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Absolute maxima and minima a. Find the critical points of

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

Solution for problem 32E 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 32E

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+ l)4/3 on[?8, 8]

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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) = 0 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) = (x + 1) 4/3 , on [-8,8].Clearly the function is polynomial 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 + 1) 4/3then differentiate the function both sides with respect to x. f (x) = dx (x + 1)4/3 = 4 (x + 1) (4/3)1 ,since d ( x ) = nx n1 and d (c) = 0, where c is 3 dx dx constant. 4 1/3 = 3 (x + 1) Since , from the definition f (c)=0 = 4 (c + 1)1/3 3 4 1/3 3 (c + 1) = 0 1/3 (c + 1) = 0 (c+1) = 0 Therefore , c = -1. Clearly ‘-1’ lies between [-8,8]. 4/3 At , x = -1 then f(-1) = ( 1 + 1) = 0. Therefore ,f(x) attains the critical value at x =-1, and the critical point is (-1,0). Step-4 b). Now , we have to determine the absolute extreme values of f on the given interval ; Here the given interval is [-8,8] , and the function attains the critical value at x= -1.So the endpoints are -8,8. since from the step -2, the absolute extreme values are ; 4/3 Therefore , at x=-1 the value of the function is f(-1) = ( 1 + 1) = 0 . 4/3 4/3 At x = -8 the value of the function is f(-8) = ( 8 + 1) = ( 7) = 13.3905 4/3 4/3 At x = 8 the value of the function is f(8) = ( 8 + 1) = ( 9) = 18.72075 Therefore , f(-8) = 13.3905 , f(-1) = 0 , and f(8) = 18.72075 Hence , the largest value of f(x) is 18.72075 ,this attains at x= 8. Therefore , the absolute maximum is f(8) = 18.72075. Hence , the smallest value of f(x) is 0 ,this attains at x=-1. Therefore , the absolute minimum is f(-1) = 0. Step-5 4/3 c). The graph of the related function f(x) = (x + 1) , on [-8,8] Hence, from the above graph all the absolute extreme values are true.

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Chapter 4.1, Problem 32E 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 critical points of