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Math Help - More help.. on functions

  1. #1
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    More help.. on functions

    Hi,

    Let f: R --> R be the function f(x) = (x + 1)^2. Compute the following sets:

    i) f [{-1}]
    ii) f^-1 [[-1, 1]]
    iii) f[ [-1, -1] ] union [1, 3] ]
    iv) f[f^-1 [ [-3, -1] ] ]
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  2. #2
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    Quote Originally Posted by jenjen View Post
    Hi,

    Let f: R --> R be the function f(x) = (x + 1)^2. Compute the following sets:

    i) f [{-1}]
    Is the set of points which are the image of all the points in {-1} under f.

    As {-1} contains but one point f({-1}) = {f(-1)} = {0}.

    ii) f^-1 [[-1, 1]]
    Is the set of all points which f maps into the closed interval [-1,1], or all
    points x such that |f(x)|<=1. As the domain of f is R and f(x)>0 for
    all x in R this is the set of all x in R such that:

    (x+1)^2 <=1

    or:

    -1<x+1<=1,

    or:

    -2<=x<=0

    which is the closed interval [-2,0].
    iii) f[ [-1, -1] ] union [1, 3] ]
    The use of brackets here is a bit confusing, but I will assume this means:

    f([-1,-1] Union [1,3]) = {f(-1)} Union f([1,3]) = {0} Union [1,16]

    iv) f[f^-1 [ [-3, -1] ] ]
    As we are working in R f^-1([-3,-1]) = NullSet (because there is no real
    x such that (x+1)^2 is negative), and f(NullSet)=NullSet, so:

    f(f^-1 ([-3, -1])) = NullSet.

    RonL
    Last edited by CaptainBlack; October 27th 2006 at 10:45 PM.
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  3. #3
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    Quote Originally Posted by CaptainBlank View Post
    As we are working in R f^-1([-3,-1]) = NullSet (because there is no real
    x such that (x+1)^2 is negative), and f(NullSet)=NullSet, so:

    f(f^-1 ([-3, -1])) = NullSet.
    I would say that such a question cannot be asked. Since the inverse image is the empty set we cannot calculate the image of that since mappings of null sets was not defined (at least not in my book).
    Last edited by ThePerfectHacker; October 29th 2006 at 04:57 AM.
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  4. #4
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    Quote Originally Posted by ImperfectHacker
    Quote Originally Posted by CaptainBlack View Post
    As we are working in R f^-1([-3,-1]) = NullSet (because there is no real
    x such that (x+1)^2 is negative), and f(NullSet)=NullSet, so:

    f(f^-1 ([-3, -1])) = NullSet.
    I would say that such a question cannot be asked. Since the inverse image is the empty set we cannot calculate the image of that since mappings of null sets was not defined (at least not in my book).
    We have extended the definition of f from its domain D to the power set of
    its domain by defining:

    f(A): {all b in the Range of f, such that there exists a in A such that f(a)=b}.

    Thus if A is the null set (which is in the power set of D), there is no b in the
    range of f which is the image of an a in A. Therefore f(NullSet)=NullSet.

    RonL
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  5. #5
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    Thank you so much for the replies ThePerfectHacker and CaptainBlack!!!!!
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