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The problem provides the equation \( 2x^3 + x^2 - 1 = 0 \) and its corresponding Newton-Raphson iterative formula:

\[
x_{n+1} = \frac{4x_n^3 + x_n^2 + 1}{6x_n^2 + 2x_n}
\]

You are asked to explain why the Newton-Raphson method cannot be used with \( x_1 = 0 \).

To understand why, let's examine the Newton-Raphson iteration formula:

\[
x_{n+1} = \frac{4x_n^3 + x_n^2 + 1}{6x_n^2 + 2x_n}
\]

If \( x_1 = 0 \), then substituting \( x_n = 0 \) into the formula gives:

\[
x_{n+1} = \frac{4(0)^3 + (0)^2 + 1}{6(0)^2 + 2(0)} = \frac{1}{0}
\]

Here, the denominator becomes zero, which results in an undefined expression. This is the main reason why the Newton-Raphson method cannot be used with \( x_1 = 0 \), as the formula would involve division by zero, making the method inapplicable.

Would you like more details or have any further questions?

### Related Questions:
1. What is the Newton-Raphson method, and how is it generally applied?
2. Can the Newton-Raphson method converge for all initial guesses?
3. What happens if the derivative in the Newton-Raphson method is zero or close to zero?
4. How do you choose a good initial guess for the Newton-Raphson method?
5. What other numerical methods can be used if the Newton-Raphson method fails?

### Tip:
Always check the derivative's value at your initial guess when applying the Newton-Raphson method to ensure it doesn't lead to division by zero or other computational issues.

The equation 2x^3 + x^2 - 1 = 0 has exactly one real root. For this equation, the Newton-Raphson formula can be written as: x_(n+1) = (4x_n^3 + x_n^2 + 1) / (6x_n^2 + 2x_n). Explain why, for this question, the Newton-Raphson method cannot be used with x_1 = 0.

Learn why the Newton-Raphson method cannot be applied to the equation 2x^3 + x^2 - 1 = 0 with an initial guess of x1 = 0. The problem explores division by zero in the iteration formula and discusses key concepts in calculus and numerical methods. Suitable for students in Grades 11-12.

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