What kinds of discontinuities are possible for a rational function?

Question

anonymous · Answer

i think um,

anonymous · Answer

um i forgot i know this somewhere,

jim_thompson5910 · Answer

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 

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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

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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

anonymous · Answer

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 $y+1$.