Locating critical points a. Find the critical | Ch 4.1 - 27E

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

Problem 27E 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 27E

Locating critical points a. Find the critical points of the following functions on the domain or on the given interval. b. Use a graphing unity In determine whether each critical point corresponds to a local minimum. local maximum. or neither. x ?x f(x) = e +e 2

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STEP_BY_STEP SOLUTION Step-1 Let f be a continuous function defined on an open interval containing a number ā€˜cā€™.The number ā€˜cā€™ is critical value ( or critical number ). If f (c) = 0 1 1 or f (c) is undefined. A critical point on that graph of f has the form (c,f(c)). Example ; Find all critical points of f(x) = x 8x . 2 Because f (x) is a polynomial function, its domain is all real numbers. 1 f (x) = 4x 16x f (x) = 0 =4x 16x 4x ( x - 4) = 0 4x(x+2)(x-2) = 0 , since a - b = (a+b)(a-b) X = 0 , x = -2 and x=2. 4 2 At x= -2 , then f(-2) = (2) 8(2) = 16 - 32 = -16. At x= 0 , then f(0) = (0) 8(0) = 0 - 0 = 0. At x= 2 , then f(2) = (2) 8(2) = 16 - 32 = -16. Hence, the critical points of f(x) are (2,16), (0,0), and (2,16). Step-2 e +e x a). The given equation is ; f(x) = 2 , here f(x) is exponential function , its domain is all real numbers .so, f(x) is continuous for all of x. e +ex Now , f(x) = 2 then differentiate the function both sides with respect to x. x x So , f (x) = d ( e +e ) dx 2...

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Chapter 4.1, Problem 27E 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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Locating critical points a. Find the critical | Ch 4.1 - 27E

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