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Answer: Absolute maxima and minima a. Find the critical

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett ISBN: 9780321570567 2

Solution for problem 33E Chapter 4.1

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Calculus: Early Transcendentals | 1st Edition

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Problem 33E

Absolute maxima and minima a. Find the critical points of f on the given interval. b. Determine the absolute extreme values off on the given interval. c. Use a graphing utility to confirm your conclusions. f? (x) = cos (x) on [0,?]

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STEP_BY_STEP SOLUTION Step-1 Let f be a continuous function defined on an open interval containing a number ‘c’.The number ‘c’ is critical value ( or critical number ). If f (c) = 01 1 or f (c) is undefined. A critical point on that graph of f has the form (c,f(c)). Step-2 When an output value of a function is a maximum or a minimum over the entire domain of the function, the value is called the absolute maximum or the absolute minimum. Let f be a function with domain D and let c be a fixed constant in D. Then the output value f(c) is the 1. Absolute maximum value of f on D if and only if f(x) f(c) , for all x in D. 2. Absolute minimum value of f on D if and only if f(c) f(x) , for all x in D. Step_3 a). The given function is f(x) = cos (x) , on [0 , ].Clearly the function is a trigonometric function and it is continuous for all of x , and also the function is even function it gives only positive values. Now , we have to find out the critical points of f on the given interval. 2 Now , f(x) = cos (x) then differentiate the function both sides with respect to x. 1 d 2 f (x) = dx(cos (x)) d = 2cos(x) dx(cos(x)) , [ since dx( x ) = nx n1 and dx (cos (x )) = -sin( x )n dx ( x )] = -2cos(x) sin(x) = -sin(2x) , since by the formula Since , from the definition f (c)=0 = - sin(2c) sin(2c) = 0 sin(2c) = sin(n), where n = 0,1,2…………… Therefore , 2 c = n c= n 2 But the given interval is [0,] Therefore , c = 0 , , . 2 Clearly x = is the only value lies between [0,] 2 2 At x= 2 , then f(2) = cos ( ) 2 0 , since cos( ) = 0 2 Therefore ,f(x) attains the critical value at x = , and the critical point is ( , 0). 2 2 Step-4 b). Now , we have to determine the absolute extreme values of f on the given interval ; Here the given interval is [0, ] , and the function attains the critical value at x = 2 .So the endpoints are 0 , . From the step-2 , the absolute extreme values are; 2 Therefore , at x = 0 ,then f(0) = cos ( 0) = 1 , since cos (0) =1 . 2 At x= , then f( ) = cos ( ) = 0 , since cos( ) = 0 2 2 2 2 2 2 At x = , then f( ) = cos ( ) = ( 1) = 1 , since cos( ) = -1 Therefore , f(0) = 1 , f( ) =0 , and f( ) = 1. 2 Hence , the largest value of f(x) is 1 ,this attains at x= 0 , Therefore , the absolute maximum is f( ) = 1, f(0) = 1. Hence , the smallest value of f(x) is 0 ,this attains at x= (/2) Therefore , the absolute minimum is f(/2) = 0. Step-5 2 c) . The graph of the related function f(x) = cos (x) , on [0 , ] Hence, from the above graph all the absolute extreme values are true.

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Chapter 4.1, Problem 33E is Solved
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Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

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Answer: Absolute maxima and minima a. Find the critical