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A bijective map Question

  1. Jun 10, 2017 #1
    1. The problem statement, all variables and given/known data
    Let f: R->R and f(x)=x3-x. By restricting the domain and range of f appropriately, obtain from f a bijective function g.

    2. Relevant equations
    x3-x=(x+1)(x-1)x
    g(x): R->R

    3. The attempt at a solution
    we can find roots from the polynomial form (x+1)(x-1)x and restrict the domain and range by avoiding intervals including more than one root. There are several ways to obtain g, because we are free to choose an interval, but what I'm interested in is this.

    if we say g:R->R such that
    x=-1 -> g(x)=-∞
    x=0 -> g(x)=0
    x=1 -> g(x)=∞

    we can define a 1 to 1 map because R is an infinite set, then can we say this is also an answer? Because it is bijective to R.
     
  2. jcsd
  3. Jun 10, 2017 #2

    andrewkirk

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    Although they do not say so explicitly, one can infer that they want the function g to match f on the shared domain.
    Since f is not injective, some of the domain of f is going to have to be removed.

    BTW, avoiding intervals including more than one root is not necessary or sufficient. What you need is an interval on which f is purely non-increasing or purely non-decreasing.
     
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