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An exotie curve (Putnam Exam 1942) Find the coordinates of

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

Solution for problem 78AE Chapter 4.3

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 78AE

An exotie curve (Putnam Exam 1942) Find the coordinates of four local maxima of the function and graph the function for 0 ? ?x? ? 10.

Step-by-Step Solution:
Step 1 of 3

Solution 78AE Step1 Given that x f(x)= 1+x sin x The domain of the graph is 0 x 10 For maxima The maximum will be when the denominator is in its minimum Now, x sin x is a square term, so, the minimum of denominator will be 1 So, for minimum 2 1+x sin x 1 (x sin x) 0 3 x sin x 0 This is possible when x=0 or sin x=0 Now, the general solution of Sin x=0 is x=n Here n is an integer and 3.14 Another solution for numerator near 4 So, x= = 3.14=0.79 4 4 Step2 Now at x=n and n=1 x==3.14 At n=2 x=2=2*3.14 = 6.28 So, the local maxima are at x=0,0.79,3.14,6.28 For co-ordinates At x=0 f(x)= 6 2 1+x sin x f(0)= 0 =0 1+0sin 0 At x=0.75 x f(x)= 1+x sin x 0.79 f(0.79)= 6 2 1+(0.79) sin (0.79) = 0.79 1+0.1215 0.70 At x=3.14 x f(x)= 1+x sin x f(3.14)= 36142 1+(3.14) sin (3.14) 3.14 = 1+0 3.14 At x=6.28 f(x)= 6x 2 1+x sin x 6.28 f(6.28)= 1+(6.28) sin (6.28) = 6.28 1+0 6.28 So, Local maxima are(0,0),(0.79,0.70),(3.14,3.14),(6.28,6.28) Step3 Below table shows values of x in range of 0 x 10 x 0 2 4 6 8 10 f(x) 0 0.04 0.0017 0.00165 3.1×10 5 3.4×10 5 Graph of given function is

Step 2 of 3

Chapter 4.3, Problem 78AE 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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An exotie curve (Putnam Exam 1942) Find the coordinates of