Question

State true or false:

By the method of Newton-Raphson, the cube root of $10$ after the first iteration is $2.167$.

• A. True
• B. False

Answer 1

The correct option is A True

Let $f\left(x\right)=x^3-10$

$f\left(x\right)=3x^2$

Letting initial guess be $x_0=2$ ($2^3=8$ which is close to $10$)

We have

$x_1=2-\left[\frac{{\left(2\right)}^3-10}{3\times {\left(2\right)}^2}\right][\because x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime }\left(x_n\right)}]$

$x_1=2-\left[\frac{-2}{12}\right]$

$x_1=2+\frac{1}{6}$

$x_1=2+0.167$

$x_1=2.167$