Question

Consider the function $f(x) = x^3 - x - 1$ for finding the root the equation $f(x) = 0$. If the initial approximation is $x_0 = 1$, then the next approximation $x_1$ using the Newton-Raphson method is

• A. 1.25
• B. 1.75
• C. 1.5
• D. 2

Hint:

The Newton-Raphson method converges quadratically when the initial guess is close to the actual root and the derivative $f'(x_0) \neq 0$.
Always double check the calculation of $f'(x)$ before substituting.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This question asks for the first iterative approximation $x_1$ of the root of the equation $f(x) = x^3 - x - 1 = 0$ using the Newton-Raphson numerical method, starting from an initial guess $x_0 = 1$.

Step 2: Key Formula or Approach:
The iterative formula for the Newton-Raphson method is:
\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] For the first iteration ($n = 0$), the formula is:
\[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} \]

Step 3: Detailed Explanation:

• Define the function and its first derivative:
\[ f(x) = x^3 - x - 1 \] \[ f'(x) = 3x^2 - 1 \]
• Evaluate the function at the initial approximation $x_0 = 1$:
\[ f(1) = (1)^3 - (1) - 1 = 1 - 1 - 1 = -1 \]
• Evaluate the derivative at the initial approximation $x_0 = 1$:
\[ f'(1) = 3(1)^2 - 1 = 3 - 1 = 2 \]
• Substitute these values into the Newton-Raphson formula to find $x_1$:
\[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1 - \frac{-1}{2} \] \[ x_1 = 1 + \frac{1}{2} = 1.5 \]

Step 4: Final Answer:
The next approximation $x_1$ is $1.5$, which corresponds to option (C).