Answered: yA y = fr) 0.5 0.5- 0.5 I 1.5 12 -0.5

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

Let ƒ(x) = 2x³ - 6x² + 4x. Use Newton’s method to find x₁
given that x₀ = 1.4. Use the graph of ƒ (see figure) and an appropriate
tangent line to illustrate how x₁ is obtained from x₀.

Expert Solution

Step 1

Consider the following function:

$f (x) = 2 x^{3} - 6 x^{2} + 4 x$

Differentiate the above function with respect to x:

$f' (x) = 6 x^{2} - 12 x + 4$

Consider the following Newton's formula:

$\begin{array}{rcl} x_{1} & = & x_{0} - \frac{f (x_{0})}{f' (x_{0})} \\ = & 1.4 - \frac{2 {(1.4)}^{3} - 6 {(1.4)}^{2} + 4 (1.4)}{6 {(1.4)}^{2} - 12 (1.4) + 4} \\ = & 1.4 - \frac{2 \times 2.744 - 6 \times 1.96 + 5.6}{6 \times 1.96 - 12 \times 1.4 + 4} \\ = & 1.4 - \frac{5.488 - 11.76 + 5.6}{11.76 - 16.8 + 4} \\ = & 1.4 - (\frac{- 0.672}{- 1.04}) \end{array}$

$\begin{array}{rcl} x_{1} & = & 1.4 - 0.646 \\ = & 0.754 \end{array}$

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