Answered: The real root of the equation x³ =…

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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Question

The real root of the equation x³ = √(x+1) to at least four decimal places by Newton's method and compare this root with the root obtained by calculator.

Expert Solution

Step 1

$g i v e n x^{3} = \sqrt{x + 1} t a k i n g s q u a r e o n b o t h s i d e x^{6} = x + 1 x^{6} - x - 1 = 0 f (x) = x^{6} - x - 1 p u t x = 0 t h e n f (0) = 0^{6} - 0 - 1 = - 1 p u t x = 1 t h e n f (1) = 1^{6} - 1 - 1 = 1 - 1 - 1 = - 1 < 0 f (2) = 2^{6} - 2 - 1 = 64 - 3 = 61 > 0 t h e r e f o r e r o o t l i e s i n b e t w e e n (1, 2) p u t x = 1.5 t h e n f (1.5) = {(1.5)}^{6} - (1.5) - 1 = 11.390625 - 1.5 - 1 f (1.5) = 8.890625 > 0 t h e r e f o r e r o o t l i e s b e t w e e n (1 t o 1.5) p u t x = 1.25 t h e n f (1.25) = {(1.25)}^{6} - (1.25) - 1 = 1.5647 t h e r e f o r e r o o t l i e s i n (1, 1.25) p u t x = 1.2 f (1.2) = {(1.2)}^{6} - 1.2 - 1 = 0.785984 > 0 t h e r e f o r e r o o t l i e s b e t w e e n (1, 1.2) p u t x = (1.1) f (1.1) = {(1.1)}^{6} - 1.1 - 1 = - 0.328439 < t h e r e f o r e r o o t l i e s b e t w e e n (1.1, 1.2) l e t x_{0} = 1.1 w e k n o w t h a t f (x_{n + 1}) = x_{n} - \frac{f (x_{n})}{f' (x_{n})} x_{1} = x_{0} - \frac{f (x_{0})}{f' (x_{0})} h e r e f' (x) = 6 x^{5} - 1 f' (1.1) = 6 {(1.1)}^{5} - 1 = 9.629366 f (1.1) = - 0.328439 x_{1} = 1.1 - \frac{- 0.328439}{9.629366} x_{1} = 1.1 + 0.03410806 = 1.134108061 x_{2} = x_{1} + \frac{f (x_{1})}{f' (x_{1})} x_{2} = 1.1341 - \frac{f (1.1341)}{f' (1.1341)} x_{2} = 1.1341 - \frac{- 0.00641}{10.2566} x_{2} = 1.1341 + 0.0006250 x_{2} = 1.1347$