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Mean value theorem section problem

  1. Sep 11, 2011 #1
    f is continues in [0,1] and differentiable in (0,1)

    f(0)=0 and for[TEX] x\in(0,1)[/TEX] [TEX]|f'(x)|<=|f(x)|[/TEX] and 0<a<1 prove:

    (i)the set [TEX]{|f(x)| : 0<=x<=a}[/TEX] has maximum

    (ii)for every [TEX]x\in(0,a][/TEX] 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]

    (iiiן)f(x)=0 for [TEX]x\in[0,1][/TEX]

    in each of the following subquestion we can use the previosly proves subquestion.



    first part i have solve by saying that f is continues in the subsection so

    by weirshtrass we have max and min

    and max|f(x)|=max{|maxf(x)|,|minf(x)|}





    in the second part

    we know that max|f(x)|>|f(x)|>=|f'(x)|

    and we take c in [0,x] a subsection of [0,a]

    |f(c)|>=|f'(c)|

    and we know that f(0)=0 so we take [0,a]

    |f'(c)|=|f(0)-f(x) /x-0 |

    |f'(c)|=|f(x)/x|

    |f(x)|>|f(x)|>=|f'(x)|

    so i got all the parts but i cant join them because its c there and not x

    c is inside point x is on the border.



    what to do?
     
  2. jcsd
  3. Sep 11, 2011 #2

    LCKurtz

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    Science Advisor
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    Use lower case in your tex tags as I have done above to make it readable. Also if you preview your posts you will see if it works. I will let others comment because I have a football game to watch :smile:
     
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