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iNeedhelpwithSchool

  • 2 years ago

What kinds of discontinuities are possible for a rational function?

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  1. AKANicole
    • 2 years ago
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    i think um,

  2. AKANicole
    • 2 years ago
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    um i forgot i know this somewhere,

  3. jim_thompson5910
    • 2 years ago
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    Case 1) The function is completely continuous over the entire domain, so there are no discontinuities Ex: y = 1/(x^2 + 1) The denominator is ALWAYS positive, so it can NEVER be zero ----> no vertical asymptotes ----> no discontinuities ------------------------------------------------------- Case 2) There is at least one asymptotic discontinuity, so it's discontinuous at this point or points. Ex: y = 1/(x+1) ... has a vertical asymptote at x = -1 ------------------------------------------------------- Case 3) There is a removable discontinuity (or a point discontinuity) Ex: y = ((x+1)(x+2))/(x+2) simplifies to y = x+1, but it has a removable discontinuity at x = -2

  4. oldrin.bataku
    • 2 years ago
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    Adding on to @jim_thompson5910's last note -- when you "simplify" \(\frac{(x+1)(x+2)}{x+2}=x+1\), you're dividing both our numerator and denominator by \(x+2\). Remember that we're never allowed to divide by \(0\), so by simplifying we're essentially assuming that \(x+2\ne0\) -- and this assumption is good for almost all \(x\) -- except for \(x=-2\). We'll just be left with a little hole there, with the rest of the function looking identical to \(x+1\), but there's still that *removable* discontinuity, removable because it's just a single little hole that we could "fill in" to be \(y+1\).

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