   Chapter 8.4, Problem 58E ### 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 = 1 2 sec x

To determine

To calculate: The relative extrema of the trigonometric function y=12secx over the interval [0,2π].

Explanation

Given Information:

The provided trigonometric function is y=12secx over the interval [0,2π].

Formula used:

Secant derivative Rule:

ddx[secu]=secutanududx

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=12secx

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

dydx=ddx[12secx]y(x)=12ddx[secx]=12[secxtanx]=12secxtanx

Apply dydx=0 in order to find critical points.

12secxtanx=0secxtanx=0

Further solve.

tanx=0

And,

secx0

Since, the value of the tanθ is zero at θ=π between [0,2π]. So,

tanx=0tanx=tanπx=π=3.141592654

Thus, critical point is x=π between [0,2π].

Now, rewrite the first derivative of the function.

y(x)=12secxtanx

At x=π4.

y(π4)=12secπ4tanπ4=12(2)(1)=12=+ve>0

At x=5π4

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