   Chapter 4.4, Problem 46E Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516

Solutions

Chapter
Section Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516
Textbook Problem

Using the Second Derivative Test InExercises 37-58, find all relative extrema of the function. Use the Second Derivative Test where applicable. f ( x ) = 2 sin x + cos 2 x ,     [ 0 , 2 π ]

To determine

To calculate: The relative extrema of the provided function f(x)=2sinx+cos2x using the second derivative test.

Explanation

Given:

The function f(x)=2sinx+cos2x over the interval [0,2π].

Formula used:

For a function f that is twice differentiable on an open interval I,

If f'(c)=0 for some c, and,

If f''(c)>0 the function f has relative minima at c

If f''(c)<0 the function f has relative maxima at c.

If f''(c)=0, the test fails.

Calculation:

First differentiate the provided function,

f'(x)=2cosx2sin2x

Equate the first derivative to zero to obtain the critical points.

2cosx2sin2x=0cosx=sin2xx=π6,π2,5π6,3π2

Now differentiate the function for the second time and replace this for x:

f''(π6)=2sin(π6)4cos(2(π6))=3<0

This implies that the function has a relative maxima at this point.

And,

f''(π2)=2sin(π2)4cos(2(π2))=2>0

This implies that the function has a relative minima at this point.

f''(5π6)=2sin(5π6)4cos(2(5π6))=3<0

This implies that the function has a relative maxima at this point

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