Can someone help me with calculus?

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mathgeek17
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Can someone help me with calculus?

Postby mathgeek17 » Tue Sep 11, 2012 10:53 pm UTC

So, my calc homework has the following problem:

lim (-3x2+5x+2)/|2-x|
x->2+

If you plug in 2, you end up with 0/0. So I decided to factor the quadratic into (x-2)(x+2/3). My memory is kind of fuzzy, but I remember doing something with the absolute values where you do the equation with the regular argument and then again with the negative argument. So I ended up with (x-2)(x+2/3)/(x-2) or (x-2)(x+2/3)/(2-x). Now, if you plug 2 back into those after canceling the (x-2)'s, you end up with 8/3 or -8/3. Now, looking at the graph, I know that the answer is -8/3. However, we're not supposed to use the graph to solve the problem. So, my question is how do I figure out whether 8/3 or -8/3 is correct without using the graph?

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Dopefish
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Re: Can someone help me with calculus?

Postby Dopefish » Tue Sep 11, 2012 11:03 pm UTC

Under what condition does |2-x|=x-2, and under what condition does |2-x|=2-x?

Related, notice also you're not taking the limit as x approaches 2, you're taking the limit as x approaches 2+, so you only need to consider one side of the limit.

mathgeek17
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Re: Can someone help me with calculus?

Postby mathgeek17 » Wed Sep 12, 2012 12:48 am UTC

OK. First off, I'm now realizing that I factored the quadratic wrong. It should be 1/3, not 2/3. But that only changes the +-8/3 to +-7/3. Anyway, since |2-x|=x-2 for all x>2 which is the same as 2+. So we use the expression with the 2-x in the denominator, leaving just -7/3 as the answer which agrees with the graph. Is this correct?

Deedlit
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Re: Can someone help me with calculus?

Postby Deedlit » Wed Sep 12, 2012 1:04 am UTC

Did you divide by 3 for some reason?

mathgeek17
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Re: Can someone help me with calculus?

Postby mathgeek17 » Wed Sep 12, 2012 1:12 am UTC

Are you referring to +-7/3? Because that's just what you get when you factor -3x2+5x+2.

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Dopefish
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Re: Can someone help me with calculus?

Postby Dopefish » Wed Sep 12, 2012 1:21 am UTC

Alright, I actually bothered to look at your factoring now, and I'm pretty sure you lost a factor when you factored the original quadratic.

mathgeek17 wrote:Anyway, since |2-x|=x-2 for all x>2 which is the same as 2+

I'm not sure I'd quite call it "the same", since the former is a statement about how aboslute values work, and the latter is (in this context) refering to approaching 2 from the positive side, but I think you've got the right idea. As you get arbitrarily close to 2 from the positive side, you consistantly stay in the region where |2-x|=x-2, so you can use that to cancel things and find the limit you're interested in.

mathgeek17 wrote:So we use the expression with the 2-x in the denominator...

This doesn't follow from what you said earlier, I'm pretty sure you mean you want to use x-2 because that's the expression thats relevant in calculating the limit from the positive side. I suspect you just took 2-x since it got you the right sign without thinking about it too much, but the negative aspect actually comes from the factor you lost earlier on. It's because of this sort of thing why you're probably better offer avoiding graphing things, as you may inadvertantly make leaps that don't actually follow from your work.
Last edited by Dopefish on Wed Sep 12, 2012 1:57 am UTC, edited 2 times in total.

Deedlit
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Re: Can someone help me with calculus?

Postby Deedlit » Wed Sep 12, 2012 1:22 am UTC

mathgeek17 wrote:Are you referring to +-7/3? Because that's just what you get when you factor -3x2+5x+2.


Try factoring it again. It should factor just fine in the integers.

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dudiobugtron
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Re: Can someone help me with calculus?

Postby dudiobugtron » Wed Sep 12, 2012 1:37 am UTC

mathgeek17 wrote:Are you referring to +-7/3? Because that's just what you get when you factor -3x2+5x+2.

Actually, it's what you get when you factor x2 - 5/3 x - 2/3
Factorising is different from finding the roots. Try expanding (x-2)(x+1/3) and see for yourself. ;)

Also, you can't just cancel (x-2) and (2-x), since they are different. You need to 'factorise' one of them a bit further first.
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mathgeek17
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Re: Can someone help me with calculus?

Postby mathgeek17 » Wed Sep 12, 2012 1:41 am UTC

OK. Maybe I shouldn't have said that -3x2+5x+2 factors into 7/3. Like I said in the original post, it's (x-2)(x+1/3) but I wasn't thinking and skipped to plugging in 2 after you cancel the (x-2)'s, if that clears up some of the confusion. If not, I have no idea what factor you guys are talking about. I tried using the quadratic equation, I tried redistributing the two factors again, and I had my calculator solve the quadratic for me and I still got 2 and -1/3. Is there something really obvious that I'm just missing?

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Dopefish
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Re: Can someone help me with calculus?

Postby Dopefish » Wed Sep 12, 2012 1:52 am UTC

How are you getting the -3x2 part from (x-2)(x+1/3)? The coefficient of the x2 term you get from that is 1, rather than -3.

Really, just multiply out (x-2)(x+1/3) and you'll find you get something that differs from -3x2+5x+2. It's 'close', but it's off by a factor.

edit: Getting a calculator to 'solve' -3x2+5x+2 assumes that it's equal to zero, which in this case it's not. It'll spit out the right roots, but you can't just get rid of constant factors when things aren't equal to 0.
Last edited by Dopefish on Wed Sep 12, 2012 1:54 am UTC, edited 1 time in total.

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dudiobugtron
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Re: Can someone help me with calculus?

Postby dudiobugtron » Wed Sep 12, 2012 1:52 am UTC

mathgeek17 wrote:I tried using the quadratic equation, I tried redistributing the two factors again, and I had my calculator solve the quadratic for me and I still got 2 and -1/3. Is there something really obvious that I'm just missing?


Yes. You are solving the equation -3x2+5x+2=0
You should not do that. You should instead factorise -3x2+5x+2

This is what you are doing:
http://www.purplemath.com/modules/solvquad.htm

And this is what you should be doing:
http://www.purplemath.com/modules/factquad.htm

They do seem pretty similar, so I can understand the confusion!
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mathgeek17
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Re: Can someone help me with calculus?

Postby mathgeek17 » Wed Sep 12, 2012 1:59 am UTC

I'M MISSING THE -3 THAT MULTIPLIES THROUGH!!!!!! FINALLY GOT IT!!!!! Thank you so much for helping me!


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