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### Course: Differential Calculus>Unit 1

Lesson 12: 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?
• 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).
https://en.wikipedia.org/wiki/History_of_mathematical_notation
• Do people use (a,b] to describe an interval which involves all the points between a and b and the point a?
• 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.
• 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!
• interval (-2,1) why does continuity exist? we have (-2,0) and (-2,-3)
• 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
• 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.
• 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).
• What do the shaded and unshaded endpoints mean again in algebraic form?
• 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
• What about the open interval (3,4) in the graph shown? Would it be continuous with the asymptote?
• Yes it would still be continuous because in that interval, 4 is excluded. However, as it approaches 4, the number will get extremely large, and only get larger and larger the closer you get to 4. If you tried to include 4 as part of the interval (3,4], then it is discontinuous at 4.
• If the interval has one end in an asymptote, like the interval (2,4), is the function still continuous?
• Yes, the function is continuous on the open interval, but not on the corresponding closed interval [2, 4].