} has a maximum.
    The answer is pretty clear - you just need to formulate a short proof.
    Get it?
  14. Sep 6, 2011 #13
    ahh now i get it
    we need to formulate some other functionand using it we prove about
    our abstract f(x)

    maybe g(X)=|f(x)|
  15. Sep 6, 2011 #14
    Yes, so prove that |f(x)| is continuous if f is.
  16. Sep 6, 2011 #15
    ionly know one way
    and it showing that [tex]lim_{x->x0}f(x)=f(x0)[/tex]

    but its not possible
    because we dont have an actual function
  17. Sep 6, 2011 #16
    Can you show |x| to be continuous??

    Using that, you can show |f(x)| to be continuous as composition of f(x) and |x|.
  18. Sep 7, 2011 #17
    You don't need to show that, it's a given.
    Like said above me, asking if the set {|f(x)| | a<=x<=b} has a maximum is just like asking "does the function |f(x)| (notice the absolute value!) get a maximum in the segment [a,b]?". You know there would be an easy answer if you'd say "|f(x)| is continuous on this closed segment and therefore gets a maximum". Therefore, you need to show that |f(x)| is a continuous.
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