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Absolute maxima and minima a. Find the | Ch 4.1 - 51E

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

Solution for problem 51E 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 51E

Absolute maxima and minima a. Find the critical points off 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) = sec ?x; [??/4, ?/4]

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Solution 51E Step-1 Critical point definition; 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 Absolute extreme values definition ; 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) = sec(x) on [ - , ].Clearly the function is a trigonometric 4 4 function . Now , we have to find out the critical points of f on the given interval. Now , f(x) = sec(x) for the critical values we have to differentiate the function both sides with respect to x. d (f(x)) = d (sec(x)) dx dx 1 d f (x) = sec(x)tan(x) ,since dx sec(x) = sec(x)tan(x). Since , from the definition f (c)=0 = sec(c)tan(c) 1 sin(c) sec(c)tan(c) = 0 cos(c) cos(c) = 0 sin(c) = 0. cos (c) That is , sin(c) = 0 c = n , here n = ….-2,-1,0,1,2……. But the given interval is [ - ,4]. 4 Therefore , c = 0 . Clearly x = 0, lies between the given interval [ - 4 ]4 At x = 0 , the value of the function is f( 0) = sec (0) = 1, since sec(0) =1. Therefore ,f(x) attains the critical value at x = 0, and the critical point is (0,1). Step-4 b) Now , we have to determine the absolute extreme values of f on the given interval ; Here the given interval is [ - 4 ]4 , and the function attains the critical value at x = 0 .So the endpoints are - , . 4 4 From the step-2 , the absolute extreme values are; Therefore , at x = - 4 then the value of the function is ; f( - ) = sec( - ) = 2 = 1.414 , sec(-x)= sec(x). 4 4 At x = 0 , the value of the function is f( 0) = sec (0) = 1, since sec(0) =1. At x = 4 then the value of the function is ; f( ) = sec( ) = 2 = 1.414 4 4 Therefore , f(- 4 = 1.414 , f(0) =1, and f( ) = 1.414 Hence , the largest value of f(x) is 1.414 ,this attains at x= - , 4 4 Therefore , the absolute maximum is f( ) = 1.414 = f( - ). 4 4 Hence , the smallest value of f(x) is ‘1’,this attains at x = 0 Therefore , the absolute minimum is f( 0) =1. Step_5 c) . The graph of the related function f(x) = sec(x) , on [ - , 4 is4 Hence, from the above graph all the absolute extreme values are true.

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Chapter 4.1, Problem 51E 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 - 51E