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Intermediate Value Theorem

  1. Sep 13, 2014 #1
    From the Intermediate Value Theorem, is it guaranteed that there is a root of the given equation in the given interval?

    cos(x)=x, (0,1)

    cos(0)= 1
    cox(1)= 0.540...

    So using intermediate value theorem, no.

    and

    x=0 can't be possible because 0 was excluded in the domain by the round bracket so 0<x<1. Therefore there can't be a root. In other words the domain makes it so the graph is discontinuous at 0 (f(0) does not exist) and if the graph/function doesn't at that point there cannot be a root there.

    The answer is yes, but I am struggling to understand why and I am not sure if the books answer is wrong or if I'm wrong.

    *I also don't know why cos(x) = x , because when x=1 they aren't equal and would only be true at 1 point in this domain (when x=~0.79).
     
    Last edited: Sep 13, 2014
  2. jcsd
  3. Sep 13, 2014 #2
    It's probably better to examine the equivalent question; Is there a solution to ##\cos x-x=0## on the interval ##(0,1)##?

    The Intermediate Value Theorem says
    If we put ##f(x)=\cos x-x##, your (revised) question is asking whether there is ##c## in ##(0,1)## satisfying ##f(c)=0##, right?

    Given that ##f(0)=\cos 0-0=1>0## and ##f(1)=\cos 1-1<0## (you should check that), do you see how the IVT guarantees a ##c## in ##(0,1)## with ##f(c)=0##? Do you see how this is the same as saying ##\cos c=c##?

    Note that you also need to say something regarding the continuity of ##f(x)=\cos x-x## on the closed interval ##[0,1]##.
     
  4. Sep 13, 2014 #3
    Oh ok, that makes sense, I thought the question was asking me to do something else.

    Yeah, that would have made it much clearer imo.
     
    Last edited: Sep 13, 2014
  5. Sep 13, 2014 #4
    For what it's worth, I'm not a big fan of the use of the word "root" in the problem statement. It'd probably be better to use the word "solution" in that context.
     
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