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Math Help - who can answer this???

  1. #1
    Junior Member
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    who can answer this???

    This is my problem: g(x)=x+1/x-1 g(0)=? g(2)=? g(2x)=? g(x.x)=? g(1/x)=? 1/g(x)=? responde me. Who Can answer this?
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  2. #2
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    Hello, blertta

    g(x)\:=\:\frac{x+1}{x-1}

    Find: . (1)\;g(0)\quad (2)\;g(2)\quad(3)\;g(2x)\quad (4)\;g(x^2)\quad(5)\;g(1/x)\quad (6)\;1/g(x)

    Who can answer this? . . . Anyone with one lesson on functions

    (1)\;g(0) \:=\:\frac{0+1}{0-1} \:=\:-1

    (2)\;g(2) \:=\:\frac{2+1}{2-1} \:=\:3

    (3)\;g(2x) \:=\:\frac{2x + 1}{2x-1}


    Get the idea?

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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by blertta View Post
    This is my problem: g(x)=x+1/x-1 g(0)=? g(2)=? g(2x)=? g(x.x)=? g(1/x)=? 1/g(x)=? responde me. Who Can answer this?
    i suppose you meant g(x) = \frac {x + 1}{x - 1} (please use the proper parentheses).

    they are not really asking for much here. just take what the put in brackets and replace the x's with it and simplify.

    example. what if they asked for g(3), we would do the following:

    g(x) = \frac {x + 1}{x - 1}

    \Rightarrow g(3) = \frac {3 + 1}{3 - 1}

    = \frac 42

    = 2
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