Find the positive real root of x³ - x - 3 = 0 using Newton-Raphson method. If the starting guess (x₀) is 2, the numerical value of the root after two iterations (x₂) is __________ (round off to two decimal places).

Answer :

Correct answer is : 1.67

f(x) = x³ - x - 3

$f^{\prime }\left(x\right)=3x^2-1$

Starting guess (x₀ = 2)

Now first iterations

$x_1=x_0-\frac{(x_0^3-x_0-3)}{3x_0^2-1}$

$x_1=x_0-\frac{f\left(x_0\right)}{f^{\prime }\left(x_0\right)}=2-\frac{{\left(2\right)}^3-2-3}{3{\left(2\right)}^2-1}=1.7273$

Second iterations

$x_2=x_1-\frac{(x_1^3-x_1-3)}{3x_1^2-1}$

x₁ = 1.7273

$x_2=x_1-\frac{f\left(x_1\right)}{f^{\prime }\left(x_1\right)}=1.7273-\frac{{\left(1.7273\right)}^3-1.7273-3}{3{\left(1.7273\right)}^2-1}=1.67369$

x₂ = 1.67