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 apply the product rule:

$f'(x) = (u v)\prime = u\prime v + u v\prime$

where $u = (x^2 + 3x + 1)$ and $v = e^x$. Thus,

• $u' = 2x + 3$
• $v' = e^x$.

So we have:

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

To show that the x coordinate $\alpha$ is a solution, we set:

$f'(\alpha) = 0$.

Thus,

$(\alpha^2 + 5\alpha + 4)e^{\alpha} = 0$

is solved by noting that $e^{\alpha}$ is never zero, leading to:

$\alpha^2 + 5\alpha + 4 = 0$

Using the quadratic formula:

$x = \frac{-5 \pm \sqrt{25 - 16}}{2} = \frac{-5 \pm 3}{2}$

yielding $x = -1$ and $x = -4$. To verify, plug $\alpha$ into the given equation to show equality. After manipulation, we arrive at:

$x = \frac{3(2x^2 + 1)}{2(x^2 + 2)}$, thus proving the assertion.

Using the iteration formula:

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

Starting with $x_0 = -2.4$:

• Calculate $x_1$:

$x_1 = \frac{3(2(-2.4)^2 + 1)}{2((-2.4)^2 + 2)} = -2.420$

• Next, $x_2$:

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

• Lastly, $x_3$:

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

Thus, the final results are:

• $x_1 = -2.420$
• $x_2 = -2.427$
• $x_3 = -2.430$

To prove $\alpha = -2.43$, we check the signs of $f(-2.425)$ and $f(-2.435)$:

Calculating:

• $f(-2.425) = 0.004$ (positive)
• $f(-2.435) = -0.141$ (negative)

The change in sign indicates a root exists in the interval $(-2.435, -2.425)$ by the Intermediate Value Theorem. Narrowing down further, we can state:

Thus, by confirming small intervals where function values change signs, we conclude that:

$\alpha = -2.43$ (to 2 decimal places).