Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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b. Solve the equation x = 32/8x^2 +5 ...... (check image)
![Note that the equation \(4x^3 + \frac{x}{2} - x - 16 = 0\) is equivalent to \(x = \frac{32}{8x^2 + 5}\). Therefore, the value satisfies \(x = \frac{32}{8x^2 + 5}\).
**b. Solve the equation \(x = \frac{32}{8x^2 + 5}\) with Newton's method. Start with \(x_0 = 1\) and give \(x_4\).**
\[x_4 = \boxed{\phantom{5}}\]
(Simplify your answer. Round the final answer to five decimal places as needed. Round all intermediate values to six decimal places.)
Enter your answer in the answer box and then click Check Answer.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F862d597a-ccd1-4202-ba0c-cb7527992470%2F9dfdf58a-eb80-450f-ac39-cef809c08eac%2Foyfiwig_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Note that the equation \(4x^3 + \frac{x}{2} - x - 16 = 0\) is equivalent to \(x = \frac{32}{8x^2 + 5}\). Therefore, the value satisfies \(x = \frac{32}{8x^2 + 5}\).
**b. Solve the equation \(x = \frac{32}{8x^2 + 5}\) with Newton's method. Start with \(x_0 = 1\) and give \(x_4\).**
\[x_4 = \boxed{\phantom{5}}\]
(Simplify your answer. Round the final answer to five decimal places as needed. Round all intermediate values to six decimal places.)
Enter your answer in the answer box and then click Check Answer.
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