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Math Help - Tricky inverse question

  1. #1
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    Tricky inverse question

    Hi everyone,

    I've just encountered a tricky inverse question. Can someone please give me a hint.

    Assume that f and g are both one-to-one functions. Express (f of g )^(-1) in terms of f^(-1) and g^(-1).

    Thanks a lot everyone.
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  2. #2
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    Re: Tricky inverse question

    you should know that for any function f ,f^-1 of f has the property f(f^-1)(x) = x or f^-1(f(x)) = x

    so suppose that "h " is the inverse function of (f\;o\;g) now just find "h" such that

     (f\; o\; g\;o\; h )(x) = x \Rightarrow \;\;f(g(h(x))) =x

    can you handle it ?
    take inverse function of "f" then for "g"
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  3. #3
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    Re: Tricky inverse question

    Quote Originally Posted by pohkid View Post
    Assume that f and g are both one-to-one functions. Express (f of g )^(-1) in terms of f^(-1) and g^(-1).
    Suppose that (x,z) \in f \circ g then \left( {\exists y} \right)\left[ {(x,y) \in g\;\& \,(y,z) \in f} \right].

    So (y,x) \in g^{ - 1} \;\& \,(z,y) \in f^{ - 1} or \left( {z,x} \right) \in g^{ - 1}  \circ f^{ - 1} .

    Thus \left( {f \circ g} \right)^{ = 1} =~?
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  4. #4
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    Re: Tricky inverse question

    another way to look at this is:

    g:x-->y f:y-->z

    fog:x-->y-->z, or just fog:x-->z for short.

    1-1 means we can "reverse the arrows" (because each y = g(x) came from a single x, and each z = f(y) came from a single y). so:

    f^-1:z-->y g^-1:y-->x

    now express a function that takes z-->x in 2 different ways: one as the inverse of fog, and another as some composition......
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