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Mean Value Theorem

  1. Aug 4, 2005 #1
    Please help me on this.

    If f`(x) is continuous on [a,b],apply the Mean Value Theorem to prove the inequalities min[f`] [tex]\leq[/tex] f(b)-f(a)[tex]/_[/tex] b-a [tex]\leq[/tex] [Max f`]
     
  2. jcsd
  3. Aug 4, 2005 #2

    lurflurf

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    well
    f'(x)=(f(b)-f(a))/(b-a)
    for some x in (a,b)
    so what are the biggest and smallest possible values of
    (f(b)-f(a))/(b-a)
     
  4. Aug 4, 2005 #3
    please lurflurf clarify more and give me the values
     
  5. Aug 4, 2005 #4

    lurflurf

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    Ok
    remember a<x<b
    x* is some number in the interval so
    f'(x*)=(f(b)-f(a))/(b-a)
    M>=f'(x) for all x
    M>=(f(b)-f(a))/(b-a)
    max(f')>=f'(x) for all x
    so
    max(f')>=f(x*)=(f(b)-f(a))/(b-a)
    the min case is analogous

    The idea is f'(x) may be greater than, less than, or equal to (f(b)-f(a))/(b-a).
    min(f')<=f'(x)<=max(f')
    so since for some x*
    f'(x*)=(f(b)-f(a))/(b-a)
    and for all x
    min(f')<=f'(x)<=max(f')
    then
    min[f`]<=f(b)-f(a) b-a <=[Max f`]
    since
    min(f')<=f'(x)<=max(f')
    is true for all x
    including
    min(f')<=f'(x*)<=max(f')
     
  6. Aug 5, 2005 #5
    Thanx a lot lurflurf
     
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