Question

Starting from $x_0=1$, one step of Newton-Raphson method in solving the equation $x^3+3x-7=0$ gives the next value $\left(x_1\right)$ as

• A. $x_1=0.5$
• B. $x_1=1.406$
• C. $x_1=1.5$
• D. $x_1=2$

Answer 1

The correct option is C $x_1=1.5$

Let $f\left(x\right)=x^3+3x-7$
$f^{\prime }\left(x\right)=3x^2+3$
Newton raphson iteration:
$x_{n+1}=x_n-\frac{x_n^3+3x_n-7}{3x_n^2+3}$
$x_{n+1}=\frac{2x_n^3+7}{3x_n^2+3}$
Given, $x_0=1$
$x_1=\frac{2\times 1^3+7}{3\times 1^2+3}=\frac{9}{6}=1.5$