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Math Help - Find the range

  1. #1
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    Find the range

    1. y=\frac{3x}{x^2+1}

    2. y=\frac{3x-5}{x^2-1}

    Answer:
    1. -\frac{3}{2} \leq y \leq \frac{3}{2}

    2. y \leq \frac{1}{2},y \geq \frac{9}{2}
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  2. #2
    MHF Contributor
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    Quote Originally Posted by cloud5 View Post
    1. y=\frac{3x}{x^2+1}

    2. y=\frac{3x-5}{x^2-1}

    Answer:
    1. -\frac{3}{2} \leq y \leq \frac{3}{2}

    2. y \leq \frac{1}{2},y \geq \frac{9}{2}

    (1) yx^2+y=3x

    yx^2-3x+y=0

    b^2-4ac>=0 .. provided that x is real .

    (-3)^2-4y^2>=0

    y^2<=9/4

    then u will get -\sqrt(9/4)<=y<=sqrt(9/4)


    same goes to the other one .
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