   Chapter 8.4, Problem 53E ### 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 = cos π x 2

To determine

To calculate: The relative extrema of the trigonometric function y=cosπx2 over the interval [0,2π].

Explanation

Given Information:

The provided trigonometric function is y=cosπx2 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=cosπx2

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

dydx=ddx[cosπx2]y(x)=(sinπx2)ddx[πx2]=(sinπx2)(π2)=π2sinπx2

Apply dydx=0 in order to find critical points.

π2sinπx2=0sinπx2=0

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

At θ=π

sinπx2=sinππx2=πx=2ππ=2

At θ=2π

sinπx2=sin2ππx2=2πx=4ππ=4

At θ=3π

sinπx2=sin3ππx2=3πx=6ππ=6

Thus, critical points are x=2, x=4 and x=5.

Now, rewrite the first derivative of the function.

y(x)=π2sinπx2

At x=1.

y(1)=π2sinπ(1)2=π2=ve<0

At x=3.

y(3)=π2sinπ(3)2=π2=+ve>0

At x=5.

y(3)=π2sinπ(5)2=π2=ve<0

At x=6.15.

y(6.15)=π2sinπ(6.15)2=1.3755=+ve>0

Thus, function y=cosπx2 over the interval [0,2π] can be defined in tabular form as:

Interval0<x<22<x<44<x<66<x<2πSign of y(x)y(x)<0y(x)>0y(x)<0y(x)>0yDecreasingIncreasingDecreasingIncreasing

From above table, y(x) is changing from negative to positive at x=2

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