   Chapter 8.4, Problem 6CP ### 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

# Find the relative extrema of y = x 2 − cos x in the interval ( 0 , 2 π ) .

To determine

To calculate: The relative extrema of the trigonometric function y=x2cosx over the interval [0,2π].

Explanation

Given Information:

The provided trigonometric function is y=x2cosx 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=x2cosx

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

dydx=ddx[x2cosx]y(x)=12(sinx)=12+sinx

Apply dydx=0 in order to find critical points.

12+sinx=0sinx=12

Since, Sine is negative in third and fourth quadrant. Thus, three are two solutions for sinx=12, which can be mathematically calculated as:

sinx=12sinx=sin7π6x=7π6

And,

sinx=12sinx=sin11π6x=11π6

Thus, critical points are x=7π6 and x=11π6.

Now, rewrite the first derivative of the function.

y(x)=12+sinx

At x=π2.

y(π2)=12+sinπ2=12+1=32=+ve>0

At x=3π2.

y(3π2)=12+sin3π2=12+(1)=12=ve<0

At x=23π12

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