Use the false position method to approximate the roots of f(x) = 2x³ - 5 at the starting interval [1,2].

The false position method (also known as the regula falsi method) is a root-finding algorithm that approximates the root of a function \( f(x) \) by iteratively refining an interval \([a, b]\) within which the root lies. Here's the step-by-step process for using the false position method to approximate the roots of \( f(x) = 2x^3 - 5 \) in the interval \([1, 2]\):

### Step-by-Step Process

1. **Define the function:**
   \[
   f(x) = 2x^3 - 5
   \]

2. **Initial interval:**
   \([a, b] = [1, 2]\)

3. **Check function values at the endpoints:**
   \[
   f(1) = 2(1)^3 - 5 = 2 - 5 = -3
   \]
   \[
   f(2) = 2(2)^3 - 5 = 16 - 5 = 11
   \]
   Since \( f(1) \) and \( f(2) \) have opposite signs, there is at least one root in the interval \([1, 2]\).

4. **Calculate the false position (x-intercept of the secant line):**
   \[
   c = b - \frac{f(b)(b - a)}{f(b) - f(a)}
   \]
   Plugging in the values:
   \[
   c = 2 - \frac{11(2 - 1)}{11 - (-3)} = 2 - \frac{11 \cdot 1}{11 + 3} = 2 - \frac{11}{14} = 2 - 0.7857 \approx 1.2143
   \]

5. **Evaluate the function at \( c \):**
   \[
   f(1.2143) = 2(1.2143)^3 - 5 \approx 2(1.7931) - 5 \approx 3.5862 - 5 = -1.4138
   \]

6. **Update the interval:**
   Since \( f(1.2143) \) and \( f(2) \) have opposite signs, we update the interval to \([1.2143, 2]\).

7. **Repeat the process:**
   - Calculate the new false position:
     \[
     c = 2 - \frac{11(2 - 1.2143)}{11 - (-1.4138)} = 2 - \frac{11 \cdot 0.7857}{12.4138} \approx 2 - 0.6967 \approx 1.3033
     \]
   - Evaluate the function at \( c \):
     \[
     f(1.3033) = 2(1.3033)^3 - 5 \approx 2(2.2164) - 5 \approx 4.4328 - 5 = -0.5672
     \]
   - Update the interval to \([1.3033, 2]\) since \( f(1.3033) \) and \( f(2) \) have opposite signs.

8. **Continue iterating:**
   Each iteration refines the interval and gets closer to the root. The process is repeated until \( f(c) \) is sufficiently close to zero or the interval \([a, b]\) is sufficiently small.

### Summary of Iterations
1. Initial interval: \([1, 2]\)
2. First approximation: \( c \approx 1.2143 \)
3. Second approximation: \( c \approx 1.3033 \)

### Stopping Criterion
You can stop the iterations when the difference between \( b \) and \( a \) is very small (e.g., \( |b - a| < \epsilon \)) or when \( |f(c)| \) is very small (e.g., \( |f(c)| < \epsilon \)) for a predefined tolerance \(\epsilon\).

Would you like more details or have any questions?

---

Here are 8 relative questions to expand your knowledge:

1. What are the main advantages of the false position method over the bisection method?
2. How does the convergence rate of the false position method compare to Newton's method?
3. Can the false position method fail to converge? If so, under what conditions?
4. What are some real-world applications where the false position method is particularly useful?
5. How would you modify the false position method to handle functions with multiple roots?
6. What are the limitations of using the false position method on non-continuous functions?
7. How can you implement the false position method algorithmically in a programming language?
8. What are some ways to improve the efficiency of the false position method?

### Tip:
Always verify that the function values at the initial interval endpoints have opposite signs; otherwise, the false position method may not work correctly.

Learn how to use the false position method to approximate roots of the polynomial function 2x³ - 5 within the interval [1, 2]. This detailed guide provides step-by-step instructions, formula applications, and iterative refinements. Suitable for advanced high school students and beyond.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation