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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:
Step 1 of 3

STEP_BY_STEP SOLUTION Step-1 Let f be a continuous function defined on an open interval containing a 1 number ‘c’.The number ‘c’ 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 1 1 1 2 1 1 At x = 2 , then f( 2) = ( 2+ cos ( 2) 1 1 1 1 1 = 2 + cos (2 ) ,since cos (-x) = cos (x) 1 1 1 = 2 + 4 , since cos ( 2 ) = 4 1 3 = 2 + 4 = 2.8562 , since = 3.14 1 1 1 2 1 1 At x = 2, then f( 2 ) = (2 + cos ( 2 ) 1 1 1 1 1 = 2 +cos ( 2) ,since cos ( 2 ) = 4 1 = 2 + 4 = 1.285 , since = 3.14 Therefore ,f(x) attains the critical values at x = 1 , 1 , and the critical 2 2 points are ( 1 , 2.8562) ( 1 , 1.285). 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 [-1,1] , and the function attains the 1 1 critical values at x = 2, 2 .So the endpoints are -1, 1. From the step-2 , the absolute extreme values are; 2 1 Therefore , at x = -1 then f(-1) = ( 1) + cos ( 1 ) 1 1 1 = 1+ cos (1) ,since cos (-x) = cos (x) = 1+ 0 , since cos (1) = 0 = 1+ = 4.14 , since = 3.14 At x = 1 , then f( 1) = ( ) + cos ( 1 1) 2 2 2 2 = 1 + cos 1( 1 ) ,since cos (-x) = cos (x)1 2 2 1 1 = 2 + 4 , since cos ( 2 ) = 4 1 3 = 2 + 4= 2.8562 , since = 3.14 1 1 1 2 1 At x = , then f( ) = () + cos ( 1 ) 2 2 2 2 1 1 1 1 1 = 2+cos (2 ) ,since cos ( 2 ) = 4 = 1+ = 1.285 , since = 3.14 2 4 At x = 1 then f(1) = (1) + cos (1) 1 = 1+ cos (1) = 1+ 0, since cos (1) = 0 = 1 1 1 Therefore , f( 2) = 2.8562 , f(-1) = 4.14 , f( 2 ) = 1.285 and f(1) = 1. Hence , the largest value of f(x) is 4.14 ,this attains at x= -1 Therefore , the absolute maximum is f( -1) = 4.14 Hence , the smallest value of f(x) is 1,this attains at x = 1 Therefore , the absolute minimum is f( 1) = 1

Step 2 of 3

Chapter 4.1, Problem 39E is Solved
Step 3 of 3

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