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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 2.6 - Problem 91ae
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Ch 2.6 - 91AE

ISBN: 9780321570567 2

Solution for problem 91AE Chapter 2.6

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition

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Problem 91AE

91AE

Step-by-Step Solution:
Step 1 of 3

STEP_BY_STEP SOLUTION Step-1 A continuous function can be formally defined as a f unction f : x y ,where the preimage of every open set in y is open in x. More concretely, a function f(x) in a single variable x is said to be continuous at point x if, 0 1. If f(x 0 is defined, so that x is in t0e domain of ‘ f’. 2. lim f(x) exists for x in the domain of f. x x 0 3. lx x(x) = f( x ). 0 0 Left continuous : lim fxa f(a) , then f(x) is called a left continuous at x=a. Right continuous : lim f(x) =+f(a) , then f(x) is called a right continuous at x=a. xa If , lim f(x) = f(a) = lim f(x) , then f(x) is called a continuous function at x=a. xa xa + If , f(x) is not continuous at x =a means , it is discontinuous at x=a. Step-2 Definition ; (com posite function): Let g be a functio n from a set A to a set B , and let f be a unction from B to a set C . Then the co mposition of functions f and g , denoted by fg , is the function from A to C that satisfies g(x) = f( g(x) ) = (fog)(x) , for all x in A . 2 2 Exam ple: Let ) = , d (x) = x + 1 . Then f( ) ) = ( x + ) . If f and g are two functions at x=c then the functions defined by : + F 1x) = f(x) g(x) , F (x)2= K f(x) , here K is any real number , and F (3) = f(x).g(x) are also continuous at x=c.Further if g(c) is not zero then F (x) = f(x) is also continuous at x=c. 3 g(x) Step-3 a). Now , we need to find the functions f and g such that each function is continuous at x=0. But (fog) is not continuous at zero. 1 Let us assume the functions f(x) = x+1and g(x) = x-1………………(1) Now, we have to check f(x) , and g(x) are continuous at x =0 or not . So, limf(x) = lim 1 = 1 , take the limits x0 x0 x+1 0+1 = 1. At , x=0 the functional value is f(0) = (1/0+1) = 1. Therefore , limf(x) = 1 = f(0) x0 Hence , f(x) is continuous at x=0. So, limg(x) = lim (x1) = (0-1) take the limits x0 x0 = -1. At , x=0 the functional value is g(0) = (0-1) = -1. Therefore , limg(x) = -1 = g(0) x0 Hence , g(x) is continuous at x=0. Now, we have to check continuity of the composite function (fog)(x). So, (fog)(x) = f(g(x)) = f(x-1) , since from(1). = x1+1, from(1). = 1. x Therefore , (fog)(x) = 1 , is a rational function and it is continuous for all values x of x, except x=/ 0. Because from step(2). So , lim(fog)(x) = lim x x0 x0 = (1/0) , take the limits. = ,since (1/0) = . Therefore , (fog)(x) is not continuous at x=0. Step_4 b). The continuity of composite functions at a point states that ; If f and g are two functions at x=c then the functions defined by : F (x) = f(x) + g(x) , F (x) = K f(x) , here K is any real number , and F (x) = f(x).g(x) are 1 2 f(x) 3 also continuous at x=c.Further if g(c) is not zero then F 3x) = g(x) is also continuous at x=c……………(a) lim(fog)(x) = f (lim(g(x))) = f(g(c))...........(b) xc xc From (a) and (b) the theorem satisfies the part (a) and not contradicts the theorem.

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Step 3 of 3

ISBN: 9780321570567

This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1. This full solution covers the following key subjects: . This expansive textbook survival guide covers 112 chapters, and 7700 solutions. Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567. Since the solution to 91AE from 2.6 chapter was answered, more than 459 students have viewed the full step-by-step answer. The answer to “91AE” is broken down into a number of easy to follow steps, and 1 words. The full step-by-step solution to problem: 91AE from chapter: 2.6 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM.

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