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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.1 - Problem 25e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.1 - Problem 25e

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# Answer: Locating critical points a. Find the critical

ISBN: 9780321570567 2

## Solution for problem 25E Chapter 4.1

Calculus: Early Transcendentals | 1st Edition

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Problem 25E

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(x) = x +1

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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 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. 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 x a). The given equation is ; f(x) = x +1 , here f(x) is a rational function , its domain is all real numbers ,since for all values of x the denominator value is non zero.. so, f(x) is continuous for all of x. Now ,f (x) = 2x , then differentiate the function both sides with respect x +1 to x. 1 d x So , f (x) = dx ( x +1) 2 d(x) d 2 d(u) d(v) ( x +1) dx xdx x +1) d u v dxu dx = (x +1) 2 , since dx ( v = v2 ( x +1)(1)x(2x+0) x +12x 2 = 2 2 = 2 2 , since d (x ) = 2x , d (c) =0, c is (x +1) (x +1) dx dx constant 1 x2 = 2 2 (x +1) 2 Since from the definition f (c) = 0 = 1 c (c +1) 2 1 c = 0 1 = c c = -1, 1. Therefore , f(x) attains the critical values at x = -1 and 1. At , x=-1 then f(-1) = (1) +1 = 2 1 1 At , x=1 then f(1) = (1) +1 = 2 = 0.5 Hence , the critical points are (-1, -0.5) and (1,0.5) Therefore local minimum attains at x = -1, and the local maximum attains at x= 1. The graph of the related function f(x) =

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