use newtons method to approximate a root of the equation $2x^3 + 2x + 4 = 0$ as follows.
let $x_1 = -1$ be the initial approximation.
the second approximation $x_2$ is ,
and the third approximation $x_3$ is .
(although these are approximations of the root, enter exact expressions for each approximation.)

Question

use newtons method to approximate a root of the equation $2x^3 + 2x + 4 = 0$ as follows.
let $x_1 = -1$ be the initial approximation.
the second approximation $x_2$ is  ,
and the third approximation $x_3$ is  .
(although these are approximations of the root, enter exact expressions for each approximation.)

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Newtons Method" ], "new_concepts": [], "current_concepts": [ "Newtons Method" ] } ## Define the function and its derivative $$ \begin{aligned} f(x) &= 2x^3 + 2x + 4 \ f'(x) &= 6x^2 + 2 \end{aligned} $$ ## Calculate the second approximation $x_2$ $$ \begin{aligned} x_1 &= -1 \ f(-1) &= 2(-1)^3 + 2(-1) + 4 = -2 - 2 + 4 = 0 \ f'(-1) &= 6(-1)^2 + 2 = 8 \ x_2 &= x_1 - \frac{f(x_1)}{f'(x_1)} = -1 - \frac{0}{8} = -1 \end{aligned} $$ ## Calculate the third approximation $x_3$ $$ \begin{aligned} x_2 &= -1 \ f(-1) &= 0 \ f'(-1) &= 8 \ x_3 &= x_2 - \frac{f(x_2)}{f'(x_2)} = -1 - \frac{0}{8} = -1 \end{aligned} $$ Use Newton's method to approximate a root of the equation $2x^3 + 2x + 4 = 0$ as follows. Let $x_1 = -1$ be the initial approximation. The second approximation $x_2$ is -1, and the third approximation $x_3$ is -1. (Although these are approximations of the root, enter exact expressions for each approximation.) { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Calculus", "Newtons Method" ] }

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Question

Explanation:

Define the function and its derivative

Calculate the second approximation \(x_2\)

Calculate the third approximation \(x_3\)

Answer: