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Sup problem

  1. Oct 15, 2009 #1
    1. The problem statement, all variables and given/known data

    sup problem
    if f is continuous on [a,b] with f(a)<0<f(b), show that there is a largest x in [a,b] with f(x)=0


    2. Relevant equations


    i think it can be done by least upper bounds, but i dun know wat is the exact prove.

    3. The attempt at a solution
     
    Last edited: Oct 15, 2009
  2. jcsd
  3. Oct 15, 2009 #2
    Start by showing that there is one x in [a,b] such that f(x) = 0. Then form the set
    [tex]S = \{x \in [a,b] | f(x) = 0\}[/tex]
    You have already shown that S is non-empty and you know that it's bounded, so it must have a supremum. Let
    [tex]x_0 = \sup\,S[/tex]
    Since [itex]x_0[/itex] is an upper bound for S, if we can show [itex]x_0 \in S[/itex] we have shown that it's the largest element in S. So all you need to do is show [itex]x_0 \in S[/itex]. The easiest way to do this is to assume [itex]x_0 \notin S[/itex], i.e. [itex]y_0 = f(x_0) \not= 0[/itex]. Now since f is continuous at [itex]x_0[/itex] we can find some [itex]\delta > 0[/itex] such that if [itex]x \in (x_0-\delta,x_0 + \delta)[/itex] then [itex]f(x) \in (0,2y_0)[/itex] (let [itex]\epsilon = |y_0|[/itex]). Now [itex]x_0 - \delta[/itex] is an upper-bound for S, but less than [itex]x_0[/itex].
     
  4. Oct 19, 2009 #3
    but how to prove x0 is in [a,b]?
     
  5. Oct 19, 2009 #4
    i seem to know...
    if x0<a, obviously wrong.
    if x0>b, we can find some x such that b-[tex]\delta[/tex]<x<b satisfying that x is upper bound of the set.
     
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