Find a root of (x) = x' - " , by Newton's method, within the accuracy of 0.1.You can use the following table to show your calculationsf' (#n)f(rn)f' (xm)f(rn)InInf (rn)f' (En)234

Consider the given function, fx=x3-e-xf'x=ddxx3-e-x=3x2-e-x-1=3x2+e-x Apply the Newton's method,…

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Find a root of (x) = x' - " , by Newton's method, within the accuracy of 0.1.

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)

6th Edition

ISBN:9781337111348

Author:Bruce Crauder, Benny Evans, Alan Noell

Publisher:Bruce Crauder, Benny Evans, Alan Noell

Chapter2: Graphical And Tabular Analysis

Section2.1: Tables And Trends

Problem 1TU: If a coffee filter is dropped, its velocity after t seconds is given by v(t)=4(10.0003t) feet per...

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Question

Find a root of (x) = x' - " , by Newton's method, within the accuracy of 0.1.
You can use the following table to show your calculations
f' (#n)
f(rn)
f' (xm)
f(rn)
In
In
f (rn)
f' (En)
2
3
4

Expert Solution

Step 1

Consider the given function,

$f (x) = x^{3} - e^{- x} f' (x) = \frac{d}{d x} [x^{3} - e^{- x}] = 3 x^{2} - e^{- x} (- 1) = 3 x^{2} + e^{- x}$

Apply the Newton's method,

$x_{n + 1} = x_{n} - \frac{f (x_{n})}{f' (x_{n})}$

Assume that $x_{0} = 1$ , then the first iteration will be:

$\begin{array}{rcl} x_{1} & = & x_{0} - \frac{f (x_{0})}{f' (x_{0})} \\ = & 1 - \frac{{(1)}^{3} - e^{- 1}}{3 {(1)}^{2} + e^{- 1}} \\ = & 0.8123 \end{array}$

Second iteration:

$\begin{array}{rcl} x_{2} & = & x_{1} - \frac{f (x_{1})}{f' (x_{1})} \\ = & 0.8123 - \frac{{(0.8123)}^{3} - e^{- 1}}{3 {(0.8123)}^{2} + e^{- 1}} \\ = & 0.74068 \end{array}$

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