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

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

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

Step 1

Find the coordinates of the turning points of C

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Answer

To find the turning points, we first need to compute the first derivative of the function given by y=x9+x2.y = \frac{x}{9 + x^2}.

Using the quotient rule, we find:

dydx=(9+x2)(1)(x)(2x)(9+x2)2=9+x22x2(9+x2)2=9x2(9+x2)2.\frac{dy}{dx} = \frac{(9 + x^2)(1) - (x)(2x)}{(9 + x^2)^2} = \frac{9 + x^2 - 2x^2}{(9 + x^2)^2} = \frac{9 - x^2}{(9 + x^2)^2}.

Setting the derivative equal to zero to find the critical points:

9x2=0x2=9x=±3.9 - x^2 = 0 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3.

Next, we find the corresponding y-values:

When (x = 3:\n y = \frac{3}{9 + 3^2} = \frac{3}{18} = \frac{1}{6}.$$

When (x = -3:\n y = \frac{-3}{9 + (-3)^2} = \frac{-3}{18} = -\frac{1}{6}.\n Thus, the turning points of the curve are at ((3, \frac{1}{6})) and ((-3, -\frac{1}{6})).

Step 2

Find the value of \frac{dy}{dx} at x = \frac{1}{2} \ln 3

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Answer

Given the function y=(1+ex)32,y = (1 + e^{x})^{\frac{3}{2}}, we will compute the derivative:

Using the chain rule:

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

Now, we substitute (x = \frac{1}{2} \ln 3:\n e^{x} = e^{\frac{1}{2} \ln 3} = \sqrt{3}.$$

Substituting this in:

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

Now simplifying,

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

This gives us the final answer.

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