   Chapter 4.5, Problem 38E ### 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 = csc x − 2sin x, 0 < x < π

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)=cscx2sinx (1)

Calculation:

a)

Calculate the domain.

The function is defined for the given interval 0<x<π . Therefore, the domain is (0,π) .

b)

Calculate the intercepts.

Calculate the value of x -intercept.

Substitute 0 for y in the equation (1).

cscx2sinx=0

Substitute 1sinx for cscx in the above equation.

1sinx2sinx=01sinx=2sinx1=2sin2x

12=sin2x12=sinxsin1(12)=xx=π4,3π4

Hence, x intercept points are (π4,0) and (3π4,0) .

Calculate the y -intercept.

Substitute 0 for x in the equation (1).

f(0)=csc(0)2sin(0)=

Therefore, there is no y -intercept.

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)=csc(1)2sin(1)=1sin(1)2sin(1)=1.188+1.683=0.49

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

f(1)=csc(1)2sin(1)=1sin(1)2sin(1)=1.1881.683=0.49

Here, the condition f(x)=f(x) is true, hence it is an odd function and the curve is symmetric about the origin.

d)

Calculate asymptotes.

Apply limit of x tends to 0 in the equation (1).

limx0y=csc(0)2sin(0)=

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

limxπy=csc(π)2sin(π)=

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

For the given domain (0,π) , there will be vertical asymptotes at x=0 and x=π .

e)

Calculate the intervals.

Differentiate the equation (1) with respect to x .

f'(x)=cotxcscx2cosx (2)

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

cotxcscx2cosx=0

Substitute cosxsinx for cotx and 1sinx for cscx in the above equation.

(cosxsinx)(1sinx)2cosx=0cosxsin2x2cosx=0cosx(1sin2x+2)=0cosx=0x=cos1(0)x=π2

Take the interval (0,π2) .

Substitute π6 for x in the equation (2).

f'(π6)=cot(π6)csc(π6)2cos(π6)

Substitute cos(π6)sin(π6) for cot(π6) and 1sin(π6) for csc(π6) in the above equation.

f'(π6)=cos(π6)sin(π6)1sin(π6)2cos(π6)=cos(π6)sin2(π6)2cos(π6)=32(12)232

f'(π6)=32(4)32=532=4.33

Here, the condition f'(x)<0 is true and hence the function f is decreasing on (0,π2) .

Take the interval (π2,π) .

Substitute 2π3 for x in the equation (2)

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