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Math Help - HeLP Extra Credit Fixed Point Problem (please check)

  1. #1
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    HeLP Extra Credit Fixed Point Problem (please check)

    A number a, is called a fixed point of a function f(a)=a. Let f(x) be a continuous function that is differentiable everywhere. However, the derivative of f(x) is never equal to one, (f ' (x) cannot equal 1), for any x in its domain. Prove that the function, f, cannot have more than one fixed point.

    i got this answer:
    In other words, fixed points of a function are the intersection of the function and the line y=x.
    Thus, the idea here is that, if f'(x) =/= 1, then either f'(x) > 1 or < 1 for all x in its domain. If it has a fixed point, then the curve will never "turn" back towards the line y=x again. Thus, it can't have more than one fixed point.

    It should be easier to recast the scenario into the number of roots of g(x) = 0 where g(x) = f(x) - x, with the same domain as f(x).
    Given f'(x) =/= 1, g'(x) = f'(x) - 1 =/= 0.
    Thus, g(x) is either a strictly increasing or strictly decreasing function ==> g(x) = 0 cannot have more than 1 root, for any x in its domain.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    That's it, good job !
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