Use Newton's method to approximate a root of the equation 3x³ + 6x + 2 = 0 as follows. Let x₁ = -2 be the initial approximation. The second approximation 2 is and the third approximation 3 is (Although these are approximations of the root, enter exact expressions for each approximation.)
Use Newton's method to approximate a root of the equation 3x³ + 6x + 2 = 0 as follows. Let x₁ = -2 be the initial approximation. The second approximation 2 is and the third approximation 3 is (Although these are approximations of the root, enter exact expressions for each approximation.)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Newton's Method for Approximating Roots**
In this example, we will use Newton's method to approximate a root of the equation \(3x^3 + 6x + 2 = 0\).
**Initial Setup:**
- Let \( x_1 = -2 \) be the initial approximation.
**Approximations:**
- The second approximation \( x_2 \) is \(\boxed{\phantom{answer}}\).
- The third approximation \( x_3 \) is \(\boxed{\phantom{answer}}\).
*(Note: Although these are approximations of the root, enter exact expressions for each approximation.)*
**Explanation:**
Newton's method is an iterative process to find successively better approximations to the roots of a real-valued function. The formula used is:
\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]
Where:
- \( f(x) = 3x^3 + 6x + 2 \)
- \( f'(x) \) is the derivative of \( f(x) \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9978ffa8-8e6a-4550-8363-e044b6a6e895%2F900bac61-c86c-4a37-b993-b22dff317975%2Fd2jjh1i_processed.png&w=3840&q=75)
Transcribed Image Text:**Newton's Method for Approximating Roots**
In this example, we will use Newton's method to approximate a root of the equation \(3x^3 + 6x + 2 = 0\).
**Initial Setup:**
- Let \( x_1 = -2 \) be the initial approximation.
**Approximations:**
- The second approximation \( x_2 \) is \(\boxed{\phantom{answer}}\).
- The third approximation \( x_3 \) is \(\boxed{\phantom{answer}}\).
*(Note: Although these are approximations of the root, enter exact expressions for each approximation.)*
**Explanation:**
Newton's method is an iterative process to find successively better approximations to the roots of a real-valued function. The formula used is:
\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]
Where:
- \( f(x) = 3x^3 + 6x + 2 \)
- \( f'(x) \) is the derivative of \( f(x) \).
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