Question

The value of the root nearest to the 2, after first iteration of the equation $x^{4} - x - 10 = 0$ by Newton-Raphson method is

• A. 2.321
• B. 2.125
• C. 1.983
• D. 1.871

Answer 1

Answer: (D)

1.871

Answer 2

Let $f(x) = x^{4} - x - 10$, then $f(1) = - 10$ and $f(2) = 4$

Thus roots lie in (1, 2). Also $|f(2)|\mspace{6mu} < \mspace{6mu}|f(1)|$, So take $x_{0} = 2$

Also $f'(x) = 4x^{3} - 1$

$f'(2) = 4(8) - 1 = 31$

$\therefore$ By Newton's rule, the first iteration,

$x_{1} = x_{0} - \frac{f(x_{0})}{f'(x_{0})} = 2 - \frac{4}{31} = 1.871$