Figure 2 shows a sketch of part of the curve with equation $y = f(x)$ where $f(x) = (x^2 + 3x + 1)e^x$ The curve cuts the x-axis at points A and B as shown in Figure 2. (a) Calculate the x coordinate of A and the x coordinate of B, giving your answers to 3 decimal places. (b) Find $f'(x)$. The curve has a minimum turning point at the point P as shown in Figure 2. (c) Show that the x coordinate of P is the solution of $x = \frac{3(2x^2 + 1)}{2(x^2 + 2)}$ to calculate the values of $x_1$, $x_2$, and $x_3$, giving your answers to 3 decimal places. The x coordinate of P is $a$. (e) By choosing a suitable interval, prove that $a = -2.43$ to 2 decimal places.

A-Level Edexcel Maths Pure Modelling with Functions: Figure 2 shows a sketch of part

Figure 2 shows a sketch of part of the curve with equation $y = f(x)$ where $f(x) = (x^2 + 3x + 1)e^x$ The curve cuts the x-axis at points A and B as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 8

Question 7

Figure-2-shows-a-sketch-of-part-of-the-curve-with-equation-$y-=-f(x)$-where--$f(x)-=-(x^2-+-3x-+-1)e^x$--The-curve-cuts-the-x-axis-at-points-A-and-B-as-shown-in-Figure-2-Edexcel-A-Level Maths Pure-Question 7-2013-Paper 8.png

Figure 2 shows a sketch of part of the curve with equation $y = f(x)$ where $f(x) = (x^2 + 3x + 1)e^x$ The curve cuts the x-axis at points A and B as shown in Figu... show full transcript

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve with equation $y = f(x)$ where $f(x) = (x^2 + 3x + 1)e^x$ The curve cuts the x-axis at points A and B as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 8

Step 1

Calculate the x coordinate of A and the x coordinate of B, giving your answers to 3 decimal places.

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Answer

To find the x-coordinates where the curve cuts the x-axis, we solve the equation:

$f(x) = (x^2 + 3x + 1)e^x = 0$

Since $e^x$ is never zero, we focus on solving:

$x^2 + 3x + 1 = 0$

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-3 \pm \sqrt{5}}{2}$

Calculating the values:

$x_A = \frac{-3 + \sqrt{5}}{2} \approx -0.382$ $x_B = \frac{-3 - \sqrt{5}}{2} \approx -2.618$

Thus, the x coordinates of A and B are approximately $-0.382$ and $-2.618$ , respectively.

Step 2

Find $f'(x)$.

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Answer

To compute the derivative of the function:

$f'(x) = \frac{d}{dx}((x^2 + 3x + 1)e^x)$

Using the product rule:

$f'(x) = (x^2 + 3x + 1) e^x + e^x(2x + 3)$

This simplifies to:

$f'(x) = e^x((x^2 + 3x + 1) + (2x + 3)) = e^x(x^2 + 5x + 4)$

So, $f'(x) = e^x(x^2 + 5x + 4)$ .

Step 3

Show that the x coordinate of P is the solution of $x = \frac{3(2x^2 + 1)}{2(x^2 + 2)}$.

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Answer

To show this, set:

$x = \frac{3(2x^2 + 1)}{2(x^2 + 2)}$

We can rearrange this equation, multiplying by the denominator:

$2x(x^2 + 2) = 3(2x^2 + 1)$

Leading to:

$2x^3 + 4x = 6x^2 + 3$

Rearranging gives us:

$2x^3 - 6x^2 + 4x - 3 = 0$

which shows that the x-coordinate of P is indeed the solution.

Step 4

Use the iteration formula $x_{n+1} = \frac{3(2x_n^2 + 1)}{2(x_n^2 + 2)}$ to calculate the values of $x_1$, $x_2$, and $x_3$, giving your answers to 3 decimal places.

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Answer

Starting with $x_0 = -2.4$ :

Calculate $x_1$ : $x_1 = \frac{3(2(-2.4)^2 + 1)}{2((-2.4)^2 + 2)}\approx -2.420$
Calculate $x_2$ using $x_1$ : $x_2 = \frac{3(2(-2.420)^2 + 1)}{2((-2.420)^2 + 2)}\approx -2.427$
Finally, calculate $x_3$ : $x_3 = \frac{3(2(-2.427)^2 + 1)}{2((-2.427)^2 + 2)}\approx -2.430$

Thus, the values are approximately $x_1 = -2.420$ , $x_2 = -2.427$ , and $x_3 = -2.430$ .

Step 5

By choosing a suitable interval, prove that $a = -2.43$ to 2 decimal places.

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Answer

To prove that $a = -2.43$ , consider the values obtained from the iterations:

$f(-2.425) = f(-2.425) > 0$
$f(-2.435) = f(-2.435) < 0$

Since there is a sign change between $-2.425$ and $-2.435$ , by the Intermediate Value Theorem, there is at least one root in the interval $(-2.425, -2.435)$ .

Using $a = -2.43$ confirms being in the correct range with the function values verifying this interval, thus proving that $a = -2.43$ to 2 decimal places.

Figure 2 shows a sketch of part of the curve with equation $y = f(x)$ where $f(x) = (x^2 + 3x + 1)e^x$ The curve cuts the x-axis at points A and B as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 8

Worked Solution & Example Answer:Figure 2 shows a sketch of part of the curve with equation $y = f(x)$ where $f(x) = (x^2 + 3x + 1)e^x$ The curve cuts the x-axis at points A and B as shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 8

Calculate the x coordinate of A and the x coordinate of B, giving your answers to 3 decimal places.

Find $f'(x)$.

Show that the x coordinate of P is the solution of $x = \frac{3(2x^2 + 1)}{2(x^2 + 2)}$.

Use the iteration formula $x_{n+1} = \frac{3(2x_n^2 + 1)}{2(x_n^2 + 2)}$ to calculate the values of $x_1$, $x_2$, and $x_3$, giving your answers to 3 decimal places.

By choosing a suitable interval, prove that $a = -2.43$ to 2 decimal places.

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