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## APยฎ๏ธ Calculus BC (2017 edition)

### Course: APยฎ๏ธ Calculus BC (2017 edition)ย >ย Unit 1

Lesson 5: Continuity

# 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 "๐ if and only if ๐". Or, I could write this symbolically as:
๐ โ ๐
It's similar to how I could write out: "๐ฅ equals three" or simply "๐ฅ = 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. -2โคf(x)โค1
• 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
• 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?
• 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 (-โ,0) U (0,โ) 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.
• 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 "+" (see in 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

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