   Chapter 4.5, Problem 2E ### 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

# Use the guidelines of this section to sketch the curve.y = 2 + 3x2 − x3

To determine

To Sketch: The curve by plotting x and y coordinates using guidelines.

Explanation

Given:

The Cartesian equation is as below.

y=2+3x2x3 (1)

Calculation:

Calculate the domain.

Refer equation (1), the function is defined for all real values of x (there is no restriction on the value of x).

Hence, the domain of f(x) is all real values of x, which is (,).

Calculate the intercepts.

Calculate the value of x-intercept.

Substitute 0 for y in the equation (1).

0=2+3x2x3

The function is written as below.

f(x)=2+3x2x3 (2)

Solve for value of x from equation (2).

Use the Newton Raphson method formula given below.

xn+1=xnf(xn)f'(xn) (3)

Substitute 0 for n in equation (3).

x0+1=x0f(x0)f'(x0)

x1=x0f(x0)f'(x0) (4)

Substitute 8 for x0 in equation (4).

x1=8f(8)f'(8)=82+3(8)2(8)36(8)3(8)2=8318144=5.791667

Calculate the value of Δx1.

Δx1=|x1x0|

Substitute 5.791667 for x1 and 8 for x0 in the above equation.

Δx2=|5.7916678|=2.2084

Substitute 1 for n in equation (3).

x1+1=x1f(x1)f'(x1)

x2=x1f(x1)f'(x1) (5)

Substitute 5.791667 for x1 in equation (5).

x2=5.791667f(5.791667)f'(5.791667)=5.7916672+3(5.791667)2(5.791667)36(5.791667)3(5.791667)2=5.79166791.64265.8802=5.7916671.391=4.40063

Calculate the value of Δx2.

Δx2=|x2x1|

Substitute 4.40063 for x2 and 5.791667 for x1 in the above equation.

Δx2=|4.400635.791667|=1.39104

Substitute 2 for n in the equation (3).

x2+1=x2f(x2)f'(x2)

x3=x2f(x2)f'(x2) (6)

Substitute 4.40063 for x2 in the equation (6).

x3=4.400632+3(4.40063)2(4.40063)36(4.40063)3(4.40063)2=5.79166725.128363.69279=5.7916670.79273=3.60790

Calculate the value of Δx3.

Δx3=|x3x2|

Substitute 3.26814 for x3 and 3.60790 for x2 in the above equation.

Δx3=|x3x2|=|3.268143.60790|=0.33976

Substitute 3 for n in equation (3).

x3+1=x3f(x3)f'(x3)

x4=x3f(x3)f'(x3) (7)

Substitute 3.60790 for x3 in equation (7).

x4=3.60790f(3.60790)f'(3.60790)=3.607902+3(3.60790)2(3.60790)36(3.60790)3(3.60790)2=3.607905.9129217.40336=3.607900.33975=3.26814

Calculate the value of Δx4.

Δx4=|x4x3|

Substitute 3.26814 for x4 and 3.60790 for x3 in the above equation.

Δx4=|3.268143.60790|=0.33976

Substitute 4 for n in equation (3).

x4+1=x4f(x4)f'(x4)

x5=x4f(x4)f'(x4) (8)

Substitute 3.26814 for x4 in the equation (8).

x5=3.26814f(3.26814)f'(3.26814)=3.268142+3(3.26814)2(3.26814)36(3.26814)3(3.26814)2=3.268140.8639112.43335=3.268140.06948=3.19865

Calculate the value of Δx5.

Δx5=|x5x4|

Substitute 3.19865 for x5 and 3.26814 for x4 in the above equation.

Δx5=|3.198653.26814|=0.33976

Substitute 5 for n in equation (3).

x5+1=x5f(x5)f'(x5)

x6=x5f(x5)f'(x5) (9)

Substitute 3.19865 for x5 in equation (9).

x6=3.19865f(3.19865)f'(3.19865)=3.198652+3(3.19865)2(3.19865)36(3.19865)3(3.19865)2=3.198650.0325211.50225=3.198650.002827=3.19583

Calculate the value of Δx6.

Δx5=|x5x4|

Substitute 3.19583 for x6 3.19865 for x5 in the above equation.

Δx6=|3.198533.19865|=0.00283

Substitute 6 for n in equation (3).

x6+1=x6f(x6)f'(x6)

x7=x6f(x6)f'(x6) (10)

Substitute 3.19583 for x6 in equation (10)

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