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### Course: Precalculus > Unit 10

Lesson 11: Confirming continuity over an interval# Continuity over an interval

A function ƒ is continuous over the open interval (a,b) if and only if it's continuous on every point in (a,b). ƒ is continuous over the closed interval [a,b] if and only if it's continuous on (a,b), the right-sided limit of ƒ at x=a is ƒ(a) and the left-sided limit of ƒ at x=b is ƒ(b).

## Want to join the conversation?

- A long time ago, when I was young, I was taught to designate "if and only if" with "iff." When did this double headed symbol which is not easy to draw displace the designation I learned in school?(19 votes)
- I use "iff" when explaining a logical biconditional
*in words*. However, in logic and mathematics, when you are not actually explaining something but writing work out you use symbols. For example, using words, I could say "𝑄 if and only if 𝑃". Or, I could write this symbolically as:

𝑃 ⇔ 𝑄

It's similar to how I could write out: "𝑥 equals three" or simply "𝑥 = 3".

As such, I am sure that the double-sided arrow biconditional operator (the symbol itself) was probably developed after logicians first studied biconditionals themselves (as formal notation is developed far after the concepts themselves are developed). However, this still means a couple centuries ago (notation is pretty old by now).

You can read more here:

https://en.wikipedia.org/wiki/History_of_mathematical_notation(17 votes)

- Do people use (a,b] to describe an interval which involves all the points between a and b and the point a?(4 votes)
- (a, b] would correspond to a < x ≤ b. The interval would NOT include a.

https://www.khanacademy.org/math/algebra/algebra-functions/domain-and-range/v/introduction-to-interval-notation

@6:40(13 votes)

- How can you prove that a function is continuous over every point in an interval? I get the concept--no picking up pencils--but I don't know how to prove it properly.(4 votes)
- You need to show that for every r > 0 there exists a d > 0 such that |x-y| < d => |f(x) - f(y)| < r.

Example, let f(x) = x. Then for any r > 0 pick r = d then |x - y < d = r => |f(x) - f(y)| = | x - y| < r.

Example, let f(x) = x^2. Then for any r > 0 observe that |x^2-y^2| = |x-y|*|x+y|. So if |x-y| < r/(1 + |x - y) then |x^2 - y^2| < r. (Note the use of 1/(1+ ..) so that if x = y I don't divide by zero).

If the function is differentiable, then it will be continuous on your open interval!(4 votes)

- interval (-2,1) why does continuity exist? we have (-2,0) and (-2,-3)(3 votes)
- Interval (-2,1) is an
**OPEN**interval and therefore end-points -2,1 are NOT included. Parentheses or round brackets () are used to show OPEN interval. -2<f(x)<1

CLOSED interval would be written like this: [-2,1] and the end-points ARE included in the interval. -2≤f(x)≤1(2 votes)

- For the graph that Sal gives, would the interval between (2, 4) or (2, 4] be continuous?

At least for the first interval (2, 4), technically you wouldn't have to "pick up your pencil" when drawing the graph. However, the limit as x approaches 4 doesn't exist, so it kind of contradicts itself when determining if the interval is continuous or not.(2 votes)- The function is continuous on (2, 4), but not on (2, 4]. The limit as x goes to 4 doesn't exist, but that doesn't matter because 4 isn't in the interval (2, 4).(3 votes)

- What do the shaded and unshaded endpoints mean again in algebraic form?(2 votes)
- The closed circle means that the function is defined at that point (in other words, that x value is in the function's domain). An open circle means that the function isn't defined.

So,I can say the domain of this function is (-infinity, 4) U (4, infinity). See that I used () to signify that the endpoints aren't in the domain. If they were in the domain, I'd use [] instead(3 votes)

- Why there are not lessons in philosophy and logic , especially mathematical logic and computational logic(1 vote)
- Okay so there is a function f(x) = 1/x and we see that the function is not defined at x=0

but why does it make it discontinuous at that point?(2 votes)- The way I understand it, that question depends on the domain under consideration. Since f(x) is not defined at x=0, it could be argued that the domain is (-∞,0) U (0,∞) in which case 0 is not part of the domain and the function is continuous over its domain. However, if you consider the domain to be all real numbers, it is not continuous. To be continuous at a point (say x=0), the limit as x approaches 0 must equal to the actual function evaluated at 0. The function f(x)=1/x is undefined at 0, since 1/0 is undefined. Therefore there is no way that the f(0) = lim x->0 f(x).(1 vote)

- how is x continuous at x=-2 when the function clearly doesn't follow all three steps of the continuity definition? The lim has to exist at a point for the function to be continuous at that point.(2 votes)
- Watch the video again closely. Sal took the value of the function over the
**open**interval (-2,1), which does not include the jump in the value at -2.(1 vote)

- This video is VERY confusing with respect to closed intvl: in the practice I took, it emphasizes that EVERY pt including the end pts must be continuous, yet this video stipulates that continuity can be continuous at "a" coming from the Left "-", or "b" coming from the right "+" (see5:57in the 9 min video. It would seem to me that every pt makes more sense with respect to closed intvl.

By the way I too much prefer the designation iff for if and only if, but then at the ripe young age of 70, I took Calc 45 yrs ago.

Sincerely Pete

.(2 votes)- yeah I got the same issue on the quiz I was taking. But referring to the video it said as you said... very confused :((1 vote)

## Video transcript

- [Instructor] What we're
going to do in this video is explore continuity over an interval. But to do that, let's refresh our memory about continuity at a point. So we say that f is continuous when x is equal to c, if and only if, so I'm gonna make these
two-way arrows right over here, the limit of f of x as x approaches c is equal to f of c. And when we first introduced this, we said, hey, this looks
a little bit technical, but it's actually pretty intuitive. Think about what's happening. The limit as x approaches c of f of x, so let's say that f of x as x approaches c is approaching some value. So if we approach, if we approach from the left, we're getting to this value. If we approach from the right,
we're getting this value. Well, in order for the
function to be continuous, if I had to draw this function
without picking up my pen, well, the value of the
function at that point should be the same as the limit. This is really just a more
rigorous way of describing this notion of not having
to pick up your pencil, this notion of connectedness, that you don't have any jumps or any discontinuities of any kind. So with that out the way, let's discuss continuity over intervals. Let me delete this really fast, so I have space to work with. So we say, so I'm gonna first
talk about an open interval, and then we're gonna talk
about a closed interval because a closed interval gets
a little bit more involved. So we say f is continuous over an open interval from a to b. So the parentheses instead of brackets, this shows that we're not
including the endpoint. So this would be all of the
points between x equals a and x equals b, but not equaling x
equals a and x equals b. So f is continuous over
this open interval, if and only if, if and only if, f is continuous, f is continuous over every point in, over every point in the interval. So let's do a couple of examples of that. So let's say we're talking
about the open interval from negative seven to negative five. Is f continuous over that interval? Let's see, we're going from negative seven to negative five, and there's a couple of
ways you could do it. There's the
not-so-mathematically-rigorous way, where you could say, hey,
look, if I start here, I can get all the way to negative five without having to pick up my pencil. If you wanted to do more rigorously and you actually had the
definition of the function, you might be able to do a proof, that for any of these
points over the interval, that the limit as x approaches
any one of these points of f of x is equal to
the value of the function at that point. It's harder to do when
you only have a graph. When you only have a graph, you can only just do it by inspection, and say, okay, I can go from
that point to that point without picking up my pencil, so I feel pretty good about it. Now let's do another interval. Let's say the, so let me
put a check mark here, that is continuous. Let's think about the
interval from negative two to positive one, the open interval. So this is interesting
because the function at negative two is up here. And so if you really wanted
to start at negative two, you would have to start here
and then jump immediately down as soon as you get slightly
larger than negative two and then keep going. But this is an open interval, so we're not actually concerned with what exactly happens at negative two, we're concerned what happens when we are all the numbers larger than negative two. So we would actually
start right over here, and then we would go to one. And once again, based on the intuitive I didn't have to pick up my pen idea, this function would be continuous over this, over this interval. So what's an example of an interval where the function
would not be continuous? Well, think about the interval from, well, this is a pretty
straightforward one, the open interval from three to five. The function is here
when x is equal to three. But if we wanted to get to five, it looks like we're asymptoting, it looks like we're
asymptoting up towards infinity and we just keep on going
for a very long time. And then we would have to pick
up our pencil and jump over, and then we would come
back down right over here. And so here we are not
continuous over that interval. So now let's think about the more, the slightly more involved interval. The slightly more involved case is when you have a closed interval. F is continuous over the closed interval from a to b. So this includes not just
the points between a and b, but the endpoints as well, if and only if, f is continuous over the open interval and the one-sided limits. Let me right this. And the limit as x approaches a from the right of f of x is equal to f of a, and the limit as x approaches b from the left, from the left of f of x is equal to f of b. Now what's going on here? Well, it's just saying
that the one-sided limit, when you're operating within the Interval, has to approach the same
value as the function. So for example, if we
said the closed interval from negative seven to negative five, well, this one is still reasonable, you know, just based on the
picking up your pencil thing. You don't have to pick up your pencil. And what you would do is at the endpoint, and at negative seven, this function is just
plain old continuous, but if it wasn't defined over here, it could still be continuous
because you would do the right-handed limit towards it. And you'd say, okay,
the right-handed limit is equal to the value of the function. And then at this endpoint,
at the second endpoint, you'd say, okay, the left-handed limit is equal to the function, even
if it wasn't defined here, even if the two-sided
limit were not defined. And so we could actually
look at an example of that. If we were looking at the interval from the closed, and you could have one
side open, one side closed, but let's just do the closed
interval from negative three to negative two. So notice I did not have
to pick up my pencil. I'm including negative three, and I'm getting all the
way to negative two. If you knew the analytic
definition of this function, you could prove that, hey, the
limit at any of these points inside, between negative
three and negative two, is equal to the value of the function. Negative three, the function
is clearly, at negative three, the function is just plain old continuous. The two-sided limit approaches
the value of the function. But at negative two, the
two-sided limit does not exist. When you approach from the left, it looks like you're approaching zero. F of x is equal to zero. When you approach from the right, it looks like f of x is
approaching negative three. So even though the two-sided
limit does not exist, we can still be good because the left-handed limit does exist. And the left-handed limit is approaching the value of the function. So we actually are continuous
over that interval. But then if we did the interval, if we did the closed
interval from negative two to negative two to one, pause the video and think about, based on what we just talked about, are we continuous over this interval? Well, we're going from negative two to one, and negative two is the lower bound. So is this right over here, is this right over here true? Is the limit as we approach
negative two from the right, is that the same thing
as f of negative two? Well, the limit as we
approach from the right seems to be approaching negative three, and f of negative two is zero. So this limit does not, this, this, these two things, the limit
as we approach from the right and the value of the
function are not the same. And so we do not have
that, I guess you could say that one-sided (laughs)
continuity at negative two. And that also makes sense. If I start at negative two, let me do this in a color you can see, if I start at negative two and I want to go the rest
of the interval to one, I have to pick up my pencil. Pick up my pencil, go here,
and then keep on going. So this is, we are not
continuous over that interval.