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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.1 - Problem 38e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.1 - Problem 38e

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# Absolute maxima and minima a. Find the | Ch 4.1 - 38E

ISBN: 9780321570567 2

## Solution for problem 38E Chapter 4.1

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition

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Problem 38E

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) = xe ?x/2 on [0,5]

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TEP_BY_STEP SOLUTION Step-1 Critical value definition; 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) = 01 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 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) = xex/2on [0,5] . Clearly the function is a exponential function and it is continuous for all of x . Now , we have to find out the critical points of f on the given interval. x/2 Now , f(x) = xe then differentiate the function both sides with respect to x. d d x/2 dxf(x) = dx (xe ) f (x) = x d (ex/2 )+ e x/2 d (x), since d (uv) = u d(v)+v d(u) dx dx dx dx dx = x( 1 ex/2 )+ e x/2(1) 2 x x/2 x/2 x/2 x = 2 e + e = e ( 2 +1) 1 c/2 c Since , from the definition f (c)=0 = e ( 2 +1) ec/2 (c +1) = 0 2 That is , e c/2 =0 and ( 2c+1) = 0 c/2 c ln( e ) =ln(0) and 2 =1 c ln(e) = undefined , and c=2. 2 Therefore, c = 2 Clearly x = 2 lies between [0,5]. 2/2 1 2 At x=2, then f(2) = 2e = 2e = e = 0.7358, since e = 2.718 Therefore ,f(x) attains the critical values at x =2, and the critical point is (2, 0.7358) Step-4 b). Now , we have to determine the absolute extreme values of f on the given interval ; Here the given interval is [ 0 ,5] , and the function attains the critical value at x = 2.So the endpoints are 0 ,5 From the step-2 , the absolute extreme values are; 0/2 Therefore , at x =0 then f(0) = 0 (e ) = 0. 2/2 1 2 At x=2, then f(2) = 2e = 2e = e = 0.7358, since e = 2.718 At x = 5, then f(5) = 5e 5/2= 5(0.082106)= 0.41053, since e = 2.718 Therefore , f(0) =0 , f(2) = 0.7358 ,and f(5) = 0.41053 Hence , the largest value of f(x) is 0.7358 ,this attains at x= 2 Therefore , the absolute maximum is f( 2) = 0.7358 Hence , the smallest value of f(x) is 0,this attains at x =0 Therefore , the absolute minimum is f( 0) = 0 Step_5 c) . The graph of the related function f(x) = xe x/2 on [0,5] Hence, from the above graph all the absolute extreme values are true.

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##### ISBN: 9780321570567

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 38E from 4.1 chapter was answered, more than 374 students have viewed the full step-by-step answer. Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567. The full step-by-step solution to problem: 38E from chapter: 4.1 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1. 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) = xe ?x/2 on [0,5]” is broken down into a number of easy to follow steps, and 42 words.

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