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Absolute maxima and minima a. Find the | Ch 4.1 - 34E

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

Solution for problem 34E 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 34E

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?2+1)2 on [?2, 2]

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 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) = x , on [-2,2].Clearly the function is rational function and (x +1) 2 it is continuous for all of x , because the denominator is not equal to zero for all values of x. Now , we have to find out the critical points of f on the given interval. Now , f(x) = x 2 then differentiate the function both sides with respect to x. (x +1) 1 d x f (x) = dx 2 2 (x +1) ( (x +1) )ddxxdx (x +1) )2 d u vdxuuddx) = 2 4 , since ( ) = 2 (x +1) dx v v 2 2 2 d 2 ( (x +1) )(1)x 2( (x +1) dx(x ) d n n1 d = (x +1)4 , since dx ( x ) = nx dx (x). ( (x +1) )(1)x 2( (x +1) (2x) = 2 4 (x +1) 2 2 2 = (x +1) ( 2x +1) 4x ) (x +1)4 2 = 12 x 3 , ince cancel out the like terms. (x +1) 1 13 c2 Since , from the definition f (c)=0 = 2 3 (c +1) 13 c2 2 2 3 = 0 1- 3c = 0 (c +1) 2 2 3c = 1 c = Therefore, c = 1 and c = 1 3 3 1 1 Clearly 3 and 3 ies between [-2,2]. 1 1 1 1 3 3 Therefore , at x = 3 then f( 3 ) = 1 2 2 = 1 2 2 ((3) +1) (3 ) +1) 1 1 3 3 1 9 = 2 = = ( 16 ) ((3) +1) ((96) 3 = 33 = -0.324759 16 1 1 At x = 1 then f( 1 ) = 3 = 3 3 3 (( 1) +1) 2 (( ) +1) 2 3 3 1 = 3 = 1 ( 9 ) ((6) 3 16 9 33 = 16 = 0.324759 Therefore ,f(x) attains the critical value at x = 1 and 1 , and the critical 3 3 1 1 points are ( 3 , -0.324759) , ( 3 , 0.324759). Step-4 b). Now , we have to determine the absolute extreme values of f on the given interval ; Here the given interval is [-2,2] , and the function attains the critical value at x= 1 1 and .So the endpoints are -2,2. 3 3 since from the step -2, the absolute extreme values are ; Therefore , at x= -2 the value of the function is f(-2) = 2 ((2) +1) 2 = 2 2 = 2 = -0.08 (4 +1) 25 1 1 At x = 1 then f( 1 ) = 3 = 3 3 3 ((1) +1) 2 (( ) +1)2 3 3 1 1 = 3 = 3 = 1 ( 9 ) (( ) +1)2 ((16) 3 16 3 9 33 = 16 = -0.324759 1 1 At x = 1 then f( 1 ) = 3 = 3 3 3 ((1) +1) 2 (( ) +1)2 3 3 1 = 3 = 1 ( 9 ) ((1) 3 16 9 = 33 = 0.324759 16 at x= 2 the value of the function is f(2) = 2 (2+1) = 2 = 2 = 0.08 (4 +1)2 25 1 1 Therefore, f(-2) = -0.08 , f( 3 ) = -0.324759 , f( 3 ) = 0.324759 and f(2) = 0.08 1 Hence , the largest value of f(x) is 0.324759 ,this attains at x= 3 1 Therefore , the absolute maximum is f( 3 ) = 0.324759 1 Hence , the smallest value of f(x) is -0.324759,this attains at x= - 3 1 Therefore , the absolute minimum is f( ) = -0.324759 3 Step-5 x c). The graph of the related function f(x) = (x +1)2 , on [-2,2]. Hence, from the above graph all the absolute extreme values are true.

Step 2 of 3

Chapter 4.1, Problem 34E is Solved
Step 3 of 3

Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567. The answer to “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?2+1)2 on [?2, 2]” is broken down into a number of easy to follow steps, and 42 words. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1. The full step-by-step solution to problem: 34E from chapter: 4.1 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM. This full solution covers the following key subjects: absolute, interval, given, graphing, determine. This expansive textbook survival guide covers 85 chapters, and 5218 solutions. Since the solution to 34E from 4.1 chapter was answered, more than 378 students have viewed the full step-by-step answer.

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Absolute maxima and minima a. Find the | Ch 4.1 - 34E