Why ?

@Khallil Exactly.

Mathematics is discovered. Only fools think math was invented.

I've just got a lot of opposing opinions on math.

All from fools.

Math is discovered and that's final.

Both. We invent the basic rules, and then discover the consequences.

Mathematics is an eternal truth, not something found by some flim-flam jokers counting their fingers.

@DanielFischer Well. NOT 50-50, however.

Mostly discovered.

What kind of math would we be doing if our rules were slightly different?

@Khallil What is the rule you have in mind? Be explicit.

Are our rules concrete, or are they merely our own perceptions of what's going on?

@Khallil Make your slightly different rules, and take a look.

@Khallil I can see you lurking on the forums. What the devil do you have in mind? ;)

MHB, @BalarkaSen?

mmhmm

I don't even have it open!

hmm?

mathhelpboards.com/online.php?who=members

I just opened it, which explains why the last time I've been seen was a minute ago, but before then, I hadn't been online!

I see all. know all. eat all.

ლ(ಠ益ಠლ

Wat?

@Khallil I saw you looking at a challenge question on the forum.

Oh, that was the link you posted on here.

I had only looked at it for a few seconds, @BalarkaSen.

Enough to be caught in the act =P

who is that lady? eughh.

She's slim, young and beautiful, with long brown curly hair

:c

*She's the antipode of

I'd prefer the anti-iPod :D

I have to differentiate $$f(x) = \frac{x^3-2x^2-1}{2x^2-x-1}$$. I smell factoring. The denominator is no problem, but how should I factor the numerator? Anything above a power of 2 and I get a bit lost, sorry

@topper You can't.

$x^3 - 2x^2 - 1$ can't be factorized in integer polynomials.

@BalarkaSen So just plug away at a very long multiplication process on the derived numerator?

I never said that.

There might be partial factorization tricks involved.

@BalarkaSen I've never even heard of that. In the spirit of trying to help you to help me, the full question asks for a) the discontinuities and their nature and b) the number of times the function crosses the x-axis. Perhaps I can't see the forest for the trees, but I'm sure I need to differentiate it for part (a)

The discontinuities occur when the denominator of the rational function is equal to 0.

@topper you don't need to differentiate a function to get the discontinuities!

FACEPALM

What about their nature, @BalarkaSen?

After this one, I'm going for a 25-minute power nap

I guess they want me to say if they are holes or asymptotes

Holes?

holes?

jinx

Let me find it in English.

$\text{Jinx}^{\text{Jinx}}$

"Point Discontinuities"

Oh, those.

Or "Jump Discontinuities"

Oh right, because they are empty circles on a sketch.

That one

Hat tip to math.brown.edu/utra/discontinuities.html

I believe all of the discontinuities are continuously attained.

i.e., "asymptotic discontinuities"

But as I recall (I have to check my notes), I don't need to differentiate to find the type of discontinuity. I think I need to check the limit as x tends to infinity or something

@topper mmhmm

You need to check if the limit from the left is equal to the limit from the right, right?

@Khallil What'd that imply?

Ack.

limits at asymptotics discontinuities might or might not have left limit = right limit

Yep, just thought of that.

Asymptotic discontinuities have infinite left and right sided limits, right?

wat

My mistake. If e.g. $x=2$ is a discontinuity, I have to check the limit of $f(x)$ tending towards $2$, and if it is infinity, it's a vertical asymptote

Like, $\tan(x)$. The limit as $x$ approaches $\frac{\pi}{2}$ from the left is $+\infty$, whereas from the right, the limit is $-\infty$, no?

Wouldn't that imply a vertical asymptote?

Yes.

That's what I meant by the question I had before.

Also, jump discontinuities have finite left and right sided limits, don't they?

Yes.

Yep, thought so. Couldn't get the thought out very clearly for a bit there.

$$ \begin{aligned} 2x^2 - x - 1 & = 2x^2 - 2x + x - 1 \\ & = 2x(x-1) + 1(x-1) \\ & = (2x+1)(x-1) \end{aligned}$$

$x = -1/2, 1$ have the discontinuities, yes.

Okay, progress. $x= -\frac12 or 1$. Now to check the limits

Cool @topper

Haha, same time. This bit might take me a while, appreciate you not posting the answer until I've at least tried it?

Agreed.

Mhm.

Okay, so first I tried substituting $1^+$, and got $\frac{somenumber}{0^+}$, which I believe tends to infinity, so I can say that $x = 1$ is a vertical asymptote

@topper I am not sure if that's enough. The function do behaves like $-2(x-1)^{-1}$ near $1^+$. What d'you think @Khallil?

I think the number on the top is negative.

Yes. $-2$

@topper Try inspecting the behavior at $x \to 1^{-}$

I have an idea of what's going on.

It behaves a bit like $- \left| \frac{1}{x} \right|$ around $x=1$

Also, @topper, that somenumber is improtant.

@Khallil Nah, man.

Quick question. When doing these substitutions with $x^+$ or $x^-$, is it correct that in the numerator I can substitute $x$ itself, but in the denominator I have to substitute with the plus or minus?

Otherwise $x = 1$ won't be a discontinuity.

@BalarkaSen Very imp ro tant indeed

No?

Oh, wait.

Gah, I can't even add and subtract anymore.

@topper It's better to think of limits as limits. Substituting stuffs around is dangerous.

Looking now at $x \to 1^-$

@BalarkaSen Believe me my friend, I am trying very hard

(to do so)

I understand =)

I have a result of $\frac{-2}{0^-}$. Now I'll wait for somebody to tell me I'm wrong :)

:)

The factorised form in the denominator really helped.

@topper You're right in one sense. Now what is this number?

i.e., negative or positive?

So what will $f(x)$ tend to as $x$ approaches $1$ from the left?

(Ninja'd by @BalarkaSen.)

I'm still assuming that $0^-$ is some number very close to $0$ and on it's left.

It definitely is.

It shouldn't be. It's an infinitesimal.

However, let's forget the formalities.

Um, tends to $-\infty$?

@topper Nope. What is the sign of $0^-$?

Yea, I didn't want to confuse (myself and) @topper.

Ah, they're both negative

@topper Yes, so the fraction will be?

What fraction?

Ah

Positive!

Precisely!

But what did you get for $x \to 1^+$?

Ah, that fraction is negative

Exactly.

So you have a vertical asymptote.

Would it not also be a vertical asymptote if both tend to $+\infty$?

@Khallil Actually, it'd.

I am just helping him state the nature of the discontinuities.

You didn't finish your first message!

"Actually, it'd ..."

Would it not be a vertical asymptote?

I actually finished it.

It would.

Oh, it would.

That's an odd use of it'd!

it's chat-use of contractions.

grammatically incorrect.

it's crack of a dawn here. i need to run. bye.

It's pretty late where I am too.

Good night!

Good luck with the prep, @topper!

@Khallil Thanks! Goodnight people

I just came up with an identity for an answer that I hadn't seen before. That makes my day :-)

@r9m: have you seen $(2)$ in this answer before?

It has the same feel as the identity I posted here earlier:

but it is more general

@robjohn this is very nice ! :-)