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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.1 - Problem 28e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.1 - Problem 28e

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# Locating critical points a. Find the critical | Ch 4.1 - 28E ISBN: 9780321570567 2

## Solution for problem 28E Chapter 4.1

Calculus: Early Transcendentals | 1st Edition

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Problem 28E

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. ? ? ? f(? ? ) = si? cos x? on [0, 2?]

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STEP_BY_STEP SOLUTION Step-1 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 1 or f (c) is undefined. A critical point on that graph of f has the form (c,f(c)). 4 2 Example ; Find all critical points of f(x) = x 8x . Because f (x) is a polynomial function, its domain is all real numbers. 1 3 f (x) = 4x 16x f (x) = 0 =4x 16x 2 4x ( x - 4) = 0 2 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. 4 2 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 a). The given equation is ; f(x) = sin(x) cos(x) , here f(x) is trigonometric function and it is continuous for all of x.The given interval is [0, 2] Now , f(x) = sin(x) cos(x) then differentiate the function both sides with respect to x. 1 d f (x) = dx(sin(x)cos(x)) = sin(x) d cos(x) +cos(x) d sin(x), since d (uv) = u d(v)+v d(u) dx dx dx dx dx d d = sin(x)(-sin(x))+cos(x)cos(x) ,since dx sin(x) = cos(x), dx (cos)=-sin(x) = cos (x) sin (x) = cos(2x) . Since , from the definition f (c)=0 = c os(2x) . That is, cos(2x) = 0. cos(2x) = cos( ),2because cos( ) = 0 2 2x = 2 x = 4 . At , x = 4 then f(x) = sin( 4cos( ) 4 =( 1 )( 1 ) 2 2 = (½) Therefore , f(x) attains the critical value at x= , and the endpoints of the interval is x= 4 0,2 . Hence, the critical point is ( , ) 4 2 Step_3 b). The graph of the related function is : f(x) = sin(x)cos(x) , on[0,2] Therefore, from the graph it is clear that the function attains neither local maximum nor local minimum at x= . 4

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##### ISBN: 9780321570567

This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1. The full step-by-step solution to problem: 28E from chapter: 4.1 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM. This full solution covers the following key subjects: critical, points, local, corresponds, domain. This expansive textbook survival guide covers 85 chapters, and 5218 solutions. The answer to “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. ? ? ? f(? ? ) = si? cos x? on [0, 2?]” is broken down into a number of easy to follow steps, and 53 words. Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567. Since the solution to 28E from 4.1 chapter was answered, more than 349 students have viewed the full step-by-step answer.

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