Results 1 to 2 of 2

Math Help - mean value theorem section problem

  1. #1
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    mean value theorem section problem

    f is continues in [0,1] and differentiable in (0,1)
    f(0)=0 and for  x\in(0,1) |f'(x)|<=|f(x)| and 0<a<1 prove:
    (i)the set {|f(x)| : 0<=x<=a} has maximum
    (ii)for every x\in(0,a] this innequality holds \frac{f(x)}{x}\leq max{|f(x)|:0<=x<=a}
    (iii)f(x)=0 for x\in[0,a]
    (iiiן)f(x)=0 for 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]
    |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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,540
    Thanks
    780

    Re: mean value theorem section problem

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

    Hint: use \{ and \} to make curly braces show in LaTeX.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: April 20th 2010, 02:57 AM
  2. [SOLVED] Problem from a section on Bezout's Identity
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 18th 2010, 06:27 PM
  3. Problem 1, Section 1...:(
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: December 20th 2009, 07:00 AM
  4. Replies: 2
    Last Post: March 30th 2008, 07:40 AM

Search Tags


/mathhelpforum @mathhelpforum