Algebra (all content)
Discontinuities of rational functions
Sal analyzes two rational functions to find their vertical asymptotes & removable discontinuities. He distinguishes those from the zeros of the functions.
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- Why does a value that makes the denominator equal to 0 count as a vertical asymptote? Shouldn't it be a removable discontinuity? And also why is it that the removable discontinuity happens to be a factor shared in both the denominator and the numerator?(70 votes)
- what is a asymptote and what is the difference between asymptote and removable discountinuity ? please help and thanks in advance !(16 votes)
- One definition of an asymptote of a curve is that it is a LINE such that the distance between the curve and the line approaches zero as they tend to infinity.
Or, in simple terms, you could think of an asymptote as a LINE that a curve approaches but never meets. (Later on in your math career, you will discover that vertical asymptotes are as I've described, but that sometimes the function does actually cross through its horizontal asymptote(s).)
A removable discontinuity is a SINGLE POINT for which the function is not defined. If you were graphing the function, you would have to put an open circle around that point to indicate that the function was not defined there.
Hope this is of some help!(31 votes)
- 06:38x=4 should it be removable discontinuity?(11 votes)
- The vertical asymptote(s) can only be found once the equation is as simplified as possible. Removable discontinuities are found as part of the simplification process. If a factor like x=4 appears in both steps the vertical 'asymptote' label is the stronger since it produces a vertical asymptote when graphed as Sal shows.(18 votes)
- Can a point have both removable discontinuity and vertical asymptote?(6 votes)
- No. A vertical asymptote is when a rational function has a variable or factor that can be zero in the denominator.
A hole is when it shares that factor and zero with the numerator.
So a denominator can either share that factor or not, but not both at the same time. Thus defining and limiting a hole or a vertical asymptote.(7 votes)
- How come he doesn't use the abc-formula to use a more certain way to find the numbers when simplifying the function?(5 votes)
- Because he just wanted to use another way of working out the values for x. Also, if he didn't factor, then he wouldn't get the part where he states that x=-4 is a removable discontinuity, but x=-6 is not.(3 votes)
- Just to Clarify, Will a removable discontinuity always contain an extraneous solution?(6 votes)
- A removable discontinuity always contain an extraneous solution.(0 votes)
- Sorry, this sounds like a dumb question but I'll ask anyway. If you have two expressions like in the video above, and they cancel each other out, then the number that makes those two equal to zero is the removable discontinuity.(4 votes)
- Yes, exactly. You can see the top answer for details:)(2 votes)
- What is a (Removable Discontinuity)?(2 votes)
- Also called a hole, it is a spot on a graph that looks like it is unbroken that actually has nothing there, a hole in the line. the simplest example is x/x. if you graphed it it would look like y=1, but if you tried to plug in 0 you would get undefined, so there is a hole at x=0, or a removable discontinuity.
Let me know if that doesn't make sense.(5 votes)
- couldn't it be argued that if you plug in 6 for x (second example in the video) it too would be undefined, and therefore a removable discontinuity?(3 votes)
- No. The function isn't undefined when x=6
(6-6)/(6+6) = 0/12 = 0
Hope this helps.(3 votes)
- At2:34, I tried graphing the function, but it didn't give me a little circle thing. Why would that exist anyways?(3 votes)
- i graphed it in desmos. it doesn't put a circle when y is undefined, but if you go to that point in x it says "(x, undefined)"(2 votes)
- [Voiceover] So we have this function, f of x, expressed as a rational expression here, or defined with a rational expression. And we are told at each of the following values of x, select whether f has a zero, a vertical asymptote, or a removable discontinuity. And before even looking at the choices, what I am going to do, because you're not going to always going to have these choices here. Sometimes you might just have to identify the vertical asymptotes, or the zeros, or the removable discontinuities, is I am going to factor this out. And hopefully, or I'm going to factor the numerator and the denominator to hopefully makes things a little bit more clear. And we are to think about what x values make the numerator and, or the denominator equal to zero. So the numerator, can I factor that? Well, let's see. What two numbers, their product is negative 24, and they add to negative two. See, that would be negative six and positive four. So we could write x minus six times x plus four. Did I do that right? Negative six times four is negative 24. Negative six x plus four x is negative two x. Yeah, that's right. Alright, and now in the denominator, let's see, six times four is 24, six plus four is 10. So we can say, six, x plus six times x plus 4. So let's look at the numerator first. The numerator zero if x is equal to positive six, or x is equal to negative four. The denominator is zero if x is equal to negative six, or x is equal to negative four. So, let's think about it. Can we simplify this, what we have up here a little bit? And I, in other words, keep in mind what I just wrote down. Well it might be, just jumping at it, you get an x plus four divided by an x plus four, why can't we simplify this expression a little bit? And just write this whole thing, write our original expression as being x minus six over x plus six, where x, if we want it to be algebraically equivalent we have to constrain the domain, where x cannot be equal to negative four. Well this is interesting because before, when x equals negative four, we had all this weird zero over zero stuff. Well we were able to remove it, and we were able to make it this kind of, this extra condition. And if you were to graph it, it would show up as a little bit of a removable discontinuity. A little bit of a point discontinuity, just one point, where the function is not actually, is actually not defined. So in this situation, and this is typical where there's a thing that you can factor out, it was making the numerator and the denominator equal zero but you can factor that out, so it no longer does it. That's going to be removable discontinuity. So x equals negative four is a removable discontinuity. And once you've factored out all the things that would make it a removable discontinuity, then you can think about what's going to be a zero and what's going to be a vertical asymptote. If it makes, once you've factored out everything that's common to the numerator and the denominator, if of what's left, something makes the numerator equal to zero, well, then it's going to make this whole expression equal to zero and so you're dealing with a zero, so x equals six. Six would make the numerator equal to zero. Six minus six is zero. So there you're dealing with a zero. And to make the denominator equal to zero, you would get x equals negative six. That would make the denominator equal zero so that's a vertical asymptote. It's called a vertical asymptote because as you approach this value, as approach negative six, either from values smaller than negative six or larger than negative six, the denominator is going to become either very, very, very, it's either going to become a very, very, very, very small positive number or a very, very, very small negative number, so we're going to approach zero either from above or below and so when you divide by that, you're going to get very large positive values or very large negative values. And so your graph does stuff like that. So you have a vertical asymptote. It might either do something like that or it might do something like this, where this is your vertical asymptote, and then when you approach from here, it goes, it bumps up like that. So either way, so that's why that's a vertical asymptote. Let's do another one. Alright, so same drill and like before, let, we'll always pause the video and see if you can work it out on your own. Let me factor it. So let's see, two numbers, product is negative 32. They're going to have different signs, eight, four. And then we want the large one to be positive since they have to add up to positive four x. So x plus eight times x minus four. Yep, I think that's right. And then divided by, this is four times four is 16. Four, negative four plus negative four is negative eight so this x minus four times x minus four. Now this is really, really interesting because you might say, oh, I have an x minus four x minus four, maybe, x equals four is a removable discontinuity. And it would have been unless you had, if you didn't have this over here, because x equals four, even after you've factored this. You still can't, the expression, the function, is still not going to be defined when x equals four. And so this function is actually equivalent to x plus eight over x minus 4. I don't have to put that extra little constraint like I did before. I don't have to put, I don't have to put this little constraint that describes this removable discontinuity because the constraint is still there, after I've factored out, after I've cancelled out this x minus four with this x minus four. And so here, we've, this is actually an algebraically equivalent expression to this one right over here. And so now, we can think about what the zeros or the vertical asymptotes or the removable discontinuities are going to be. Well, something that makes the numerator zero without doing it to the denominator is going to be a zero. And x equals negative eight makes the numerator equal to zero without doing anything to, without making the denominator equal to zero. It would make the denominator to negative 12. So you could literally just evaluate h of negative eight is equal to zero over negative 12, which is equal to zero. So that's why we call it a zero. What about x equals four? Well, x equals four is going to make the denominator only equal to zero, so that's going to give us a vertical asymptote. And we're all done.