   Chapter 4.5, Problem 33E ### 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 = sin3x

To determine

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

Explanation

Given:

The Cartesian equation is as below.

y=sin3x (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).

sin3x=0sinx=0x=sin1(0)x=nπ

Hence, x -intercept is (nπ,0) .

Calculate the y -intercept.

Substitute 0 for x in the equation (1).

f(0)=sin3(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)=sin3(1)=0.0053

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

f(1)=sin3(1)=0.0053

Here the condition f(x)=f(x) is true, it is an odd function and the graph is symmetrical about origin.

Also f(x) is periodic with periodicity 2π .

sin3(x+2π)=(sin(x+2π))3=sinx

d)

Calculate asymptotes.

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

limxy=

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

limxy=

Here, the value of limit gets infinity. This implies that there is no horizontal asymptote or vertical asymptote.

Here f(x) is bounded function. Its range is (1,1) .

e)

Calculate the intervals.

Differentiate the equation (1) with respect to x .

Apply the chain rule as below.

df(u)dx=dfdududx

Substitute u3 for f and sinx for u in the above equation.

df(u)dx=ddu(u3)ddx(sinx)=3u31(cosx)=3u2(cosx)

Substitute sinx for u in the above equation.

f'(x)=3(sinx)2(cosx)

f'(x)=3sin2xcosx (2)

Here, the expression f'(x) increases, when the value of cosx is positive and decreases when cosx value is negative.

Take the interval as (0,π2) .

Substitute π4 for x in the equation (2).

f'(π4)=3sin2(π4)cos(π4)=3×0.5×(12)=1.0607

Here the function f'(π4)>0 is true and the function f(x) is increasing on (0,π2) .

Take the interval as (π2,3π2) .

Substitute 5π4 for x in the equation (2).

f'(5π4)=3sin2(5π4)cos(5π4)=3×(12)2×(12)=3×12×(12)=1.0607

Here the function f'(5π4)<0 is true and the function f(x) is decreasing on (π2,3π2) .

Take the interval as (3π2,2π) .

Substitute 7π4 for x in the equation (2).

f'(7π4)=3sin2(7π4)cos(7π4)=3×(12)2×12=3×12×12=1.061

Here, the condition f'(7π4)>0 is true and function f(x) is increasing on (3π2,2π) .

Hence, the function f(x) is increasing in the interval (0,π2) and (3π2,2π) . The function f(x) is decreasing in the interval (π2,3π2) .

f)

Calculate the local maximum and minimum value.

Differentiate the equation (2) with respect to x .

Apply the product rule.

(uv)'=uv'+vu'

Substitute sin2x for u and cosx for v in the above equation.

3(sin2xcosx)'=3(sin2x(sinx)+cosx(2sinxcosx))=3(sin3x+2cos2xsinx)=3sinx(sin2x+2cos2x)f''(x)=3sinx(2cos2xsin2x)

Substitute (1sin2x) for cos2x in the above equation

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