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Thread: mean value theorem section problem

  1. #1
    MHF Contributor
    Nov 2008

    mean value theorem section problem

    f is continues in [0,1] and differentiable in (0,1)
    f(0)=0 and for$\displaystyle x\in(0,1)$ $\displaystyle |f'(x)|<=|f(x)|$ and 0<a<1 prove:
    (i)the set $\displaystyle {|f(x)| : 0<=x<=a}$ has maximum
    (ii)for every $\displaystyle x\in(0,a]$ this innequality holds $\displaystyle \frac{f(x)}{x}\leq max{|f(x)|:0<=x<=a}$
    (iii)f(x)=0 for $\displaystyle x\in[0,a]$
    (iiiן)f(x)=0 for $\displaystyle x\in[0,1]$
    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]
    and we know that f(0)=0 so we take [0,a]
    |f'(c)|=|f(0)-f(x) /x-0 |
    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?
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  2. #2
    MHF Contributor
    Oct 2009

    Re: mean value theorem section problem

    By MVT, $\displaystyle \frac{f(x)-f(0)}{x-0}=\frac{f(x)}{x}=f'(c)$ for some $\displaystyle c\in(0,x)$. Now, $\displaystyle f'(c)\le|f'(c)|\le |f(c)|$ by assumption, and $\displaystyle |f(c)|\le\max\{|f(x)|:0\le x\le a\}$ since $\displaystyle x\in(0,a]$.

    Hint: use \{ and \} to make curly braces show in LaTeX.
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