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The curve C has equation y= rac{x}{9+x^2} - Edexcel - A-Level Maths Pure - Question 6 - 2007 - Paper 6

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The curve C has equation y= rac{x}{9+x^2}. Use calculus to find the coordinates of the turning points C. Given that y=(1+e^{x})^{ rac{3}{2}}, find the value of ... show full transcript

Worked Solution & Example Answer:The curve C has equation y= rac{x}{9+x^2} - Edexcel - A-Level Maths Pure - Question 6 - 2007 - Paper 6

Step 1

i) Use calculus to find the coordinates of the turning points C.

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Answer

To find the turning points of the curve, we need to differentiate the given function and set the derivative equal to zero.

  1. Differentiate the function:

    We have:

    y=x9+x2y = \frac{x}{9+x^2}

    To find \frac{dy}{dx}, we can use the quotient rule:

    dydx=(9+x2)(1)x(2x)(9+x2)2\frac{dy}{dx} = \frac{(9+x^2)(1) - x(2x)}{(9+x^2)^2} Simplifying this expression gives:

    dydx=9x2(9+x2)2\frac{dy}{dx} = \frac{9 - x^2}{(9+x^2)^2}

  2. Set the derivative to zero:

    We therefore set:

    9x2=09 - x^2 = 0 This yields:

    x2=9x=±3x^2 = 9 \Rightarrow x = \pm 3

  3. Find the corresponding y-coordinates:

    For (x = 3):

    y=39+32=318=16y = \frac{3}{9 + 3^2} = \frac{3}{18} = \frac{1}{6}

    For (x = -3):

    y=39+(3)2=318=16y = \frac{-3}{9 + (-3)^2} = \frac{-3}{18} = -\frac{1}{6}

Thus, the coordinates of the turning points are ((3, \frac{1}{6})) and ((-3, -\frac{1}{6})).

Step 2

ii) find the value of dy/dx at x=1/2 ln 3.

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Answer

We need to find (\frac{dy}{dx}) for the function (y = (1 + e^x)^{\frac{3}{2}}).

  1. Differentiate the function:

    Using the chain rule, we have:

    dydx=32(1+ex)12ex\frac{dy}{dx} = \frac{3}{2}(1+e^x)^{\frac{1}{2}} \cdot e^x

  2. Substitute (x = \frac{1}{2} \text{ln} 3):

    First, evaluate (1 + e^{\frac{1}{2} \text{ln} 3}):

    (e^{\frac{1}{2} \text{ln} 3} = \sqrt{3}), hence:

    1+e12ln3=1+31 + e^{\frac{1}{2} \text{ln} 3} = 1 + \sqrt{3}

    Now substitute into the derivative:

    dydx=32(1+3)12e12ln3=32(1+3)123\frac{dy}{dx} = \frac{3}{2}(1 + \sqrt{3})^{\frac{1}{2}} \cdot e^{\frac{1}{2} \text{ln} 3} \quad = \quad \frac{3}{2}(1 + \sqrt{3})^{\frac{1}{2}} \cdot \sqrt{3}

  3. Calculate the final value:

    As a final step:

    dydx=32(1+3)123=332(1+3)12\frac{dy}{dx} = \frac{3}{2}(1 + \sqrt{3})^{\frac{1}{2}} \cdot \sqrt{3} = 3 \cdot \frac{\sqrt{3}}{2}(1 + \sqrt{3})^{\frac{1}{2}}

    Using approximations, this yields approximately 18.

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