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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 2.6 - Problem 90ae
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 2.6 - Problem 90ae

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# Ch 2.6 - 90AE ISBN: 9780321570567 2

## Solution for problem 90AE Chapter 2.6

Calculus: Early Transcendentals | 1st Edition

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Problem 90AE

90AE

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 0 is in the domain of ‘ f’. 2. lim f(x) exists for x in the domain of f. x x0 3. lim f(x) = f( x ). x x0 0 Left continuous : lim f(x) = f(a) , then f(x) is called a left continuous at x=a. xa Right continuous : lim xa)+= f(a) , then f(x) is called a right continuous at x=a. 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 ; (composite function): Let g be a function from a set A to a set B , and let f be a function from B to a set C . Then the composition of functions f and g , denoted by fg , is the function from A to C that satisfies fg(x) = f( g(x) ) = (fog)(x) , for all x in A . 2 2 Example: Let f(x) = x , and g(x) = x + 1 . Then f( g(x) ) = ( x + 1 ) . Step-3 Given that g(x) is continuous function at x=a , and f(x) is continuous function at g(a) then it can be written as ; xag(x) = g(a) . ……………(1) and lim f(x) = f(g(a) )……………………(2) xa Now , we need to prove that the composite function (fog)(x) is continuous at x=a. Thus , lim(fog)(x) = lim(fg(x)) , since by the definition in step 2 . xa xa = lim(f(g(x))) xa = f( xa(g(x)) = f(g(a) ) , since from(1) = (fog)(a). Therefore , xa(fog)(x) = (fog)(a) . Hence , (fog)(x) is continuous at x=a , since by the definition in step_1.

Step 2 of 3

Step 3 of 3

##### ISBN: 9780321570567

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