   Chapter 4.5, Problem 36E ### 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 = 2x − tan x, − π/2 < x < π/2

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)=2xtanx (1)

Calculation:

a)

Calculate the domain.

The function is defined for the given interval π2<x<π2 . So the domain is (π2,π2)

b)

Calculate the intercepts.

Calculate the value of x -intercept.

Substitute 0 for y in the equation (1).

2xtanx=0

Thus,

f(x)=2xtanx (2)

Solve for x from the equation (2).

Using Newton Rapson method.

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

Substitute 0 for n in the equation (3).

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

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

Substitute 1.1 for x0 in the equation (4).

x1=1.1f(1.1)f'(1.1)=1.12(1.1)tan(1.1)2sec2(1.1)=1.12.2tan(1.1)2(1cos2(1.1))

x1=1.12.21.96482(1(0.4536)2)=1.1822

Substitute 1 for n in the equation (3).

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

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

Substitute 1.1822 for x1 in the equation (5).

x2=1.1822f(1.1822)f'(1.1822)=1.18222(1.1822)tan(1.1822)2sec2(1.1822)=1.18222(1.1822)2.44272(1(0.3789)2)=1.1664

Substitute 2 for n in the equation (3).

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

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

Substitute 1.1664 for x2 in the equation (6).

x3=1.1664f(1.1664)f'(1.1664)=1.16642(1.1664)tan(1.1664)2sec2(1.1664)=1.16642.3328tan(1.1664)2(1cos2(1.1664))=1.16642.33282.33652(1(0.3935)2)=1.1655

Substitute 3 for n in the equation (3).

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

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

Substitute 1.1655 for x3 in the equation (7).

x4=1.1655f(1.1655)f'(1.1655)=1.16552(1.1655)tan(1.1655)2sec2(1.1655)=1.16552.331tan(1.1655)2(1cos2(1.1655))=1.16642.33282.33072(1(0.3943)2)=1.1655

Hence, x -intercept point is (1.1655,0) .

Calculate the y -intercept.

Substitute 0 for x in the equation (1).

f(0)=2(0)tan(0)=0

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

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)=2(1)tan(1)=2(1.5574)=0.443

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

f(1)=2(1)tan(1)=2(1.5574)=3.557

Here the condition f(x)f(x) is true, hence it is an odd function and the graph is rotational symmetry about the origin.

d)

Calculate asymptotes.

Apply limit of x tends to π2 in the equation (1).

limxπ2y=2(π2)tan(π2)=π()=

Apply limit of x tends to π2 in the equation (1).

limxπ2y=2(π2)tan(π2)=π=

Here, the value of limit gets infinity; this implies there is no horizontal asymptote.

For the given domain (π2,π2) , there will be vertical asymptotes at x=±π2

e)

Calculate the intervals.

Differentiate the equation (1) with respect to x .

f'(x)=2sec2x (8)

Substitute 0 for f'(x) in the equation (8).

2sec2x=0sec2x=2secx=2

Substitute 1cosx for secx in the above equation.

1cosx=2cosx=±12x=cos1(12)x=±π4

Take the interval (π2,π4) .

Substitute π3 for x in the equation (8).

f'(π3)=2sec2(π3)=21cos2(π3)=21(0

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Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th 