Can you solve this 'simple' calculus problem?

Can you solve this calculus problem that even stomped a math PhD? The problem is: find all the x such that the function f(x) above is discontinuous. Can you find all of them? Watch this video and find out

[blackpenredpen]

Oooh yea I know this problem and that professor hahaha

[thedoublehelix]

Rational functions always set off alarm bells in my head because dividing by 0

[williamadams]

I could immediately figure out x = 1 and x = 2 are discontinuous points, but I missed x = 0. 🥲

[benjaminmordaunt]

Usually a fan of these vids, but that seemed blatantly obvious - especially for a 6 min vid :/ I was expecting something strange to appear.

[ucr]

If I were a Calculus 1 or an AP Calc AB professor/teacher, I would put this problem on a Calculus 1 midterm. Heck, you do not even need Calculus to solve this actually simple Calculus problem. All you need is algebra to solve this problem. That being said, Calculus 1 and beyond is basically where someone's algebra and trigonometry skills could fail them. If they have not mastered algebra and trigonometry at a high level, then they will struggle with Calculus 1 and beyond (particularly Calculus 2, Calculus 3, Linear Algebra and Differential Equations). My advice to a student who is planning to take Calculus 1 and beyond in the future is to brush up on their algebra (particularly with finding a function's domain and with logarithms and sequences and series), and trigonometry (particularly with trigonometric identities and formulas such as the law of sines and cosines) skills. Calculus 1 is not that difficult compared to higher levels of math but it is very important to learn in anyone's STEM (particularly with Mathematics, Engineering and Physics) career.

[TheVianzo]

Okay i'm 0:40 into my video,
My bet is it's discontinuous in 0, 1 AND 2.

[Jaeghead]

I disagree with the given solution. Like some other commenters have pointed out, it doesn't make sense to talk about (dis)continuity at a point outside the domain. The function x |-> 1/x on its natural domain is continuous, it doesn't have a discontinuity at 0.

[cepatwaras]

I immediately see it is not continuous at X = {1, 2} but I failed to see it is also not continuous at X = 0. If we worked together in exam, we would help correct each other 😄

[bogdanenescu]

At a point where a function is not defined it is meaningless to check continuity. Hence saying that the given function is not continuous at 1 and 2 is simply wrong.

[nournote]

Your professor seems not to have studied formal proofs and axiomatic math.
One should do it by the book.
How do you define point of discontinuity of a function?
As far as I can remember my maths studies (from 20 years ago), there is no such definition. On the contrary there is a defintion of continuity on point. The latter implies (1) function DEFINED on that point and (2) the limit on that point exists and is equal to the value of the funtion on that same point.

All in all, it's all a matter of rigour-lacking than a 'tricky/show-off questio' by a professor.

[wgregor]

The only point of discontinuity is 0. At the points 1 and 2, f is simply udefined. Continuity and discontinuity are properties that apply only to points in the domain of f. At a point where a function is not defined, we cannot ask whether the function is continuous or discontinuous, because the function simply does not exist there. Thus, if we had g(x)=sqrt(1-x^2), we would not say that g is discontinuous at x=-2, we just say that it is not defined there. Now, if the given function f was defined somehow at x=1 and x=2, these would become points of discontinuity. But as defined, the points x=1 and x=2 are not points of discontinuity, just points outside of the domain of f (just like x=-2 is outside of the domain of g).

[mdperpe]

Can a function be discontinuous at a point where it is not even defined?

The topological definition continuity of f : X -> Y is that f^{-1}(V) is open in X whenever V is open in Y.
By this definition, f(x) = 1/x, x != 0, is continuous. If f is continuous, can f really have a discontinuity?

[garyhuntress]

for once I got one right straight outta the box!!

[malaymukherjee]

Big fan of yours, Dr. Peyam!!! You and bprp are two of my greatest inspirations for doing maths😊😊😊

[MichaelRothwell]

Dear Dr. Peyam, you were right the first time - the function is continuous everywhere in its domain except for x=0, and the notion of continuity/discontinuity only applies to points in the domain of a function. Indeed, the two branches are elementary functions, and so continuous in their natural domains. The question is misleadingly set, as it suggests that the domain of the function is R, when in fact it is R\{1, 2}. For the function actually to be defined on R, it would have to be given values at 1 & 2, and in this case whatever values given we would get a discontinuous function. To say the function is "discontinuous" at x=1 & x=2 is just as wrong as to say 1/x is discontinuous at x=0.

[JohnSmith-rftx]

I can guess why so many of the students erred on this question. It's because there are 2 different ways to make a careless mistake if your brain is sort of on autopilot. One group of students will have done exactly what Dr. Peyam did. Recognize the obvious necessity to check for continuity at x=0, but be so locked into that idea that they forget to check that the function is continuous elsewhere. The other group will have looked at it and immediately recognized that in the domain x>=0, the function is undefined where the expression in the denominator is equal to zero. But they got so focused on that part of the question forgot to check whether the function is continuous where the domain changes.

[user]

I am sensitive at these "obviously factor-able" polynomials, and so before reading the question I usually attempt to factorize them first ;) Then I noticed the (x-1)(x-2)
That's how I didn't get fooled by these questions lol

[Mercurc]

I've just been using the Residue theorem heaps, so finding singularities was the first thing I did haha

[otakurocklee]

But x=1 and x=2 are not part of the domain. f(x) does not exist at x=1 or x=2, so I don't see how we can say they it is continuous or discontinuous there. The function just does not exist at those points. Continuity or discontinuity don't apply.

[willFALL]

Ahh yes, always gotta factor them rational polynoniminals and find those pesky zeros 😎