Chapter 4.6, Problem 40E

Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516

Chapter
Section

Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516
Textbook Problem

Analyzing the Graph of a Trigonometric Function In Exercises 35-42, analyze and sketch a graph of the function over the given interval. Label any intercepts, relative extrema, points of inflection, and asymptotes. Use a graphing utility to verify your results.Function Interval y = sec 2 π x 8 − 2 tan π x 8 − 1 − 3 < x < 3

To determine

To graph: The function y=sec2πx82tanπx81

Explanation

Given: The function y=sec2Ï€x8âˆ’2tanÏ€x8âˆ’1 over the interval âˆ’3â‰¤xâ‰¤3.

Graph: The domain of the function has been provided as [âˆ’3,3].

Now find the x and y intercepts by equating y and x to zero respectively to obtain:

The function x-intercepts as (0,0),(2.82,0) and y-intercept as (0,0).

The function has no vertical or horizontal asymptotes.

Now, differentiate the function with respect to x and equate it to zero to obtain the critical points.

Ï€4sec2Ï€x8(tanÏ€x8âˆ’1)=0(tanÏ€x8âˆ’1)=0tanÏ€x8=1x=2

This gives two test intervals (âˆ’3,2),(2,3).

Let 0âˆˆ(âˆ’3,2).

f'(0)=Ï€4sec2Ï€(0)8(tanÏ€(0)8âˆ’1)<0

The function is decreasing on this interval.

Let 2.5âˆˆ(2,3).

f'(2.5)=Ï€4sec2Ï€(2.5)8(tanÏ€(2.5)8âˆ’1)>0

The function is increasing on this interval.

This implies that the function has a relative minimum at 2.

Now obtain the value of the function at these critical points.

f(2)=sec2Ï€(2)8âˆ’2tanÏ€(2)8âˆ’1=âˆ’1

Thus the relative minimum is (2,âˆ’1)

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