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Proving f(x) intersects a graph when given an interval

  1. Oct 18, 2004 #1
    Hi,

    Does anyone here have any pointers on how to even start the following question?

    Suppose for all x element [0, 1] we have f(x) element [0, 1]. Prove that the graph f(x) must intersect the line y = 1 - x

    Any help would be great thanks.
     
  2. jcsd
  3. Oct 18, 2004 #2

    Tide

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    Look at the difference between f(x) and 1-x. At x = 0, the difference will either be zero or nonzero. If it is zero then you have an intersection. If the difference is not zero then it is either positive or negative but must change sign somewhere between x = 0 and x = 1 since f(0) < 1 and f(x) must reach 1 somewhere in the interval while 1-x goes to 0 at the right boundary. Since the sign of the difference changes at some point in the interval then there must be an x' for which f(x') = 1 - x'.
     
  4. Oct 19, 2004 #3

    HallsofIvy

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    I might point out that Tide's solution works (and, indeed, the statement is true) only if f is a continuous function on [0,1] which was not stated in the original problem:
    If f(x)= 1/4 for x<= 1/2 and 3/4 for x> 1/2 then its graph does NOT intersect the line y= 1-x.
     
  5. Oct 19, 2004 #4
    Thanks for the help!
     
  6. Oct 19, 2004 #5
    i know this one!, btw its called the Intermediate value theroem
     
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