To find a positive root of the equation $x^3 - 4x + 1 = 0$ using the Method of False Position (also known as the Regula Falsi Method), we follow these steps:

Steps:

1. Identify the interval $[a, b]$ where the root lies, such that $f(a)$ and $f(b)$ have opposite signs. This ensures that there is a root in the interval.

2. Apply the formula for the False Position Method to estimate the root: $c = b - \frac{f(b)(b - a)}{f(b) - f(a)}$ where $f(c)$ is the value of the function at $c$.

3. Check the sign of $f(c)$:

   • If $f(a) \times f(c) < 0$, then the root lies between $a$ and $c$, so update $b = c$.
   • If $f(b) \times f(c) < 0$, then the root lies between $c$ and $b$, so update $a = c$.

4. Repeat the process until the desired level of accuracy is achieved (i.e., until $|f(c)|$ is sufficiently small).

Step-by-Step Calculation:

1. Initial Interval: We need to find $a$ and $b$ such that $f(a) \times f(b) < 0$. Let's test a few values.

   • For $x = 0$: $f(0) = 0^3 - 4(0) + 1 = 1$
   • For $x = 1$: $f(1) = 1^3 - 4(1) + 1 = -2$
   • For $x = 2$: $f(2) = 2^3 - 4(2) + 1 = 1$

   The function changes sign between $x = 0$ and $x = 1$, so we choose the interval $[0, 1]$.

2. First Iteration:

   • $a = 0$, $b = 1$
   • $f(a) = 1$, $f(b) = -2$
   • Using the formula: $c = 1 - \frac{-2(1 - 0)}{-2 - 1} = 1 - \frac{-2}{-3} = 1 - \frac{2}{3} = \frac{1}{3}$
   • $f(c)$ for $c = \frac{1}{3}$: $f\left(\frac{1}{3}\right) = \left(\frac{1}{3}\right)^3 - 4\left(\frac{1}{3}\right) + 1 = \frac{1}{27} - \frac{4}{3} + 1 = \frac{1 + 27 - 108}{27} = \frac{-80}{27} \approx -2.96$ Since $f(c) < 0$, update the interval to $[0, \frac{1}{3}]$.

3. Second Iteration:

   • $a = 0$, $b = \frac{1}{3}$
   • $f(a) = 1$, $f(b) = -2.96$
   • Using the formula: $c = \frac{1}{3} - \frac{-2.96\left(\frac{1}{3} - 0\right)}{-2.96 - 1} = \dots \ (Iterate until convergence)$

Final Answer:

Iterate this process until you reach the required accuracy.

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Would you like me to complete more iterations, or explain anything further?

5 Related Questions:

1. How does the False Position method differ from the Bisection Method?
2. What is the convergence rate of the False Position method?
3. Can you apply the method to another cubic equation?
4. How can you improve the efficiency of the False Position method?
5. What are the advantages of using the False Position method over other methods?

Tip: Always double-check the function signs to ensure the interval has a root!