f(x) = 3x^3 - 2x - 6 (a) Show that f(x) = 0 has a root, α, between x = 1.4 and x = 1.45 - Edexcel - A-Level Maths Pure - Question 1 - 2008 - Paper 5

We start from the equation f(x) = 0: $3x^3 - 2x - 6 = 0$ Rearranging the equation gives us: $3x^3 = 2x + 6$ Dividing both sides by 3 results in: $x^3 = \frac{2}{3}x + 2$ Now, solving for x: $x = \sqrt{\frac{2}{x} + \frac{2}{3}}$

1. Calculate x₁: $x_1 = \sqrt{\frac{2}{1.43} + \frac{2}{3}} = 1.4371$ (to 4 decimal places).

2. Calculate x₂: $x_2 = \sqrt{\frac{2}{x_1} + \frac{2}{3}} = \sqrt{\frac{2}{1.4371} + \frac{2}{3}} = 1.4374$ (to 4 decimal places).

3. Calculate x₃: $x_3 = \sqrt{\frac{2}{x_2} + \frac{2}{3}} = \sqrt{\frac{2}{1.4374} + \frac{2}{3}} = 1.4355$ (to 4 decimal places).

To show that α = 1.435 is correct to 3 decimal places, we can choose the interval (1.4345, 1.4355).

1. Calculate f(1.4345): $f(1.4345) ext{ results in a value slightly less than 0 (f(1.4345) = -0.01).}$

2. Calculate f(1.4355): $f(1.4355) ext{ results in a value slightly greater than 0 (f(1.4355) > 0).}$

Since f(1.4345) < 0 and f(1.4355) > 0, we see a change of sign in the interval (1.4345, 1.4355). Therefore, by the Intermediate Value Theorem, α = 1.435 is correct to 3 decimal places.