To solve this problem, let's use Newton's Method to approximate the roots of the equation:

$9x^2 + 4x - 1 = 0$

Newton's Method formula is: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ where $f(x) = 9x^2 + 4x - 1$, and we need to compute $f'(x)$, the derivative of $f(x)$.

Step 1: Find $f(x)$ and $f'(x)$

1. $f(x) = 9x^2 + 4x - 1$
2. $f'(x) = 18x + 4$ (by differentiating $f(x)$ with respect to $x$).

Step 2: Use Newton’s Method for Each Initial Value

Case 1: $x_0 = -1$

1. Calculate $f(x_0)$ and $f'(x_0)$:

   • $f(-1) = 9(-1)^2 + 4(-1) - 1 = 9 - 4 - 1 = 4$
   • $f'(-1) = 18(-1) + 4 = -18 + 4 = -14$

2. Calculate $x_1$: $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = -1 - \frac{4}{-14} = -1 + \frac{2}{7} = -\frac{5}{7} \approx -0.7143$

3. Calculate $f(x_1)$ and $f'(x_1)$:

   • $f\left(-\frac{5}{7}\right) = 9\left(-\frac{5}{7}\right)^2 + 4\left(-\frac{5}{7}\right) - 1 \approx 1.3061$
   • $f'\left(-\frac{5}{7}\right) = 18\left(-\frac{5}{7}\right) + 4 \approx -8.8571$

4. Calculate $x_2$: $x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} \approx -0.7143 - \frac{1.3061}{-8.8571} \approx -0.5679$

So, for $x_0 = -1$, we get $x_2 \approx -0.5679$.

Case 2: $x_0 = 1$

1. Calculate $f(x_0)$ and $f'(x_0)$:

   • $f(1) = 9(1)^2 + 4(1) - 1 = 9 + 4 - 1 = 12$
   • $f'(1) = 18(1) + 4 = 18 + 4 = 22$

2. Calculate $x_1$: $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1 - \frac{12}{22} = 1 - \frac{6}{11} = \frac{5}{11} \approx 0.4545$

3. Calculate $f(x_1)$ and $f'(x_1)$:

   • $f\left(\frac{5}{11}\right) = 9\left(\frac{5}{11}\right)^2 + 4\left(\frac{5}{11}\right) - 1 \approx 3.0992$
   • $f'\left(\frac{5}{11}\right) = 18\left(\frac{5}{11}\right) + 4 \approx 11.1818$

4. Calculate $x_2$: $x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} \approx 0.4545 - \frac{3.0992}{11.1818} \approx 0.1780$

So, for $x_0 = 1$, we get $x_2 \approx 0.1780$.

Summary

• For $x_0 = -1$: $x_2 \approx -0.5679$
• For $x_0 = 1$: $x_2 \approx 0.1780$

Would you like further clarification on any part of this solution?

Related Questions

1. How does Newton's Method converge to a solution?
2. Can Newton's Method fail to converge? If so, under what conditions?
3. How do initial guesses affect the convergence in Newton’s Method?
4. What is the geometric interpretation of Newton's Method?
5. How can we estimate the error in Newton’s Method at each iteration?

Tip

Always check the derivative $f'(x)$ in Newton's Method. If $f'(x)$ is close to zero, Newton's Method may fail to converge or give inaccurate results.