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 2 - 2013 - Paper 8

To find the derivative $f'(x)$, we use the product rule:

$f'(x) = u'v + uv'$

where $u = x^2 + 3x + 1$ and $v = e^x$:

1. Differentiate $u$: $u' = 2x + 3$
2. Differentiate $v$: $v' = e^x$

Thus,

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

Factoring out $e^x$ gives:

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

To show this, we start by identifying point P where the minimum occurs, which requires setting $f'(x) = 0$:

Set:

$x^2 + 5x + 4 = 0$

Factoring gives:

$(x + 4)(x + 1) = 0$
Thus:

• $x = -4$
• $x = -1$

Next, we need to evaluate:

$f(-2.4) = (2.4^2 + 3(2.4) + 1)e^{-2.4}$

We then consider the iteration formula:

$x_{n+1} = \frac{3(2x_n^2 + 1)}{2(x_n^2 + 2)}$

and set up the iterations to demonstrate convergence towards the correct solution.

We start with $x_0 = -2.4$. Using the iteration formula:

1. Calculate $x_1$: $x_1 = \frac{3(2(-2.4)^2 + 1)}{2((-2.4)^2 + 2)} = \frac{3(2(5.76) + 1)}{2(5.76 + 2)} = \frac{3(11.52 + 1)}{2(7.76)} = \frac{3(12.52)}{15.52} \approx -2.420$

2. Calculate $x_2$: $x_2 = \frac{3(2(-2.420)^2 + 1)}{2((-2.420)^2 + 2)} \approx -2.427$

3. Calculate $x_3$: $x_3 = \frac{3(2(-2.427)^2 + 1)}{2((-2.427)^2 + 2)} \approx -2.430$

Thus, the approximate values are:

• $x_1 \approx -2.420$
• $x_2 \approx -2.427$
• $x_3 \approx -2.430$.

To prove that $a = -2.43$, we will examine the values of $f(x)$ around $x = -2.43$.

1. Calculate $f(-2.425)$ and $f(-2.435)$:

   • $f(-2.425) = (2.425^2 + 3(2.425) + 1)e^{-2.425}$
   • $f(-2.435) = (2.435^2 + 3(2.435) + 1)e^{-2.435}$

2. Check the signs of the results:

   • If $f(-2.425) > 0$ and $f(-2.435) < 0$, it indicates a root exists in the interval.

3. Therefore, as $f'(-2.43) = 0$ indicates a turning point, we conclude:

   • Thus, we verify that $a = -2.43$ to 2 decimal places.