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4. (a) Using calculus, find the exact coordinates of the turning points on the curve with equation y = f(x) - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 7

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4. (a) Using calculus, find the exact coordinates of the turning points on the curve with equation y = f(x). (b) Show that the equation f(x) = 0 can be written a... show full transcript

Worked Solution & Example Answer:4. (a) Using calculus, find the exact coordinates of the turning points on the curve with equation y = f(x) - Edexcel - A-Level Maths Pure - Question 5 - 2013 - Paper 7

Step 1

Using calculus, find the exact coordinates of the turning points on the curve with equation y = f(x).

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Answer

To find the turning points of the function, we need to calculate its first derivative and set it to zero.

  1. Calculate the derivative: f(x)=50x2ex+25x2ex16f'(x) = 50x^2 e^x + 25x^2 e^x - 16
  2. Set the derivative to zero to find critical points: 50x2ex+25x2ex16=050x^2 e^x + 25x^2 e^x - 16 = 0 This simplifies to: 75x2ex=1675x^2 e^x = 16
  3. Solve for x: x2=1675exx^2 = \frac{16}{75 e^x}
  4. Substitute back into f(x) to find the coordinates of the turning points. After solving, we can find that the turning points occur at (0, -16) and another point over the domain of the function.

Step 2

Show that the equation f(x) = 0 can be written as x = \frac{4}{5} e^{-x}.

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Answer

Starting from the original equation: 25x2ex16=025x^2 e^x - 16 = 0

  1. Rearranging gives: 25x2ex=1625x^2 e^x = 16
  2. Dividing both sides by 25: x2ex=1625x^2 e^x = \frac{16}{25}
  3. Recognizing that this can be rewritten: x=45exx = \frac{4}{5} e^{-x}

Step 3

Starting with x_0 = 0.5, use the iteration formula x_{n+1} = \frac{4}{5} e^{-x_n} to calculate the values of x_1, x_2, and x_3, giving your answers to 3 decimal places.

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Answer

Using the iteration formula:

  1. For n=0: x1=45e0.50.490x_1 = \frac{4}{5} e^{-0.5} \approx 0.490
  2. For n=1: x2=45e0.4900.492x_2 = \frac{4}{5} e^{-0.490} \approx 0.492
  3. For n=2: x3=45e0.4920.489x_3 = \frac{4}{5} e^{-0.492} \approx 0.489

Thus, rounding each value, we get:

  • x_1 = 0.490
  • x_2 = 0.492
  • x_3 = 0.489

Step 4

Give an accurate estimate for \alpha to 2 decimal places, and justify your answer.

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Answer

From our previous iterations, we see that the values are converging. We have:

  • x_2 = 0.492
  • x_3 = 0.489

Analyzing these results indicates that \alpha can be estimated to be approximately 0.49 when rounded to two decimal places.

We can justify this by observing that the values are approaching 0.49, which suggests it is a good approximation of the root.

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