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Function Question

  1. Sep 14, 2016 #1
    1. The problem statement, all variables and given/known data
    Let f: ℝxℝ → ℝ and g: ℝ → ℝxℝ be given by f(x,y) = x and g(x) = (x,0) Compute f°g and g°f (This means for each function find the domain and codomain, and show what each function evaluates at each point in the domain.

    2. Relevant equations


    3. The attempt at a solution
    g°f ⇒ f(g(x)) or f(x,y) = x → g(x) = (x,0). Therefore x ∈ ℝ and y ∈ ℝ is the domain and x ∈ ℝ and y ∈ 0 is part of the codomain. When placing x ∈ ℝ and y ∈ ℝ in the function the output will be (x,0) iff (x,y)

    That is for g°f. Not going to spend another 3 minutes typing out f°g because I think it would be a waste of mine and your time. I am just really confused about everything on this section as a whole and hopefully your insights will help. Please ... Please point out anything that you don't like about the way I write or my notation or really anything. Thanks
     
  2. jcsd
  3. Sep 14, 2016 #2

    SammyS

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    g°f is often expressed as g(f( argument of f )) in this case the argument of f is an element of ℝ2, often written as an ordered pair .

    f(g(x)) gives f°g rather than g° .

    g(x) = (x,0) .
    Therefore, f(g(x)) = f((x,0)) = x . Your result was correct for f°g, but the intermediate step was wrong.
     
  4. Sep 15, 2016 #3

    PeroK

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    This may or may not help.

    Think of a function as a process that takes an input and changes it to an output according to some rule. And, the composition of two functions is like a mini production line.

    For this problem you could imagine that you are ##f## and a friend is ##g##. If I give you an input ##(x, y)## you transform that into an output according to the rules of ##f## and pass the output to your friend. They apply the rules of ##g## to that. And, as there only these two steps in the production line, that's the final output which gets handed back to me.

    So, if I give you ##(x,y)## what do I get back from your friend?

    That's g°f. Which means ##f## first then ##g##. You'll have to remember this.

    We can also set it up the other way round. For f°g, I give ##x## to your friend who does ##g## and passes the result to you, who does ##f## and gives the final output to me.

    So, if I give your friend ##x## what do I get back from you.

    Finally, I never liked the term codomain. I use the terms domain and range. In any case, these define the input and output for a function. The set of all allowed inputs is the domain. And the set of all possible outputs is the codomain /range.

    In this case, ##f## takes ordered pairs of real numbers and outputs a single real number, while ##g## does the opposite.
     
  5. Sep 15, 2016 #4
    Is this true: f is surjective, g is injective, g°f is neither,f°g is bijective?
     
  6. Sep 15, 2016 #5

    PeroK

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    What do you think?
     
  7. Sep 15, 2016 #6
    Do you want me to explain why i think it is true?
     
  8. Sep 15, 2016 #7

    PeroK

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    Yes, that's all true. Although, f o g is more than a bijection: It's the identity function.
     
  9. Sep 15, 2016 #8
    Thank you
     
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