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MVT Proof, please help

  1. Dec 20, 2006 #1
    Use the Mean Value Theorem to show that:

    a)Suppose f is a diferentiable function on the interval a < b, and suppose f '(x) is not equal to 0 for all x Element Symbol (a,b). Show that f is one-to-one on the interval (a,b).

    b) Assume that |f ' (x)| < C < 1 for all x. Show that f (x) = x has at most one solution.



    So far I got-> Suppose f is not one-one on the interval then there exists u, v in (a,b) such that f(u)=f(v).
    Then by the mean value theorem there exists a point w in (u,v) such that f'(w)=0, a contradiction.

    I don't know where to go after that.
     
  2. jcsd
  3. Dec 20, 2006 #2

    quasar987

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    You got a).

    For b), suppose there is one point such that f(x) = x. Then for all y, there exists a c such that

    [tex]f'(c)=\frac{f(y)-x}{y-x}[/tex]

    You go from there..
     
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