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Tricky Continuity

  1. Nov 14, 2008 #1
    1. The problem statement, all variables and given/known data

    1. show there is some point x in the interval [0,pi/2] so that x = cos(x)^2

    2. let f:R-> be continuous at c and suppose f(c) =1. show that there is some a > 0 such
    that f(x) > 1/2 whenever |x-c| < a

    2. Relevant equations

    intermediate value theorem. maximum minum

    3. The attempt at a solution

    1. We know that f(x) = cos(x)^2 assumes its max value at f(0) = 1 and its min. at f(pi/2) = 0

    by the int. value theoreom 0 < pi/2 implies f(pi/2) < y < f(0)

    implies for some z in [0,pi/2] f(z) = y

    the only conclusion I can come up with is that since the interval [0,pi/2] is greater than the distance between |f(0)-f(pi/2)| there has to be at least one x in [0,pi/2] where f(x) = x. Is this plausible?

    b) Unsure as to how to properly approach this problem.

    since f(x) > 1/2

    |f(x) - f(c)| = |1/2 - 1| <= |1/2| < e/2

    let a equal e/2?

    I truly apologize for this poor approach. If someone could just direct me in the right direction it would be greatly appreciated.
  2. jcsd
  3. Nov 14, 2008 #2


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    Why not just the basic definition of continuity?

    If f is continuous at c, then, given any [itex]\epsilon> 0[/itex], there exist [itex]\delta>0[/itex] such that if [itex]|x- c|< \delta[/itex], then [/itex]|f(x)- f(c)|< \epsilon[/itex].

    Take [itex]\epsilon= 1/2[/itex].
  4. Nov 14, 2008 #3
    was part 1. correct? I still feel a bit insecure about my answer. thanks
  5. Nov 14, 2008 #4


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    Staff Emeritus
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    Yes, part 1 is correct.
  6. Nov 14, 2008 #5


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    Homework Helper

    Your proof of part 1 is a little confusing. Consider the function g(x)=x-cos(x)^2. Can you show g(x) has a zero in [0,pi/2]?
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