$g(x) = 2x^3 - 6x^2 + 3$

Show that $g (x)$ has a root between 0.8 and 0.9

Let $\alpha\,$ be the root of $g (x) = 0$ between 0.8 and 0.9. Show that $α=36−2α\displaystyle \alpha = \sqrt{\frac{3}{6-2\alpha}}$ .

Use the iteration formula $xn+1=36−2xn\displaystyle x_{n+1} = \sqrt{\frac{3}{6-2x_n}}$ with $x_0 = 0.8$ to find, to 3 decimal places, the values of $x_1$ , $x_2\,$ and $x_3$ .

By choosing a suitable interval, prove that $α=0.832\alpha = 0.832$ to 3 decimal places

$g(x) = 2x^3 - 6x^2 + 3$

Show that $g (x)$ has a root between 0.8 and 0.9

Let $\alpha\,$ be the root of $g (x) = 0$ between 0.8 and 0.9. Show that $α=36−2α\displaystyle \alpha = \sqrt{\frac{3}{6-2\alpha}}$ .

Use the iteration formula $xn+1=36−2xn\displaystyle x_{n+1} = \sqrt{\frac{3}{6-2x_n}}$ with $x_0 = 0.8$ to find, to 3 decimal places, the values of $x_1$ , $x_2\,$ and $x_3$ .

By choosing a suitable interval, prove that $α=0.832\alpha = 0.832$ to 3 decimal places

1 Pure Mathematics

Question 14

Markscheme

1 Pure Mathematics

Question 14

Markscheme

1 Pure Mathematics

Question 14

1 Pure Mathematics Questions

1 Pure Mathematics

Question 14

1 Pure Mathematics Questions