Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 12 - Sequences and Series - 12.6 Newton's Method - 12.6 Exercises - Page 652: 8

Answer

$x=-2.363$ is a reasonably accurate solution of $f(x)=3x^4+4x^3-6x^2-2x-12$ in the interval $[-3,-2]$ $x=1.496$ is a reasonably accurate solution of $f(x)=3x^4+4x^3-6x^2-2x-12$ in the interval $[1,2]$

Work Step by Step

We are given $3x^4+4x^3-6x^2-2x-12=0$ which is also $f(x)=3x^4+4x^3-6x^2-2x-12$ $f'(x)=12x^3+12x^2-12x-2$ Since $f(-3)$ and $f(-2)$ have opposite signs, there is a solution for the equation in the interval $[-3,-2]$ As an initial guess, let $c_1=-3$ $c_2$ can be found as follows: $c_2=c_1-\frac{f(c_1)}{f'(c_1)} \approx -2.5879$ In the same way, $c_3=-2.403$ $c_4=-2.3648$ $c_5=-2.363$ $x=-2.363$ is a reasonably accurate solution of $f(x)=3x^4+4x^3-6x^2-2x-12$ in the interval $[-3,-2]$ Since $f(1)$ and $f(2)$ have opposite signs, there is a solution for the equation in the interval $[1,2]$ As an initial guess, let $c_1=1$ $c_2$ can be found as follows: $c_2=c_1-\frac{f(c_1)}{f'(c_1)} = 2.3$ In the same way, $c_3=1.831$ $c_4=1.581$ $c_5=1.5032$ $c_6=1.496$ $c_7=1.496$ $x=1.496$ is a reasonably accurate solution of $f(x)=3x^4+4x^3-6x^2-2x-12$ in the interval $[1,2]$
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