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Expanding Composite Functions( help needed)

  1. Nov 18, 2004 #1
    Expanding Composite Functions(urgent help needed)

    I had to solve an assignment question, we are asked to find the formula for g^-1, except once i plug n do everything, i come up with x=f(g^-1(x)+c), which is x=f(g^-1(x)) + f(c), now i m suppose to put the formula in terms of g^-1=......, but i don't know how i can expand the function f(g^-1(x)). So how do i expand composite functions?
     
  2. jcsd
  3. Nov 18, 2004 #2

    cepheid

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    g-1(x) is just the argument you are passing into the function f. So treat it no differently than you would if you were passing in x, or t, or y etc:

    E.g. if f(x) = x2, then f(g-1(x)) = [g-1(x)]2
     
  4. Nov 18, 2004 #3
    Also, we haven't learned the chain rule or derivatives yet, so it doesn't involve any of that in the solution.
     
  5. Nov 18, 2004 #4

    cepheid

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    :confused: There was no differentiation of any kind in my last post.

    If you could post the expression for f(x), that would be helpful.
     
  6. Nov 18, 2004 #5
    well that is obvious to me, its just that, i dont know how to expand.
    This is the question.
    Suppose f is a one-one function. g(x)=f(x + c) for all x s.t. x +c (element)dom f , now prove that g is one-one and find the formula for g^-1.
     
  7. Nov 18, 2004 #6
    so basically, any one-one function f, can possibly work, dont you think?
     
  8. Nov 18, 2004 #7

    Hurkyl

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    Why do you think that?


    Anyways, what about using f^-1 somewhere?
     
  9. Nov 18, 2004 #8
    I realized though, that since the squaring funcion is one-one. That means i can suppose for f that it is f(x)=x^2.
    So then, x=g(g^-1(x))=f(g^-1(x)) + f(c)=[g^-1(x)]^2 + c^2.
    So x - c^2=[g^-1(x)]^2.
    So g^-1(x)=sq.root(x - c^2).
    Plz tell me if this is right.
     
  10. Nov 18, 2004 #9

    cepheid

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    f(x) = x^2 is not one to one (unless you restrict the domain of the inverse function appropriately).
     
  11. Nov 18, 2004 #10
    so if i restrict the domain of the inverse, then it becomes one-one and then my solution will be correct right.
     
  12. Nov 18, 2004 #11
    btw can someone give me a proof of the function f(x)=x^2, as being one-one.
     
  13. Nov 18, 2004 #12

    Hurkyl

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    You cannot; you're reversing the statement.

    The statement "the squaring function is one-one" does not mean you can say "a one-one function is the squaring function".

    (And, of course, you have cephid's observation that the squaring function is not one-one)
     
  14. Nov 18, 2004 #13
    i c, i realized that i should have used the cubic function, since that is one-one.
     
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