Q3Advanced Engineering Mathematics I

Question

Using Newton-Raphson's method, find the root of $x^{4} - 12 x + 7 = 0$ which is near to $x = 2$ .

Answer 1

Given $f (x) = x^{4} - 12 x + 7 = 0$ . The derivative is $f^{'} (x) = 4 x^{3} - 12$ .

Newton-Raphson formula: $x_{n + 1} = x_{n} - \frac{f ( x _{n} )}{f ^{'} ( x _{n} )}$

Initial guess $x_{0} = 2$ . $f (2) = (2)^{4} - 12 (2) + 7 = 16 - 24 + 7 = - 1$ $f^{'} (2) = 4 (2)^{3} - 12 = 32 - 12 = 20$

$x_{1} = 2 - \frac{- 1}{20} = 2 + 0.05 = 2.05$ .

So the root near $x = 2$ is approximately $2.05$ .

Question