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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

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Chapter
Section
BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

Use the guidelines of this section to sketch the curve.

y = x + cos x

To determine

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

Explanation

Given:

The Cartesian equation is as below.

y=f(x)=x+cosx (1)

Calculation:

a)

Calculate the domain.

From the 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 .

For the polynomial the domain is (,) .

b)

Calculate the intercepts.

Calculate the value of x -intercept.

Substitute 0 for y in the equation (1).

x+cosx=0

Solve for x from the equation (1).

Use the Newton Rapson method with trial and error procedure to find the value of x .

xn+1=xnf(xn)f'(xn) (2)

Differentiate equation (1) with respect to x .

f'(x)=1sinx

Substitute 0 for n in the equation (2).

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

x1=x0f(x0)f'(x0) (3)

Substitute 0 for x0 in the equation (3).

x1=0f(0)f'(0)=00+cos(0)1sin(0)=011=1

Calculate the value of increment in value of x Δx1 .

Δx1=|x1x0|

Substitute -1 for x1 and 0 for x0 in the above equation.

Δx1=|10|=1

Substitute 1 for n in the equation (2).

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

x2=x1f(x1)f'(x1) (4)

Substitute -1 for x1 in the equation (5).

x2=1f(1)f'(1)=11+cos(1)1sin(1)=11+0.54031(0.8415)=0.75036

Calculate the value of increment in value of x Δx2 .

Δx2=|x2x1|

Substitute -0.75036 for x2 and -1 for x1 in the above equation.

Δx2=|0.75036(1)|=0.24964

Substitute 2 for n in the equation (2).

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

x3=x2f(x2)f'(x2) (5)

Substitute the value of -0.75036 for x2 in the equation (5).

x3=0.75036f(0.75036)f'(0.75036)=0.750360.75036+cos(0.75036)1sin(0.75036)=0.75036(0.01125)=0.73911

Calculate the value of increment in value of x Δx3 .

Δx3=|x3x2|

Substitute -0.73911 for x3 and -0.75036 for x2 in the above equation.

Δx3=|x3x2|=|0.73911(0.75036)|=0.01125

Substitute 3 for n in the equation (2).

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

x4=x3f(x3)f'(x3) (6)

Substitute the value of -0.73911 for x3 in the equation (6).

x4=0.73911f(0.73911)f'(0.73911)=0.739110.73911+cos(0.73911)1sin(0.73911)=0.7390

Calculate the value of Δx4 .

Δx4=|x4x3|

Substitute -0.7390 for x4 and -0.73911 for x3 in the above equation.

Δx4=|0.7390(0.73911)|=0.00011

Substitute 4 for n in the equation (2).

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

x5=x4f(x4)f'(x4) (7)

Substitute the value of -0.7390 for x4 in the equation (7).

x5=0.7390f(0.7390)f'(0.7390)=0.73900.7390+cos(0.7390)1sin(0.7390)=0.7390

Calculate the value of Δx5 .

Δx5=|x5x4|

Substitute -0.7390 for x5 and -0.7390 for x4 in the above equation.

Δx5=|0.7390(0.7390)|=0

Hence, x -intercept point is (0.739,0) .

Calculate the y-intercept.

Substitute 0 for x in the equation (1).

f(0)=0+cos0=1

Therefore, the y -intercept point is (0,1) .

c)

Calculate the symmetry.

Apply negative and positive values for x in the equation (1).

Substitute 1 for x in the equation (1).

f(1)=1+cos(1)=1+0.5403=0.4597

Substitute +1 for x in the equation (1).

f(1)=1+cos(1)=1+0.5403=1.5403

Here the condition f(x)-f(x) is false, hence it has no symmetry

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