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onegirl
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Find all discontinuities of f(x). For each discontinuity that is removable, define a new function that removes the discontinuity. f(x) = x  1/x^2  1
 one year ago
 one year ago
onegirl Group Title
Find all discontinuities of f(x). For each discontinuity that is removable, define a new function that removes the discontinuity. f(x) = x  1/x^2  1
 one year ago
 one year ago

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tkhunny Group TitleBest ResponseYou've already chosen the best response.0
First, please remember your Order of Operations. You have NOT written \(\dfrac{x1}{x^{2}1}\). Give it another go and use more parentheses. Denominator = 0  Discontinuity. Is it an Asymptote or NonRemovable Discoutinuity? Numerator = 0 AT THE SAME PLACE, this it's Removable and NOT an Asymptote.
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
non removable
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.0
?? There are two. 1) Which one are you talking about. 2) What's the other one?
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
removable discontinuity or infinite
 one year ago

onegirl Group TitleBest ResponseYou've already chosen the best response.0
so it will be replacing 0 will give you discontity right? and not any other number?
 one year ago

tkhunny Group TitleBest ResponseYou've already chosen the best response.0
Please make a better effort to use complete sentences and to be substantially more clear. I'll do a quick example. Sentences, paragraphs, examples, order. Working with the Denominator: \(x^{2}  1 = (x+1)(x1)\) This denominator takes on the value zero at x = 1 and x = 1. These values are NOT in the Domain and are discontinuities. We do not yet know what kind of discontinuity. Working with the Numerator x  1 = 0 when x = 1 This is enough information. x = 1 makes both Numerator and Denominator zero. This is, therefore, a removable discontinuity. Our original expression is equivalent to 1/(x+1) everywhere EXCEPT x = 1. x = 1 makes only the denominator zero. This is, therefore, an asymptote or infinite discontinuity. Now, we are done.
 one year ago
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