Question

Find the positive real root of \( x^3 - x - 3 = 0 \) using Newton-Raphson method. If the starting guess \( x_0 \) is 2, the numerical value of the root after two iterations \( (x_2) \) is \(\underline{\hspace{1cm}}\) (round off to two decimal places).

Hint:

Newton-Raphson method is an iterative method for finding roots of real-valued functions. The accuracy increases with each iteration.

Answer 1

Correct Answer: 1.66 - 1.68

To find the positive real root of the equation \( x^3 - x - 3 = 0 \) using the Newton-Raphson method, we begin by defining:

The function \( f(x) = x^3 - x - 3 \) and its derivative \( f'(x) = 3x^2 - 1 \).

Starting with an initial guess \( x_0 = 2 \), the Newton-Raphson iteration formula is given by:

\( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \).

Step 1:

Calculate \( x_1 \):

• \( f(2) = 2^3 - 2 - 3 = 3 \).
• \( f'(2) = 3(2)^2 - 1 = 11 \).
• \( x_1 = 2 - \frac{3}{11} \approx 1.7273 \).

Step 2:

Calculate \( x_2 \):

• \( f(1.7273) = (1.7273)^3 - 1.7273 - 3 \approx -0.4438 \).
• \( f'(1.7273) = 3(1.7273)^2 - 1 \approx 7.9486 \).
• \( x_2 = 1.7273 - \frac{-0.4438}{7.9486} \approx 1.67\).

Rounding \( x_2 \) to two decimal places, we get \( \underline{1.67} \).