   Chapter 8.4, Problem 57E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Finding Relative Extrema In Exercises 53-62, find the relative extrema of the trigonometric function in the interval ( 0 ,   2 π ) . Use a graphing utility to confirm your results.See Examples 6 and 7. y = cos 2 x

To determine

To calculate: The relative extrema of the trigonometric function y=cos2x over the interval [0,2π].

Explanation

Given Information:

The provided trigonometric function is y=cos2x over the interval [0,2π].

Formula used:

Cosine derivative Rule:

ddx[cosu]=sinududx

General power derivative rule:

ddxxn=nxn1

Condition for critical points:

dydx=f(x)=0

If function f(x) is increasing, then f(x)>0 and if function is decreasing, then f(x)<0.

From first derivative Test, if f(x) changes positive to negative, then function f(x) has relative maxima and, if f(x) changes negative to positive then function f(x) has relative minima.

In order words, if function f(x) changes from increasing function to decreasing at any point critical, then function f(x) will have relative maxima at that point, and if function f(x) changes from decreasing function to increasing at any critical point, then function f(x) will have relative minima at that point.

Calculation:

Consider the provided trigonometric function:

y=cos2x

Differentiate the above function with respect to x using Cosine and Power derivative rules.

dydx=ddx[cos2x]y(x)=(2cosx)ddx[cosx]=(2cosx)[sinx]=2sinxcosx

Apply dydx=0 in order to find critical points.

2sinxcosx=02sinxcosx=0sin2x=0

Since, the value of the sinθ is zero at θ=π, θ=2π, and θ=3π.

So,

sin2x=sinπ2x=πx=π2=1.57079632

And,

sin2x=sin2π2x=2πx=π=3.14159265

And,

sin2x=sin3π2x=3πx=3π2=4.71238898

Thus, critical points are x=π2, x=π and x=3π2.

Now, rewrite the first derivative of the function.

y(x)=2sinxcosx

At x=π4.

y(π4)=2sinπ4cosπ4=2(12)(12)=1=ve<0

At x=3π4.

y(3π4)=2sin3π4cos3π4=2(12)(12)=1=+ve>0

At x=5π4.

y(5π4)=2sin5π4cos5π4=2(12)(12)=1=ve<0

At x=7π4

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