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Definition of Funtion

  1. May 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Let ## C= \{ x \in R : x \geq 1 \} ## and ## D = R^+ ##
    For each f defined below, determine ## f(C), f^{-1}(C), f^{-1}(D), f^{-1} (\{1\}) ##

    a.) ## f: R -> R ## is defined by ## f(x) =x^2##
    I have problems with the definitions

    3. The attempt at a solution

    a.)
    ## f(C)= { 1 , 4, 9, 16, ...} ## according to the definitions of x in C, x belongs to reals and it's greater or equal to 1. Then, 2 should be in f(C), but it's not. However, in the solutions I see this.

    ##f(C)= C, f^{-1}(C)=C \cup \{x \in R: -x \in C\} , f^{-1}(D) = R - \{0\} , f^{-1}(\{1\})= \{1, -1\}##

    -----
    About , ## f^{-1} (C) ## should we exclude C, since it is the inverse of this? How do they find this?
    About the inverse of f(D), why do they exclude only 0, shouldn't they exclude all positive Reals?
    The only one I understand is ## f^{-1} (\{1\}) = \{1, -1\} ## :/

    The last one is true since we could do this:

    f(x)= y =x^2
    change vars.

    ## f^{-1} (x) =x = f(x)^2 ##
    Then, x can be -1 or 1
     
    Last edited: May 7, 2014
  2. jcsd
  3. May 7, 2014 #2

    Dick

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    C isn't only integers {1,2,3,...}, C is all real numbers greater than or equal to 1. So that includes sqrt(2). f(sqrt(2))=2. So 2 is in f(C).
     
  4. May 7, 2014 #3
    Thank you, you are right.
    How do I explain ## f^{-1} (C)## ? Why does it include all R+ too?
     
  5. May 7, 2014 #4

    SammyS

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    What is the definition of ## f^{-1} (C)\ ?##


    In more descriptive terms, it's the pre-image of set C .


    So, what values of ##\ x\ ## will produce values of ##\ f(x)\ ## which are in C ?
     
  6. May 8, 2014 #5

    HallsofIvy

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    It doesn't. What is f(1/2)? Is that in C?
     
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