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Absolute maxima and minima a. Find the | Ch 4.1 - 37E

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

Solution for problem 37E 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 37E

31–42. Absolute maxima and minima

a. Find the critical points of f on the given interval.

b. Determine the absolute extreme values of f on the given interval.

c. Use a graphing utility to confirm your conclusions

\(f(x)=(2 x)^{x}\) on [0.1, 1]

Step-by-Step Solution:
Step 1 of 3

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 be a funct ion withdomain D and let c be afixed constant in D . Then he 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) = (2x) , on [0.1,1]. 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 Now , f(x) = (2x) Now, take log on both sides .Because the base and power contains the algebraic functions .So in this cases we can take log. x m ln(f(x)) = ln((2x) ) = x ln(2x), since ln(a ) = m ln(a) then differentiate the function both sides with respect to x. d d dx ( ln(f(x)) ) = dx(x ln(2x)) 1 d (f(x)) = x d ln(2x) + ln(2x) d (x)...............(1) f(x) dx dx dx d d(v) d(u) d 1 since dx (uv)= u dx +v dx and dx (ln(x)) = x f (x) 2x from(1), f(x) = 2x+ln(2x)(1) = 1+ln(2x) 1 Hence, f (x) = f(x) (1+ln(2x)) x = (2x) (1+ln(2x)) 1 c Since , from the definition f (c)=0 = (2c) (1+ln(2c)) (2c) (1+ln(2c)) =0 1+ln(2c) =0 ln(2c) = -1 1 2c = e 1 Therefore, c = e 1 2 Clearly c = e lies in the interval [0.1,1] 2 e1 e1 1 1/2e 1 1/2e e1 1 At x = 2 , then f( 2 ) = (2 2e = ( )e , since 2 = 2e 0.179856 = (0.3679) = 0.8354 , since ‘e’ = 2.718 e1 e1 Therefore ,f(x) attains the critical values at x = 2 , and the critical point is ( 2 0.8354) Step-4 b). Now , we have to determine the absolute extreme values of f on the given interval ; Here the given interval is [ 0.1 , 1] , and the function attains the critical e1 value at x = 2 .So the endpoints are 0.1, 1. From the step-2 , the absolute extreme values are; 0.1 0.1 Therefore , at f(0.1) = (2(0.1)) = (0.2) = 0.8513 1 1 1/2e 1/2e 1 At x = e 2 , then f( e2 ) = (2 2e = ( e , since e2 = 2e 0.179856 = (0.3679) = 0.8354 , since ‘e’ = 2.718 1 1 At f(1) = (2(1)) = (2) = 2. e1 Therefore , f( 2 ) = 0.8354 , f(0.1) = 0.8513 ,and f(1) = 2 Hence , the largest value of f(x) is 2 ,this attains at x= 1 Therefore , the absolute maximum is f( 1) = 2. e1 Hence , the smallest value of f(x) is 0.8354,this attains at x = 2 e1 Therefore , the absolute minimum is f( 2 ) = 0.8354 Step_5 x c) . The graph of the related function f(x) = (2x) , on [0.1,1]. Hence, from the above graph all the absolute extreme values are true.

Step 2 of 3

Chapter 4.1, Problem 37E is Solved
Step 3 of 3

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 - 37E