## Could someone explain this to me? (continuous function)

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pietertje
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### Could someone explain this to me? (continuous function)

I have a calculus final coming up in a few days and I know for a fact that an assignment like the one below is going to be in it. The problem is that I have no idea how to solve it, I've literally spend 2 hours thumbing through my calculus textbook trying to find an example or something similar on how to solve it, but I found nothing.

So if someone could explain it to me, or show me how to solve it, I'd be really grateful.

$f(x) = \left\{ \begin{array}{ll} c^2 - x^2 & \mbox{if x < 0};\\ 2(x-c)^2 & \mbox{if x \geq 0}.\end{array} \right.$

If it is of any relevance, I'm using Calculus early transcendentals seventh edition by Edwards & Penney

firechicago
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### Re: Could someone explain this to me? (continuous function)

It would be very helpful if you could post the instructions for the problem. As it is, it's not clear whether you're supposed to find the solutions to the function in terms of c, find the value(s) of c that make the function continuous, assume the function is continuous and then find its solutions or do something else entirely.

pietertje
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### Re: Could someone explain this to me? (continuous function)

firechicago wrote:It would be very helpful if you could post the instructions for the problem. As it is, it's not clear whether you're supposed to find the solutions to the function in terms of c, find the value(s) of c that make the function continuous, assume the function is continuous and then find its solutions or do something else entirely.

Oops, forgot to add that; Find a value for constant c so that the function f(x) is continuous for all of x

Surg
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### Re: Could someone explain this to me? (continuous function)

Hint: For the two functions to be continuous, seeing that they're both independently continuous, they only need to agree at the point they "join" at (x = 0)

pietertje
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### Re: Could someone explain this to me? (continuous function)

Surg wrote:Hint: For the two functions to be continuous, seeing that they're both independently continuous, they only need to agree at the point they "join" at (x = 0)

I get c=0;

my method:
$c^2-x^2 = 0$
x = c or x = -c

substituting x for -c

$2(x-c)^2 = 0$$2(-2C)^2 = 0$$8c^2=0$$c=0$

Does that make any sense?

jestingrabbit
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### Re: Could someone explain this to me? (continuous function)

Your answer is correct, but your reasoning is unclear or wrong. You appear to have assumed that f(0)=0. What you want is to assume
$\lim_{x\to 0^-} f(x) = \lim_{x\to 0^+} f(x)$
as this will guarantee continuity at the one value of x where there might be a problem.
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firechicago
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### Re: Could someone explain this to me? (continuous function)

[hint]The answer c = 0 is also incomplete. There's no particular reason that the only place that the two sides of the function can meet is at f(0) = 0 [/hint]

Mindworm
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### Re: Could someone explain this to me? (continuous function)

firechicago wrote:[hint]The answer c = 0 is also incomplete. There's no particular reason that the only place that the two sides of the function can meet is at f(0) = 0 [/hint]

c²=2(-c)² has only one solution.
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Eastwinn
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### Re: Could someone explain this to me? (continuous function)

EDIT: Nevermind, I was waaay confused about the question. Too early in the morning to be reading math.
Last edited by Eastwinn on Wed Dec 22, 2010 5:50 pm UTC, edited 3 times in total.
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Yakk
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### Re: Could someone explain this to me? (continuous function)

Eastwinn wrote:They need to have the same value at x=0, yes, but there is something else they should share at x=0 as well. It's possible for them to meet there but not be continuous, making the answer "no solution".
The above is incorrect in this case.
Remember that differentiability implies continuity...
The above is correct, yet irrelevant. A->B does not mean B->A.
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Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

firechicago
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### Re: Could someone explain this to me? (continuous function)

Mindworm wrote:
firechicago wrote:[hint]The answer c = 0 is also incomplete. There's no particular reason that the only place that the two sides of the function can meet is at f(0) = 0 [/hint]

c²=2(-c)² has only one solution.

Wait, you mean 2x2 is different from 2x2x2?

Right, that's what I get for trying to do algebra in my head after getting up early to drive my gf to work. Or to put it the way my high school Calc teacher did whenever he added 2 and 2 and got 5: "It's always the higher math that gets me."

Eastwinn
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### Re: Could someone explain this to me? (continuous function)

EDIT: See above post. OP, ignore this discussion.
Last edited by Eastwinn on Wed Dec 22, 2010 5:44 pm UTC, edited 1 time in total.
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Yakk
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### Re: Could someone explain this to me? (continuous function)

Eastwinn:
Spoiler:
No. You do not need differentiability to generate continuity. Period. Stop confusing the OP.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Dark Avorian
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### Re: Could someone explain this to me? (continuous function)

True, yet I would suspect that if this is on a calc exam they are almost certainly expecting him to also make it fully differentiable.

(I'm not sure)

In this case... it is just by virtue of making c=0, but in the actual exam it might require more jiggering or have a second constant to solve for.
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gorcee
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### Re: Could someone explain this to me? (continuous function)

Dark Avorian wrote:True, yet I would suspect that if this is on a calc exam they are almost certainly expecting him to also make it fully differentiable.

(I'm not sure)

In this case... it is just by virtue of making c=0, but in the actual exam it might require more jiggering or have a second constant to solve for.

Maybe yes, maybe no.

The math education system in the US (assuming the OP is American) is pretty shitty, so topics like continuity are often skipped in pre-calculus education, and so get covered in a calculus course instead.

Yakk
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### Re: Could someone explain this to me? (continuous function)

original poster, clarifying original question wrote:Find a value for constant c so that the function f(x) is continuous for all of x

The student was not told to find a c such that it is differentiable (or continuously differentiable) for all x. So stop telling the original poster to do it.

Please, please stop telling the original poster to answer the wrong question. Please, please stop justifying it. I don't care if you hate the American education system. Unless you are trying to harm the original poster, don't confuse the original poster.

---

In any case, for a function to be continuous, all you need is that the limit as you approach each point equals the value of the function at that point.

When you "stitch together" functions like that, away from the stitching point the continuity of the "stitched together" function remains the same as the functions you stitched it together from.

So to answer this question, you'd first want to make sure that (c^2 - x^2) for x < 0 is continuous (do you know how to justify that?) Is there any limitation of what c can be, and this function on that domain is continuous?

Then you'd want to justify that 2(x-c)^2 is continuous for all x>=0 (do you know how to justify that?)

The next step is to deal with the "stitching point" -- namely, x = 0 and surrounds. A number of posters have talked about left and right limits being equal -- did you understand that? Ignore anyone who talked about differentiability (they are busy trying to answer a different question).
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Dark Avorian
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### Re: Could someone explain this to me? (continuous function)

Yeah okay. So depending on level of rigor you'll either just be wanting to assume that polynomials are continuous, or you could do delta-epsilon.

Then you just need to get the limits to exist at 0 from both sides.

lim x-> 0- f(x) = f(0) = lim x -> 0+ f(x)

Again, you may just be able to say that because polynomials are continuous this is continuous, but if the course is at a certain level of rigour you may be expected to do a delta epsilon proof.
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gorcee
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### Re: Could someone explain this to me? (continuous function)

Yakk wrote:
original poster, clarifying original question wrote:Find a value for constant c so that the function f(x) is continuous for all of x

The student was not told to find a c such that it is differentiable (or continuously differentiable) for all x. So stop telling the original poster to do it.

Please, please stop telling the original poster to answer the wrong question. Please, please stop justifying it. I don't care if you hate the American education system. Unless you are trying to harm the original poster, don't confuse the original poster.

Woah, let's all take a deep breath here. There are a lot of decaffeinated brands on the market today that are just as tasty as the real thing. If you take a deep breath and read what I wrote, I was arguing that just because this is being covered in a calculus course, we shouldn't jump to the conclusion that it somehow magically includes differentiability. It might, although the OP didn't specifically say this.

The answers in this thread drifted off track a bit, partially due to misinterpretation, and partially because the OP didn't frame the question very well (this isn't a bad thing, OP, so please don't take this statement personally; it's not intended to be an insult). But let's summarize the results.

To show continuity everywhere of a piecewise function, you must show:

1.) That the first part of the function (call it part A) is continuous everywhere on its domain.
2.) That the second part of the function (part B) is continuous everywhere on its domain.
3.) That the limit from the left of part A equals the limit from the right of part B at the "stitching point," which is where the domains of part A and part B meet.

I think that this explanation is sufficiently clear. Problem solved. If the OP does not know how to take limits, then s/he can ask about that. I think it should be safe to assume that he knows how to do this.

Now that we've solved the original problem in a clear and unambiguous manner, I think it's safe to discuss the next step, which for the sake of this discussion involves a discussion of continuous differentiability. Maybe such a discussion will answer the OP's next question (the OP might not even know that s/he needed to ask such a thing!). As long as we're being clear in the context of our discussion, I don't think it's harmful at all. We do it in other threads all the time. I think it would be helpful to do it in "homework" threads as well, because often times the question being asked by the OP isn't the same thing as the question that the OP needs answered. Again: only as long as we're being clear about the context.

gmalivuk
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### Re: Could someone explain this to me? (continuous function)

gorcee wrote:Woah, let's all take a deep breath here. There are a lot of decaffeinated brands on the market today that are just as tasty as the real thing.
Cut the patronizing shit, please. Isn't it possible Yakk wasn't replying directly to you?
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gorcee
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### Re: Could someone explain this to me? (continuous function)

gmalivuk wrote:
gorcee wrote:Woah, let's all take a deep breath here. There are a lot of decaffeinated brands on the market today that are just as tasty as the real thing.
Cut the patronizing shit, please. Isn't it possible Yakk wasn't replying directly to you?

Considering that I'm the only one in the thread that mentioned anything about the quality of the educational system, I'm pretty sure at least part of his response was directed at me.

gorcee wrote:The math education system in the US (assuming the OP is American) is pretty shitty, so topics like continuity are often skipped in pre-calculus education, and so get covered in a calculus course instead.

Yakk wrote:Please, please stop justifying it. I don't care if you hate the American education system.

And since you apparently missed the movie reference: http://www.imdb.com/title/tt0089886/quotes

Yakk
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### Re: Could someone explain this to me? (continuous function)

Nope, still missed the reference.

And that line was addressed to you. The rest was more ... vaguely frustratingly aimed.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

gorcee
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### Re: Could someone explain this to me? (continuous function)

Yakk wrote:Nope, still missed the reference.

And that line was addressed to you. The rest was more ... vaguely frustratingly aimed.

Just to be clear, I wasn't trying to antagonize you. I was making a funny.

Now, the fact that you've never seen Real Genius... this is a serious offense. I might have to revoke your math license. This, coming from the guy who's never seen The Godfather.

Eastwinn
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### Re: Could someone explain this to me? (continuous function)

Yakk wrote:Nope, still missed the reference.

And that line was addressed to you. The rest was more ... vaguely frustratingly aimed.

Oops, looks like I caused a mess. Yesterday I read "continuous" as "smooth" and wrote "smooth" as "continuous". Suffice to say it was not one of my better word swaps. Sorry about the confusion, OP!
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### Re: Could someone explain this to me? (continuous function)

gorcee wrote:Now, the fact that you've never seen Real Genius... this is a serious offense.
I have seen it, but I didn't think it was that great and I still didn't get the reference.
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### Re: Could someone explain this to me? (continuous function)

Eastwinn wrote:Oops, looks like I caused a mess. Yesterday I read "continuous" as "smooth" and wrote "smooth" as "continuous". Suffice to say it was not one of my better word swaps. Sorry about the confusion, OP!

There is no choice of c which makes the function smooth anyways.
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Eastwinn
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### Re: Could someone explain this to me? (continuous function)

skeptical scientist wrote:There is no choice of c which makes the function smooth anyways.

That's true, c=0 only earns it once-differentiability at x=0. Right? (I'm fairly new to the more rigorous aspects of calculus, I'm just taking it this year at school. Edit: In fact, I shouldn't have even been trying to answer this question. Misplaced overconfidence.)
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vilidice
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### Re: Could someone explain this to me? (continuous function)

Actually, I'm certain that it's only once differentiable at the point x=0, it has to do with analysis stuff at that juncture; however picturing the way the limit works out it my head;the derivative ends up being a modified absolute value function, which is not differentiable at 0. That having been said the derivative is continuous, and as the derivative exists, and isn't undefined at the point it's a safe bet to say that the function is sufficiently smooth as well, but that's just happenstance because it's two polynomials which are linearly dependent (one can be expressed as a constant multiple of the other).
$-x^2 = -1/2(2x^2)$
$2x^2 = -2(-x^2)$

(note that this will always only have a first derivative, unless the two are identical)

In nearly every case continuity stitching like this won't produce a smooth or differentiable function, in fact, I haven't actually seen a general proof of the polynomial thing, but I can picture the proof for the even ones a bit, so I wouldn't even assume that much unless someone else comments to the affirmative.