   Chapter 8.4, Problem 62E ### 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 = e − x sin x

To determine

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

Explanation

Given Information:

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

Formula used:

Sine derivative Rule:

ddx[sinu]=cosududx

Product rule of differentiation:

ddx[a(x)b(x)]=a(x)b(x)+b(x)a(x)

Exponential rule of differentiation:

ddx[eu]=eududx

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=exsinx

Apply the product rule of differentiation to find the derivative of the above function.

dydx=exddx[sinx]+sinxddx[ex]

Differentiate the above function using general power rule, cosine and exponential differentiation rules.

dydx=ex(cosx)+sinx(ex)y(x)=ex(cosxsinx)

Apply dydx=0 in order to find critical points.

excosxexsinx=0ex(cosxsinx)=0

Further solve.

cosxsinx=0sinx=cosxsinxcosx=1tanx=1

And,

ex0

Since, the value of the tanθ is zero at θ=π4, and θ=5π4.

So,

tanx=1tanx=tanπ4x=π4

And,

tanx=1tanx=tan5π4x=5π4

Thus, critical points are x=π4 and x=5π4.

Now, rewrite the first derivative of the function.

y(x)=excosxexsinx

At x=π6.

y(π6)=eπ6cosπ6eπ6sinπ6=0.61788=+ve>0

At x=π2.

y(π2)=eπ2cosπ2eπ2sinπ2=4.8104=ve<0

At x=3π2.

y(3π2)=e3π2cos3π2e3π2sin3π2=111

### 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 