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Math Help - Find x-coordinates

  1. #1
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    Find x-coordinates

    Find the x-coordinates of the two stationary points on the graph of y = (2x-1)/(x^2+1).

    How is this question solved. As far as I can get is using the quotient rule to differentiate y to give:
    = 2(x^2+1)-2x(2x-1) All over (x^2+1)^2

    How do I find the stationary points from this?
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  2. #2
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    Quote Originally Posted by haku View Post
    Find the x-coordinates of the two stationary points on the graph of y = (2x-1)/(x^2+1).

    How is this question solved. As far as I can get is using the quotient rule to differentiate y to give:
    = 2(x^2+1)-2x(2x-1) All over (x^2+1)^2

    How do I find the stationary points from this?
    y = \frac{2x - 1}{x^2 + 1}

    y^{\prime} = \frac{2 \cdot (x^2 + 1) - (2x - 1)(2x)}{(x^2 + 1)^2}

    y^{\prime} = \frac{2x^2 + 2 - 4x^2 + 2x}{(x^2 + 1)^2}

    y^{\prime} = \frac{2(-x^2 + x + 1)}{(x^2 + 1)^2}

    Stationary points are where the derivative is equal to 0. So set this equal to 0, then solve.

    -Dan
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  3. #3
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    Would you use the quandratic formual to solve this? If so how would you find the values of a, b and c from the equation:

    2(-x^2+x+1)=(x^2+1)^2

    Would you apply the formula to (-x^2+x+1) ?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by haku View Post
    Would you use the quandratic formual to solve this? If so how would you find the values of a, b and c from the equation:

    2(-x^2+x+1)=(x^2+1)^2
    It's simpler than that. You made a slight error in your equation:
    \frac{2(-x^2 + x + 1)}{(x^2 + 1)^2} = 0

    Multiply both sides by (x^2 + 1)^2:
    2(-x^2 + x + 1) = 0

    Can you take it from here?

    -Dan
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  5. #5
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    I get 1 + (root5)/2 Or 1 - (root5)/2

    Is this correct?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by haku View Post
    I get 1 + (root5)/2 Or 1 - (root5)/2

    Is this correct?
    If this is meant to be
    \frac{1 \pm \sqrt{5}}{2}
    then, yes, you have it right. (Please use parenthesis!)

    -Dan
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