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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 2.6 - Problem 89ae
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 2.6 - Problem 89ae

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# Ch 2.6 - 89AE

ISBN: 9780321570567 2

## Solution for problem 89AE Chapter 2.6

Calculus: Early Transcendentals | 1st Edition

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Problem 89AE

89AE

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 if0 1. If f(x 0 is defined, so that x is 0n the domain of ‘ f’. 2. lim f(x) exists for x in the domain of f. x x0 3. lx x(x) = f( x ).0 0 Left continuous : lim f(x = f(a) , then f(x) is called a left continuous at x=a. xa Right continuous : lim f(x) = f(a) , then f(x) is called a right continuous at x=a. xa+ If , limf(x) = f(a) = lim +(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 x 4x +4x The given function is h(x) = , at x=1. x(x1) The given function is rational function , and it continuous for all values of x except x = / 0and 1. That is h( x) is discontinuous at x = 0 and 1. Note : If f(x) = p(x)is called a rational function , where q(x) = / 0. q(x) If q(x) =0 , then f(x) is discontinuous at these points. Step_3 a). Now , we need to check the discontinuity at x=0. The given function is a rational function , and as x approaches to zero, then the denominator value of h(x) is also approaches to zero. So, the numerator can be written as x 4x +4x = x(x -4x+4) 2 3 2 x 4x +4x x(x 4x+4) Hence , h(x) = x(x1) = x(x1) (x 4x+4) = (x1) , cancel out the like terms. (x2) = ……………….(1) (x1) 2 So, limh(x) = lim (x2) , from(1) x0 x0 (x1) 2 = (02) , take the limits. (01) 2 = (2) = -4 , since (x) = (x) 2 1 Therefore , limh(x) = -4 . x0 (x2)2 At , x=0 then the functional value is h(x) = (x1)= -4. Hence h(x) is removable discontinuity at x=0. Note: Removable discontinuities are where the graph of a function has a hole. This will occur in rational functions at an x-value that makes both the denominator and the numerator equal zero. For example, the graph of the function defined by x 4 f(x) = (x2), will have a hole at x=2 since both the denominator and numerator are zero for this x-value. x 4 f(x) = = (x2)(x+2= (x+2). (x2) (x2) Now , we can find the f(x)-value of the hole by substituting 2 for x in the simplified function. f(2)=2+2=4 See the image below. To remove the discontinuity, we will redefine the function at x = 2. This will effectively "fill in the hole." See the new image below. Step_4 b). The given function is a rational function , and as x approaches to one , then the denominator value of h(x) is also approaches to zero. So, the numerator can be written as x 4x +4x = x(x -4x+4) 2 x 4x +4x 2 Hence , h(x) = = x(x 4x+4) x(x1) x(x1) (x 4x+4) = (x1) , cancel out the like terms. (x2) = (x1) 2 Therefore , limh(x) = lim (x2) x1 x1 (x1) (12) = (11) , take the limits. 2 = (1) 0 = . Therefore , lix1x) = . Hence , h(x) has infinite discontinuity at x=1. Definition The function at the singular point goes to infinity in different directions on the two sides. Example

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##### ISBN: 9780321570567

Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567. The answer to “89AE” is broken down into a number of easy to follow steps, and 1 words. This full solution covers the following key subjects: . This expansive textbook survival guide covers 85 chapters, and 5218 solutions. Since the solution to 89AE from 2.6 chapter was answered, more than 378 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 89AE from chapter: 2.6 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1.

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