   Chapter 8.4, Problem 60E ### 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
1 views

# 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 = 2 cos x + cos 2 x

To determine

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

Explanation

Given Information:

The provided trigonometric function is y=2cosx+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=2cosx+cos2x

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

dydx=ddx[2cosx+cos2x]y(x)=2ddx[cosx]+ddx[cos2x]=2[sinx]+[sin2x]ddx[2x]=2sinxsin2x

Further solve.

y(x)=2sinx2sin2x

Apply dydx=0 in order to find critical points.

2sinx2sin2x=02(sinx+sin2x)=0sinx+sin2x=0sinx+2sinxcosx=0

Further solve.

sinx(1+2cosx)=0

Further solve.

sinx=0

And,

1+2cosx=02cosx=1cosx=12

Since, the value of the sinθ is zero at θ=π, while the value of the cosθ is 12 at θ=2π3, and θ=2π3.

So,

sinx=0sinx=sinπx=π

Also,

cosx=12cosx=cos2π3x=2π3

And,

cosx=12cosx=cos4π3x=4π3

Thus, critical points are x=2π3, x=π and x=4π3.

Now, rewrite the first derivative of the function.

y(x)=2sinx2sin2x

At x=π2.

y(π2)=2sin(π2)2sin2(π2)=2(1)2(0)=2=ve<0

At x=5π6.

y(5π6)=2sin(5π6)2sin2(5π6)=2(12)2(0.866)=0.732=+ve>0

At x=5π4.

y(5π4)=2sin(5π4)2sin2(5π4)=2(0.707)2(1)=0.586=ve<0

At x=5π3.

y(5π3)=2sin(5π3)2sin2(5π3)=2(0

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

#### Find the derivative of the function. y=1+xe2x

Single Variable Calculus: Early Transcendentals, Volume I

#### Convert the expressions in Exercises 6584 to power form. 23x1.2x2.13

Finite Mathematics and Applied Calculus (MindTap Course List)

#### In Exercises 49-62, find the indicated limit, if it exists. 56. limx2x+2x2

Applied Calculus for the Managerial, Life, and Social Sciences: A Brief Approach

#### In Problems 7-34, perform the indicated operations and simplify. 18.

Mathematical Applications for the Management, Life, and Social Sciences

#### The quadratic approximation for at a = 1 is:

Study Guide for Stewart's Multivariable Calculus, 8th 