   Chapter 4.8, Problem 32E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# (a) Use Newton’s method with x1 = 1 to find the root of the equation x3 − x = 1 correct to six decimal places.(b) Solve the equation in part (a) using x1 = 0.6 as the initial approximation.(c) Solve the equation in part (a) using x1 = 0.57.(You definitely need a programmable calculator for this part.)(d) Graph f(x) = x3 − x − 1 and its tangent lines at x1 = 1, 0.6, and 0.57 to explain why Newton’s method is so sensitive to the value of the initial approximation.

(a)

To determine

To find: The root of the equation x3x=1 correct to six decimal places.

Explanation

Formula used:

Newton’s method

xn+1=xnf(xn)f(xn)

Given:

f(x)=x3x1 with x1=1

Calculation:

f(x)=x3x1

Differentiate with respect to x ,

f(x)=3x21

xn+1=xnf(xn)f(xn)=xnxn3xn13xn21

x1=1

Substitute n=1 in xn+1=xnxn3xn13xn21

x2=x1x13x113x121

Substitute x1=1 in x2=x1x13x113x121

x2=113113×121=112=1+12=1.5

Substitute n=2 in xn+1=xnxn3xn13xn21

x3=x2x23x213x221

Substitute x2=1.5 in x2=x1x13x113x121

x3=1.51.531.513×1.521=1

(b)

To determine

To find: The root of the equation x3x=1 correct to six decimal places.

(c)

To determine

To find: The root of the equation x3x=1 correct to six decimal places.

(d)

To determine

To sketch: the graph f(x)=x3x1 and its tangent lines at 1, 0.6, 0.57 to explain the dependency of initial approximation in Newton’s method.

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