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Absolute maxima and minima a. Find the | Ch 4.1 - 49E

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

Solution for problem 49E 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 49E

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)=2x? sin ?x;[?2, 6]

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Solution 49E Step-1 Critical point definition; Let f be a continuous function defined on an open interval containing a number 1 1 ‘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 be a fun ction wit h domain D and let c be a fixed constant in D . Th en 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 x a). The given functio n is f(x) = 2 sin(x) on [ -2,6].Clearly the function contains both trigonometric, and algebraic functions 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) = 2x sin(x) for the critical values we have to differentiate the function both sides with respect to x. d (f(x)) = d (2 sin(x) ) dx dx 1 x x f (x) = sin(x) dx (2 ) +2 dxsin(x) , since dx(uv) = u dx(v)+v dx(u) x x d x x d = sin(x)(2 (ln(2) )+2 cos(x) , since dx (2 ) = 2 (ln(2)), dx sin(x) = cos(x). = 2 (ln(2))sin(x) +2 cos(X) = 2 (ln(2)sin(x)+cos(x)) 1 c Since , from the definition f (c)= 0 = 2 (ln(2)sin(c)+cos(c)) c 2 (ln(2)sin(c)+cos(c)) = 0 That is , 2 = 0 and (ln(2)sin(c)+cos(c)) = 0 . c So, 2 = 0 for this take log on both sides then c ln( 2 ) = ln(0) C (ln(2)) = ln(0), since ln( a ) = m(ln(a)) ln(0) C = ln(2) = undefined , since ln(0) is undefined. The second result is (ln(2)sin(c)+cos(c)) = 0 . ln(2) sin(c) = -cos(c) ln(2) = -cot(c) , since cos(c)= cot(c) sin(c) tan(c) = - (1/ln(2)) 1 1 That is , c = tan (1/ln(2)) = tan ( 0. 69314) Therefore, C = -0.96 , 2.18 ,and 5.32 , since from the general solution. Therefore , the function has critical values at c = -0.96 ,2.18, 5.32 are lies in the given interval. Step-4 b). Now , we have to determine the absolute extreme values of f on the given interval ; Here the given interval is [ -2,6], and the function has critical values at x = -0.96, 2.18,5.32.So the endpoints are -2,6. From the step-2 , the absolute extreme values are; Therefore , at x = -2, the value of the function is ; 2 sin(2) f(-2) = 2 in(-2) = 4 , since sin(-x) = -sin(x) and 2 = 0.044899 = -0.0087248 At x = -0.96, the value of the function is ; 0.96 0.96 f(-0.96) = 2 in(-0.96) , since sin(-x) = -sin(x) and 2 =0.514, sin(-0.96) = 0.819 = (0.514)(0.819) = 0.420966 At x =2.18, the value of the function is ; 2.18 2.18 f(2.18) = 2 in(2.18) , since 2 =4.53153, sin(2.18) = 0.82 = (4.53153)(0.82) = 3.71585 At x =5.32, the value of the function is ; f(5.32) = 2 5.32 in(5.32) , since 25.32=39. 94657, sin(5.32) = -0.821 = (39.94657)(-0.821) = -32.79614 Therefore , f(-0.96 ) = 0.420966 , f(2.18) =3.71585 , and f(5.32) = -32.79614, f(-2) = -0.0087248 Hence , the largest value of f(x) is 3. 71585 ,this attains at x= 2.18 Therefore , the absolute maximum is f( 2.18) = 3.71585 Hence , the smallest value of f(x) is -32.79614,this attains at x = 5.32 Therefore , the absolute minimum is f( 5.32) = -32.79614,. c) . The graph of the related function f(x) = 2 sin(x) on [ -2,6]. Hence, from the above graph all the absolute extreme values are true.

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Chapter 4.1, Problem 49E 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 - 49E