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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 6 - 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 ... 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 6 - 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 first need to determine the critical points by finding the derivative of the function and setting it to zero:

  1. Compute the first derivative: f(x)=50x2ex+25x2ex.f'(x) = 50x^2 e^{x} + 25x^2 e^{x}.

  2. Set the first derivative equal to zero: 50x2ex+25x2ex=0.50x^2 e^{x} + 25x^2 e^{x} = 0.

  3. Factor out the common terms: 25x2ex(2+1)=0.25x^2 e^{x}(2 + 1) = 0.

  4. Solve for xx:

    • The factor 25x2ex=025x^2 e^{x} = 0 gives x=0x = 0.
    • Substitute xx back into the original function: f(0)=25(0)2e016=16.f(0) = 25(0)^2 e^{0} - 16 = -16.

Thus, the turning point is at (0,16)(0, -16).

Step 2

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

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Answer

Start with the equation: 25x2ex16=025x^2 e^{x} - 16 = 0 Rearranging gives: 25x2ex=1625x^2 e^{x} = 16 Dividing both sides by 25 yields: x2ex=1625x^2 e^{x} = \frac{16}{25} Taking square roots results in: xex/2=45x e^{x/2} = \frac{4}{5} Thus, we can express it as: x=45ex.x = \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

Substituting x0=0.5x_0 = 0.5 into the iteration formula:

  1. Calculate x1x_1: x1=45e0.50.485.x_1 = \frac{4}{5} e^{0.5} \approx 0.485.

  2. Calculate x2x_2: x2=45e0.4850.492.x_2 = \frac{4}{5} e^{0.485} \approx 0.492.

  3. Calculate x3x_3: x3=45e0.4920.489.x_3 = \frac{4}{5} e^{0.492} \approx 0.489.

Thus, we have:

  • x10.485x_1 \approx 0.485
  • x20.492x_2 \approx 0.492
  • x30.489x_3 \approx 0.489.

Step 4

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

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Answer

By observing the values calculated previously:

  • From the iterations, we note that x20.492x_2 \approx 0.492 and x30.489x_3 \approx 0.489 are converging.
  • The root appears close to 0.490.49.

Thus, the accurate estimate for α\alpha to two decimal places is: α0.49.\alpha \approx 0.49. Justification can be given since x3x_3 is stable around this value.

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