   Chapter 8.4, Problem 56E ### 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 = 3 2 x + cos x

To determine

To calculate: The relative extrema of the trigonometric function y=32x+cosx over the interval [0,2π].

Explanation

Given Information:

The provided trigonometric function is y=32x+cosx 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=32x+cosx

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

dydx=ddx[32x+cosx]y(x)=32ddx[x]+ddx[cosx]=32+[sinx]=32sinx

Apply dydx=0 in order to find critical points.

32sinx=0sinx=32sinx=32

Since, the value of the sinθ is zero at θ=π3, and θ=2π3. So,

sinx=32sinx=sinπ3x=π3=1.04719755,

And,

sinx=32sinx=sin2π3x=2π3=2.09439510

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

Now, rewrite the first derivative of the function.

y(x)=32sinx

At x=π4.

y(π4)=32sinπ4=0.1589=+ve>0

At x=π2.

y(π2)=32sinπ2=0.134=ve<0

At x=π.

y(π)=32sinπ=0.866=+ve>0

Thus, function y=32x+cosx over the interval [0,2π] can be defined in tabular form as:

Interval0<x<π3π3<x<2π32π3<x<2πSign of y(x)y(x)>0y(x)<0y(x)>0yIncreasingDecreasingIncreasing

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

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