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Absolute maxima and minima a. Find the | Ch 4.1 - 39E

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

Solution for problem 39E 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 39E

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. 2 ?1 ? f(x) = x + cos (x) on [-1,1]

Step-by-Step Solution:
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STEP_BY_STEP SOLUTION Step-1 Let f be a continuous function defined on an open interval containing a 1 number ā€˜cā€™.The number ā€˜cā€™ is critical value ( or critical number ). If f (c) = 0 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 func tion with domain D and let c be a fixed constant in D . Then the output value f( ) 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) = x + cos (x) on [-1,1] .Clearly the function contain both algebraic and inverse trigonometric function and it is continuous for all of x . Now , we have to find out the critical points of f on the given interval. Now , f(x) = x + cos (x) then differentiate the function both sides with respect to x. d d 2 1 dx f(x) = dx ( x + cos (x) ) 1 f (x) = d (x ) + d (cos (x)), since d (u+v) = d (u) -+ d (v) dx dx dx dx dx 1 d n n1 d 1 1 = 2x + ( 1x2) , since dx (x ) = nx , dx(cos (x)) = ( 1x2 1 1 Since , from the definition f (c)=0 = 2c + ( 1c2) 2c - ( 1 ) =0 1c2 2c = 1 1c2 2c( 1 c ) = 1 Squaring on both sides , then 2c( 1 c ) = 1 becomes 4c (1- c ) =1 2 4c - 4c +1 = 0 That is , (2c 1) 2 = 0 2 (2c 1) = 0 2 2 1 2c =1 c = 2 1 1 Therefore, c = 2, 2 Clearly c = 1 , 1 lies between [-1,1] 2 2...

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Chapter 4.1, Problem 39E 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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Absolute maxima and minima a. Find the | Ch 4.1 - 39E

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