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Math Help - prove g is 1-1

  1. #1
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    prove g is 1-1

    g(x) = x/(1-|x|)

    I began by assuming not

    so y/(1-|y|) = x/(1-|x|)

    but came unstuck when dealing with the modulus signs - do I need to deal with the case where x<0, y>0. If so, how do I do this?

    thanks
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  2. #2
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    You must have copied it incorrectly.
    As written it is not one-to-one.
    g\left( 3 \right) = g\left( {\frac{{ - 3}}<br />
{5}} \right) = \frac{{ - 3}}<br />
{2}
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  3. #3
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    ahh sorry - its defined from -1 to 1
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  4. #4
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    In that case they cannot have different signs.
    Note \left| x \right| < 1\; \Rightarrow \;1 - \left| x \right| > 0.
    So the numerators must have the same sign.
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  5. #5
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    of course - I missed the fact it was between -1 and 1

    I am also asked to find g((-1,1))

    what does this mean? is it g(-1) and g(1)?

    I also need inverse of g.. do I need to do this with separate cases for x>0, x<0?

    many thanks
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  6. #6
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    Quote Originally Posted by James0502 View Post
    I am also asked to find g((-1,1))
    what does this mean? is it g(-1) and g(1)?
    First, did you graph the function?
    g[(-1,1)] is simply the range of the function.
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  7. #7
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    yes, I did.. so the range is infinity, since the graph goes from -inf to + inf?
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