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Math Help - Bijection

  1. #1
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    Bijection

    Determine if this is a bijection and find the inverse function.f : R − {−2} → R − {1} where f(x) = (x + 1)=(x + 2).




    I have the ans sheet: It's a bijection with inverse f-1(x) = (2x 1)=(1 x).
    But i dont know how to get the ans. Please help me
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  2. #2
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    Re: Bijection

    Quote Originally Posted by ZGOON View Post
    Determine if this is a bijection and find the inverse function.f : R − {−2} → R − {1} where f(x) = (x + 1)=(x + 2).
    I have the ans sheet: It's a bijection with inverse f-1(x) = (2x − 1)=(1 − x).
    But i dont know how to get the ans. Please help me
    Please check your post. You have too many "='s" in it.

    Do you mean f(x)=\frac{x+1}{x+2}~?
    Last edited by Plato; March 21st 2013 at 06:26 AM.
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  3. #3
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    Re: Bijection

    Sorry for the error.

    f : R − {−2} → R − {1} where f(x) = (x + 1)/(x + 2).

    inverse f-1(x) = (2x − 1)/(1 − x).
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  4. #4
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    Re: Bijection

    Quote Originally Posted by ZGOON View Post
    Sorry for the error.
    f : R − {−2} → R − {1} where f(x) = (x + 1)/(x + 2). inverse f-1(x) = (2x − 1)/(1 − x).

    If t\in\mathbb{R}\setminus\{1\} is it possible to solve t=\frac{x+1}{x+2}~? (i.e. get x in terms of t)

    If so, compare the answer.
    Last edited by Plato; March 21st 2013 at 06:24 AM.
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  5. #5
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    Re: Bijection

    How you get x-2???

    It is function since every x in the domain has an image. 1-1: f(a) = f(b) implies
    (a+1)/(a+2) = (b+1)/(b+2). This yields a = b. Onto: Let y ∈ R−{1}. Set f(x) = y.
    Solving, we get x = (2y − 1)/(1 − y). Thus If (2y − 1)/(1 − y) = −2, then we have 1 = 2
    which is impossible. Thus (2y − 1)/(1 − y) is never equal to −2 and is thus a preimage.
    The above calculations also show that f-1(x) = (2x − 1)/(1 − x).
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  6. #6
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    Re: Bijection

    Quote Originally Posted by ZGOON View Post
    How you get x-2???
    Typo
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