Author Topic: [MATH] Calculus Review Packet  (Read 9025 times)

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Offline Newby

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[MATH] Calculus Review Packet
« on: October 24, 2006, 11:58:17 pm »
I need help.

Quote
If f(x1) + f(x2) = f(x1 + x2), for all real numbers x1 and x2, which of the following could define it?

A. f(x) = x + 1
B. f(x) = 2x
C. f(x) = 1 / x
D. f(x) = ex
E. f(x) = x2

The answer is B. But how would I prove that? I tried plugging in infinite for x1 and x2, and got f(infinite + infinite) which = f(2*infinite), which could be f(2x); is that right? Or are my math skills shit? :p

EDIT -- After looking over my review packet, I realized plugging in random numbers was not this one; it was some limit that I was not sure about.
« Last Edit: October 25, 2006, 12:00:47 am by Newby »
- Newby
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[17:32:45] * xar sets mode: -oooooooooo algorithm ban chris cipher newby stdio TehUser tnarongi|away vursed warz
[17:32:54] * xar sets mode: +o newby
[17:32:58] <xar> new rule
[17:33:02] <xar> me and newby rule all

I'd bet that you're currently bloated like a water ballon on a hot summer's day.

That analogy doesn't even make sense.  Why would a water balloon be especially bloated on a hot summer's day? For your sake, I hope there wasn't too much logic testing on your LSAT. 

Offline Sidoh

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Re: [MATH] Calculus Review Packet
« Reply #1 on: October 25, 2006, 12:04:48 am »

Offline Chavo

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Re: [MATH] Calculus Review Packet
« Reply #2 on: October 25, 2006, 09:42:32 am »
f(infinite + infinite) which = f(2*infinite), which could be f(2x);
2*infinity is just infinity
« Last Edit: October 25, 2006, 12:35:00 pm by unTactical »

Offline Sidoh

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Re: [MATH] Calculus Review Packet
« Reply #3 on: October 25, 2006, 11:11:06 am »
You guys both suck at spelling infinity.

Offline Chavo

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Re: [MATH] Calculus Review Packet
« Reply #4 on: October 25, 2006, 12:35:14 pm »
who what where?  :-*

Offline Sidoh

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Re: [MATH] Calculus Review Packet
« Reply #5 on: October 25, 2006, 12:35:42 pm »
who what where?  :-*

lol.  cheating bastard.

Offline Newby

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Re: [MATH] Calculus Review Packet
« Reply #6 on: October 25, 2006, 09:05:39 pm »
f(infinite + infinite) which = f(2*infinite), which could be f(2x);
2*infinity is just infinity

Yeah. Today I realized it was dumb.

And it turns out I had to pick through the answers and try each one. so eh! :)
- Newby
http://www.x86labs.org

Quote
[17:32:45] * xar sets mode: -oooooooooo algorithm ban chris cipher newby stdio TehUser tnarongi|away vursed warz
[17:32:54] * xar sets mode: +o newby
[17:32:58] <xar> new rule
[17:33:02] <xar> me and newby rule all

I'd bet that you're currently bloated like a water ballon on a hot summer's day.

That analogy doesn't even make sense.  Why would a water balloon be especially bloated on a hot summer's day? For your sake, I hope there wasn't too much logic testing on your LSAT. 

Offline Sidoh

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Re: [MATH] Calculus Review Packet
« Reply #7 on: October 25, 2006, 09:18:30 pm »
Not to mention , and are undefined. ;)

Offline dark_drake

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Re: [MATH] Calculus Review Packet
« Reply #8 on: October 25, 2006, 11:57:20 pm »
Not to mention . . . undefined.
Isn't it 0? I only ask because that's what we do in calculus.
errr... something like that...

Offline Chavo

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Re: [MATH] Calculus Review Packet
« Reply #9 on: October 26, 2006, 12:11:26 am »
Not to mention . . . undefined.
Isn't it 0? I only ask because that's what we do in calculus.
Pretty sure its undefined.  Back when I took calc2, we always used 'no solution' as the answer to any problem that winded up with a 1/infinity in it.

Offline dark_drake

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Re: [MATH] Calculus Review Packet
« Reply #10 on: October 26, 2006, 12:50:39 am »
Not to mention . . . undefined.
Isn't it 0? I only ask because that's what we do in calculus.
Pretty sure its undefined.  Back when I took calc2, we always used 'no solution' as the answer to any problem that winded up with a 1/infinity in it.
According to wikipedia, .
errr... something like that...

Offline Rule

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Re: [MATH] Calculus Review Packet
« Reply #11 on: October 26, 2006, 01:08:52 am »
?

So if you were asked what the limit of a/x is, as x --> infinity, you would write 'undefined'?  That's horrible, and wrong: your teacher should be disciplined :P. A constant over infinity is definitely zero, and even in rigorous mathematics we sometimes define 0 * infinity = 0 (and this is certainly less obvious).

Offline Chavo

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Re: [MATH] Calculus Review Packet
« Reply #12 on: October 26, 2006, 01:12:11 am »
I must be thinking of something else.  I'm not a math nerd like certain other individuals around these parts :P

Offline Sidoh

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Re: [MATH] Calculus Review Packet
« Reply #13 on: October 26, 2006, 02:08:22 am »
?

So if you were asked what the limit of a/x is, as x --> infinity, you would write 'undefined'?  That's horrible, and wrong: your teacher should be disciplined :P. A constant over infinity is definitely zero, and even in rigorous mathematics we sometimes define 0 * infinity = 0 (and this is certainly less obvious).

No, I would write 0.  Does imply that the denominator is some arbitrary variable approaching infinity, then?

Offline Rule

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Re: [MATH] Calculus Review Packet
« Reply #14 on: October 26, 2006, 02:28:21 am »
?

So if you were asked what the limit of a/x is, as x --> infinity, you would write 'undefined'?  That's horrible, and wrong: your teacher should be disciplined :P. A constant over infinity is definitely zero, and even in rigorous mathematics we sometimes define 0 * infinity = 0 (and this is certainly less obvious).

No, I would write 0.  Does imply that the denominator is some arbitrary variable approaching infinity, then?

It depends.  Most people who write 1/infinity mean more precisely lim y--> infinity 1/y, so the answer is "yes" in most cases.  However, I think saying "yes" for all cases would be over-restrictive, unless we add all sorts of conditions about the "rate" of approaching infinity, etc.  Regardless of whether you think of infinity as some 'closed entity' (for lack of a better term) or not, (and some would argue that this is a bad interpretation, although I think it does have some merit), we can see that in this particular case 1/'infinity' and lim x--> infinity 1/x are equivalent, as both equal zero.