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Math Help - surjective, bijective, invertible

  1. #1
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    surjective, bijective, invertible

    Find if functions f, g, g(f) are injective, surjective, invertible and find inverse functions if they exist.
    Determine the function f(g).
    These are functions:
    f(x)=(2x,x-1,x+1) g(x,y,z)=-x+2y+z
    please, help.
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  2. #2
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    Hello MarkoM

    Welcome to Math Help Forum!
    Quote Originally Posted by MarkoM View Post
    Find if functions f, g, g(f) are injective, surjective, invertible and find inverse functions if they exist.
    Determine the function f(g).
    These are functions:
    f(x)=(2x,x-1,x+1) g(x,y,z)=-x+2y+z
    please, help.
    I'll give you some help with f and g(f), and then I'll let you have a further think about g itself.

    First, it's often a good idea to put some numbers in, and see what comes out. I assume that f : \mathbb{R} \to \mathbb{R}^3 - which you don't tell me, but we need to know. In other words, we can put real numbers in (the domain) and get ordered triples of real numbers out (the range). For example:
    f(2) = (4,1,3)

    f(-10) = (-20,-11,-9)

    ...etc
    Now a function f is injective (one-to-one) whenever f(x_1) = f(x_2) \Rightarrow x_1 = x_2. In other words, you can't have two different elements in the domain mapped onto the same element in the range.

    So the first question you need to answer is: can you put two different real numbers into f and get the same ordered triple out? If the answer is no, then f is injective.

    The next question is: is f surjective? It's surjective if the codomain (the things that actually come out of f) is the whole of the range; in other words, can we find a real number to put in to f that will produce any ordered triple we like, as output?

    Well can you? Can you, for instance, find an x that will produce (0,0,0) as output?

    Is f invertible? In other words, for a given output is there one and only one input? If f is injective, then the answer is yes.

    Finally, g(f) takes the output of f, and feeds it through g. Try a number and see what happens:
    f(-10) = (-20,-11,-9)

    \Rightarrow g(f(-10))= -(-20)+2(-11)+(-9) = -11
    Now try it with x:
    f(x) = (2x, x-1,x+1)

    \Rightarrow g(f(x) = ... ?
    OK, over to you. See what you can come up with.

    Grandad
    Last edited by Grandad; February 11th 2010 at 05:16 AM. Reason: Fixed a typo
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  3. #3
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    Thanks for help.
    Yes, I forgot to mention: f:R->R^3 and g:R^3->R
    g(f(x))=-2x+2(x-1)+x+1=x-1
    The explanation was useful.
    Just, I`m not sure how to prove that functions are surjective, etc., without putting an example, like the one you mentioned (0,0,0), putting just variable that represents any real number.
    I can do it when dimension of domain and range is 1, f:R->R.
    Last edited by MarkoM; February 13th 2010 at 09:12 AM.
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  4. #4
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    Recall that any linear function mx+b,~m\ne 0 is a bijection.
    Is each component in f linear? So is f injective?

    Counter examples show a function fails to have a certain property: g(1,0,1)=?~\&~g(1,1,-1)=?.

    Is the function g \circ f linear?
    Last edited by Plato; February 11th 2010 at 10:29 AM.
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