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Chapter 4, Problem 54RE
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### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516

#### Solutions

Chapter
Section
BuyFindarrow_forward

### Calculus: Early Transcendental Fun...

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

# h ( x ) = x − 2 cos x ,       [ 0 , 4 π ] Using the Second Derivative Test In Exercises 49-56, find all relative extrema of the function. Use the Second Derivative Test where applicable.

To determine

To calculate: The relative extremas of the function h(x)=x2cosx using the second derivative test.

Explanation

Given:

The function h(x)=xâˆ’2cosx over the interval [0,4Ï€].

Formula Used:

For a function f that is twice differentiable on an open interval I, if f'(c)=0 for some c, then,

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

Calculation:

First differentiate the provided function,

h'(x)=1+2sinx

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

1+2sinx=0sinx=âˆ’12x=7Ï€6,11Ï€6,19Ï€6,23Ï€6

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

h''(7Ï€6)=2cos(7Ï€6)=âˆ’1.732<0

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

And,

h''(11Ï€6)=2cos(11Ï€6)=1.732>0

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

h''(19Ï€6)=2cos(19Ï€6)=âˆ’1

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