Question

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

• A. 0.5
• B. 1.415
• C. 1.5
• D. 1.9

Hint:

The Newton-Raphson method quickly converges to the root of an equation, and it is important to carefully apply the formula with correct derivatives at each iteration.

Answer 1

The Correct Option is C

To solve the problem using the Newton-Raphson method, we need to find the next value of \( x \) after starting from \( x_0 = 1 \) for the equation \( x^3 + 3x - 7 = 0 \).

1. Define the Function and Its Derivative:
Let \( f(x) = x^3 + 3x - 7 \).
The derivative of \( f(x) \) is \( f'(x) = 3x^2 + 3 \).

2. Newton-Raphson Formula:
The Newton-Raphson formula is given by:

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

3. Substitute \( x_0 = 1 \) into the Formula:
Calculate \( f(1) \) and \( f'(1) \):

\[ f(1) = 1^3 + 3(1) - 7 = 1 + 3 - 7 = -3 \]
\[ f'(1) = 3(1)^2 + 3 = 3 + 3 = 6 \]

4. Calculate the Next Value \( x_1 \):
Substitute \( f(1) \) and \( f'(1) \) into the Newton-Raphson formula:

\[ x_1 = 1 - \frac{-3}{6} = 1 + \frac{3}{6} = 1 + 0.5 = 1.5 \]

Final Answer:
The next value of \( x \) is \( {1.5} \).