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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

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BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

(a) Use Newton’s method to find the critical numbers of the function

f ( x ) = x 6 x 4 + 3 x 3 2 x

correct to six decimal places.

(b) Find the absolute minimum value of f correct to four decimal places.

(a)

To determine

To find: the critical numbers of f(x)=x6x4+3x32x correct to six decimal places.

Explanation

Formula used:

Newton’s method

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

Given:

f(x)=x6x4+3x32x

Calculation:

f(x)=x6x4+3x32x

Differentiate with respect to x

f(x)=6x54x3+9x22

Differentiate with respect to x

f(x)=30x412x2+18x

Find the critical numbers of the function f(x)=x6x4+3x32x by obtaining roots of f(x)=6x54x3+9x22 using Newton’s method.

Then by Newton’s method on f(x),

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

Hence,

xn+1=xn6xn54xn3+9xn2230xn412xn2+18xn

In Figure 1, the curves intersect at about 1.3.

x1=1.3

Substitute n=1 in, xn+1=xn6xn54xn3+9xn2230xn412xn2+18xn,

x2=x16x154x13+9x12230x1412x12+18x1

Substitute x1=1.3 in x2=x16x154x13+9x12230x1412x12+18x1,

x2=(1.3)6(1.3)54(1.3)3+9(1.3)2230(1.3)412(1.3)2+18(1.3)=1.293344

Substitute n=2 in,xn+1=xn6xn54xn3+9xn2230xn412xn2+18xn,

x3=x26x254x23+9x22230x2412x22+18x2

Substitute x2=1.293344  in x3=x26x254x23+9x22230x2412x22+18x2,

x3=(1.293344)6(1.293344)54(1.293344)3+9(1.293344)2230(1.293344)412(1.293344)2+18(1.293344)=1.293227

Substitute n=3 in, xn+1=xn6xn54xn3+9xn2230xn412xn2+18xn,

x4=x36x354x33+9x32230x3412x32+18x3

Substitute x3=1.293227 in x4=x36x354x33+9x32230x3412x32+18x3,

x4=(1.293227)6(1.293227)54(1.293227)3+9(1.293227)2230(1.293227)412(1.293227)2+18(1.293227)=1.293227

The 1st 8 decimal places repeat at the 5th iteration.

So one root is x=1.293227

In Figure 1, the curves intersect at about 0.5.

x1=0.5

Substitute n=1 in, xn+1=xn6xn54xn3+9xn2230xn412xn2+18xn,

x2=x16x154x13+9x12230x1412x12+18x1

Substitute x1=1.3 in x2=x16x154x13+9x12230x1412x12+18x1,

x2=(0.5)6(0.5)54(0.5)3+9(0.5)2230(0.5)412(0.5)2+18(0.5)=0.444444

Substitute n=2 in,xn+1=xn6xn54xn3+9xn2230xn412xn2+18xn,

x3=x26x254x23+9x22230x2412x22+18x2

Substitute x2=0

(b)

To determine

To find: The absolute minimum value of f(x) correct to four decimal places.

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