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Absolute maxima and minima a. Find the | Ch 4.1 - 41E

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

Solution for problem 41E 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 41E

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.

Step-by-Step Solution:
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STEP_BY_STEP SOLUTION Step-1 Critical value 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 value 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 afunction with domain D and let 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 x a). The given function is f(x) = 2 on (-2,2) . Clearly the function is a rational function 4x and here the denominator is not equal to zero . So , 2 - x 2 > 0 2 That is , x < 2 Therefore , -2 < x< 2. Therefore, the given function is continuous on (-2,2) Now , we have to find out the critical points of f on the given interval. Now , f(x) = x for the critical values we have to differentiate the function both 4x 2 sides with respect to x. 1 d x f (x) = dx ( 2) 4x d(x) d d(u) ( 4 x )dxx dx 4x ) d u v dxu = ( 4x )2 , since dx ( v = v2 ( 4 x )(1) x d( 4x ) 2 4x dx d = 2 2 , since dx ( ) = 2x ( 4x ) 2 x ( 4 x )2 4x2 (2x) 4x +x 2 = 2 = 2 2 4x (4 x ) 4x 4 = 2 2 (4 x ) 4x 1 Since , from the definition f (c) = 0 = 4 (4 c ) 4c2 Therefore , 4 = 0 which is not possible This means the function has no critical points. Therefore , there are no critical points for this function . Step-4 b) . The given function is a rational function , and its domain is (-2,2) .So the function is continuous in this interval . Hence , the function does not have a minimum and maximum in this interval .Because from the step-2 , f is continuous on a closed interval [a,b] , then f has both minimum and maximum values at this interval . Therefore , the given function f(x) = x has no absolute extreme values 4x2 in given interval (-2,2) . Step-5 c). The graph of the related function f(x) = x on (-2,2) 4x2

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Chapter 4.1, Problem 41E 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 - 41E