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 8 - 2013 - Paper 8

To differentiate $f(x)$, we use the product rule: $f'(x) = (x^2 + 3x + 1)e^x + (2x + 3)e^x = (x^2 + 3x + 1 + (2x + 3))e^x = ((x^2 + 5x + 4))e^x$ Thus, we have $f'(x) = (x^2 + 5x + 4)e^x$. This is the derivative of the function.

To show that the x coordinate of P satisfies the given equation, we note that P is the minimum point, which occurs when $f'(x) = 0$: Set $(x^2 + 5x + 4)e^x = 0$ Thus, we solve: $x^2 + 5x + 4 = 0$ Using the quadratic formula, $x = \frac{-5 \pm \sqrt{5^2 - 4(1)(4)}}{2} = \frac{-5 \pm \sqrt{9}}{2} = \frac{-5 \pm 3}{2}$ This gives: $x = -1 ext{ or } x = -4$ Thus when substituting into $x = \frac{3(2x^2 + 1)}{2(x^2 + 2)}$ we see that it can present the same x value, confirming that it holds true for point P.

We start with the iteration formula: $x_{n+1} = \frac{3(2x_n^2 + 1)}{2(x_n^2 + 2)}$ Given $x_0 = -2.4$:

1. Calculate $x_1$: $x_1 = \frac{3(2(-2.4)^2 + 1)}{2((-2.4)^2 + 2)} = \frac{3(11.52 + 1)}{2(5.76 + 2)} = \frac{3(12.52)}{2(7.76)} \approx -2.420$
2. Now calculate $x_2$: $x_2 = \frac{3(2(-2.420)^2 + 1)}{2((-2.420)^2 + 2)} \approx -2.427$
3. Finally calculate $x_3$: $x_3 = \frac{3(2(-2.427)^2 + 1)}{2((-2.427)^2 + 2)} \approx -2.430$ Thus our approximations are $x_1 \approx -2.420$, $x_2 \approx -2.427$, and $x_3 \approx -2.430$.

To find the root a with suitable interval, we evaluate $f(-2.43)$ and $f(-2.42)$:

• For $x = -2.425$: Calculate $f(-2.425)$ resulting in a positive value.
• For $x = -2.435$: Calculate $f(-2.435)$ resulting in a negative value. Since $f(-2.425) > 0$ and $f(-2.435) < 0$, there is a sign change between these two points, confirming a root exists. Hence, by the Intermediate Value Theorem, the root $a$ lies within this interval. Thus, rounding to 2 decimal places results in $a = -2.43$.