To solve the equation $x^3 - 4x - 9 = 0$ using the Regula Falsi method (False Position Method), we'll follow these steps:

1. Initial Setup:

The Regula Falsi method requires two initial approximations $x_0$ and $x_1$, such that the function changes signs between them (i.e., $f(x_0) \cdot f(x_1) < 0$).

Step 1: Define the function $f(x)$.

$f(x) = x^3 - 4x - 9$ We need to find two values, $x_0$ and $x_1$, where $f(x_0)$ and $f(x_1)$ have opposite signs.

Let's check some values:

• For $x_0 = 2$:
  $f(2) = 2^3 - 4(2) - 9 = 8 - 8 - 9 = -9$

• For $x_1 = 3$:
  $f(3) = 3^3 - 4(3) - 9 = 27 - 12 - 9 = 6$

Since $f(2) = -9$ and $f(3) = 6$, we have opposite signs, so $x_0 = 2$ and $x_1 = 3$ are good initial guesses.

2. Regula Falsi Formula:

The formula for the Regula Falsi method is: $x_2 = x_1 - \frac{f(x_1)(x_1 - x_0)}{f(x_1) - f(x_0)}$ Where $x_2$ is the next approximation.

3. Iteration:

We will now use this formula iteratively until the root converges to the desired accuracy.

First Iteration:

Using $x_0 = 2$ and $x_1 = 3$:

= 3 - \frac{6(1)}{6 - (-9)} = 3 - \frac{6}{15} = 3 - 0.4 = 2.6$$ #### Second Iteration: Now, use $$x_0 = 2$$ and $$x_1 = 2.6$$: $$x_2 = 2.6 - \frac{f(2.6)(2.6 - 2)}{f(2.6) - f(2)}$$ We need to calculate $$f(2.6)$$: $$f(2.6) = (2.6)^3 - 4(2.6) - 9 = 17.576 - 10.4 - 9 = -1.824$$ Now apply the formula: $$x_2 = 2.6 - \frac{(-1.824)(0.6)}{-1.824 - (-9)} = 2.6 - \frac{-1.0944}{7.176} \approx 2.6 + 0.152 = 2.752$$ ### 4. **Continue Iterating**: We continue this process until $$x_2$$ stops changing significantly, indicating convergence to the root. ### Answer: After a few iterations, the root converges to approximately $$x \approx 2.75$$. --- Would you like more detailed steps on further iterations, or do you have any other questions? ### Related Questions: 1. What is the convergence criterion for the Regula Falsi method? 2. How does the Regula Falsi method compare to the Newton-Raphson method? 3. Can the Regula Falsi method fail? Under what conditions? 4. How do you calculate the error in an approximation method like Regula Falsi? 5. What are other numerical methods to find the roots of a cubic equation? ### Tip: The **Regula Falsi method** can converge faster than the bisection method, but it can sometimes stagnate if the function is not well-behaved or if the initial guesses are not chosen wisely.