   Chapter 8.4, Problem 55E ### 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, 2n). Use a graphing utility to confirm your results. See Examples 6 and 7. y = x − 2 sin x

To determine

To calculate: The relative extrema of the trigonometric function y=x2sinx over the interval [0,2π].

Explanation

Given Information:

The provided trigonometric function is y=x2sinx over the interval [0,2π].

Formula used:

Sine derivative Rule:

ddx[sinu]=cosududx

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=x2sinx

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

dydx=ddx[x2sinx]y(x)=ddx[x]2ddx[sinx]=12cosx

Apply dydx=0 in order to find critical points.

12cosx=02cosx=1cosx=12

Since, the value of the cosθ is 12 at θ=π3 and θ=5π3 over the interval [0,2π]. So,

cosx=12cosx=cosπ3x=π3

And,

cosx=12cosx=cos5π3x=5π3

Thus, critical points are x=π3 and x=5π3.

Now, rewrite the first derivative of the function.

y(x)=12cosx

At x=π6.

y(π6)=12cosπ6=11.732=0.732=ve<0

At x=π.

y(π)=12cosπ=1(1)=2=+ve>0

At x=11π6.

y(11π6)=12cos11π6=11.732=0.732=ve<0

Thus, function y=x2sinx over the interval [0,2π] can be defined in tabular form as:

Interval0<x<π3π3<x<5π35π3<x<2πSign of y(x)y(x)<0y(x)>0y(x)<0yDecreasingIncreasingDecreasing

From above table, y(x) is changing from negative to positive at x=π3

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