# The roots of the curve, y = ( x − 1 ) 2 . ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 4.7, Problem 31E
To determine

## To calculate: The roots of the curve,  y=(x−1)2 .

Expert Solution

The roots are

x10.4

x20.410126

x30.410245

### Explanation of Solution

Given information:

The parabola y=(x1)2 that is closest to the origin .

Formula used:

Newton’s Method:

We seek a solution of f(x)=0 , starting from an initial estimate x=x1 .

For x=xn , compute the next approximation xn+1 by

xn+1=xnf(xn)f'(xn) and so on.

Distance Formula :- D=(x2x1)2+(y2y1)2

Calculation:

Consider the parabola ,

y=(x1)2

Let the point on the parabola is (x,(x1)2) .

The distance formula from point (x,(x1)2) to origin (0,0) is

D=(x0)2+((x1)20)2D=(x)2+((x1)2)2D=x2+(x1)4

The minimum distance without the square root will be same as the minimum distance with square root.

=D=x2+(x1)4=D'=2x+4(x1)3

Therefore, g(x)=2x+4(x1)4 used to find out roots using Newton’s Method.

g(x)=2x+4(x1)3g'(x)=2+12(x1)2

Now, let initial approximation be x1=0.4

For n=1

x2=x1g(x1)g'(x1)

x2=x12x1+4(x11)32+12(x11)2

x2=(0.4)2(0.4)+4(0.41)32+12(0.41)2x2=(0.4)0.80.8642+4.32

x2=0.4(0.064)6.32x2=0.4+0.010126582x2=0.410126582

The second approximation is x2=0.410126 .

For n=2

x3=x2g(x2)g'(x2)

x3=x22x2+4(x21)32+12(x21)2

x3=(0.410126)2(0.410126)+4(0.4101261)32+12(0.4101261)2x3=(0.410126)0.8202520.8209897852+4.175416031

x3=0.410126(0.000737785)6.175416031x3=0.410126+0.000119471303x3=0.410245471

The third approximation is x3=0.410245 .

Hence , the roots of the curve are :-

x10.4

x20.410126

x30.410245

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