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Math Help - Finding the Domain of Square Root

  1. #1
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    Finding the Domain of Square Root

    f(x) = \sqrt{x^2-5x+6}

    f(x) = \sqrt{(x-2)(x-3)}

    (x-2)(x-3) \geq 0

    x-2 \geq 0 \Longrightarrow x \geq 2

    x-3 \geq 0 \Longrightarrow x \geq 3

    I know from the graph that it should be x \leq 2

    So what am I doing wrong?
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  2. #2
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    Re: Finding the Domain of Square Root

    Your workng looks good to give you some bounds on the solution

    Consider \displaystyle (-\infty ,2]\cap[3,\infty)
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  3. #3
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    Re: Finding the Domain of Square Root

    Quote Originally Posted by Gfan2010 View Post
    f(x) = \sqrt{x^2-5x+6}

    f(x) = \sqrt{(x-2)(x-3)}

    (x-2)(x-3) \geq 0

    x-2 \geq 0 \Longrightarrow x \geq 2

    x-3 \geq 0 \Longrightarrow x \geq 3

    I know from the graph that it should be x \leq 2

    So what am I doing wrong?
    Case 1: x - 2 >= 0 AND x - 3 >= 0 => x >= 2 AND x > 3 => x >= 3.

    Case 2: x - 2 =< 0 AND x - 3 =< 0 => .......

    Personally, I think it's much better to draw a graph of y = (x - 2)(x - 3) and then use it to see the values of x for which y >= 0.
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