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Math Help - Derivitive and function help.

  1. #1
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    Derivitive and function help.

    Hey guys, I'm new, but this is the first site that I found...I'm good with numbers, but, when it comes to cal and stuff..I hate variables..and I just have a way harder time understanding..math in chem and physics..easy for me..

    The page is here..
    ImageShack - Hosting :: mathup5.jpg
    Limits I got (I think) a person just went over like..10 minutes with me..a few basic things..and I understood it completly (only like 1 or 2 practice probs in book I got wrong, well..only can check evens..I'm wanting to go to the library and look at instructors solution manual)

    1a 12
    b 1/2
    c 1/2
    d -4
    e 0? Friend says 0..I thought it might be 1..but 2 people say 0..so maybe I'm missing something for e..
    f sqrt of 8

    2..man..I have no clue really..other then maybe stick 5 in a..

    3. I kind of know (I think) but..I'm not sure what to do with it.. I'm assuming its not cont at x = 2? because when I solve it..as x approaches from right, is 1 while as it approaches from the left is -11

    4, I have the answer, 1/2 but that doesn't really help me in how to derive it..I guess with the 1 in there kinda confusing..if some1 could maybe just post what the first step looks like with the 1/x^2 +1?

    5 I got, I think..

    Its off from -infinity to -3, on from -3 at y = 9 till x=2 y=4 and off again from 2 till infinity (for the parabola..it dips down near 0 right? so I get a dip near x=0? the examples we did just basically had it on only 1 side of the graph..so it was just a slightly curved line..

    I dunno..I really wanna do well..as I like science..but most of what I want needs calc..I think some of it I have down is right..but..I'm never sure if what I calc is right..1 q..in the heavyside..if it was like..
    x^2[h(x+3) + h(x-2)] where its a bolded plus instead of a minus sign..would that indicate below the graph? They really didn't do a whole lot on these..and mostly just straight lines..I dunno..thx for any1 that helps or even looks..I might just save up and get a tutor for a bit..maybe just that little bit of showing how to do it will help..

    Edit: The plus isn't bolding..but..the sign between h(x+3) (plus) h(x-2)
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  2. #2
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    1.)

    a is correct.
    b is -1/2, not 1/2
    c I can't even read the terms...
    d is 4

    And I'll stop with those because it's insanely hard to read for 1, and I got 3 classes worth of math assignments to complete for myself . But it seems you get the basic picture.
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  3. #3
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    Bleh..1 friend got -1/2 for b and 4 for d..but others when I showed em said 1/2 and -4..(at least b mostly) can u just explain briefly how how I got opposite signs?

    c is 2x^2 - 3x + 6/sqrt(16x^4 - 1) as limit approaches infinity

    I made a crude graph of my heaviside..but..it sucks cause my mouse..and..it only goes up 2 8 (I didnt leave enough room to put the 9th tick)

    ImageShack - Hosting :: heavisidedy7.jpg
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  4. #4
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    c is correct, it is 1/2.

    As for the 1 - sqrt(x+1) problem, if you use L'Hopital's rule, you get -1/(2*sqrt(x+1))/1, which then if you take the limit you have -1/2; 1/(sqrt(x+1))... I pulled out the -1/2 and then you're taking the limit of 1/(sqrt(x+1)), which of course will be 0. Thus, you're left with -1/2.
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  5. #5
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    For d, if I can read it correctly, you have

    lim (x->2) (x^3 - 2*x^2)/(x-2) ?

    What happens when you reduce (x^3 - 2*x^2)/(x-2) ?

    You will notice you get x^2. Can you do it now?
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  6. #6
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    Yea, mine looks the same (well original, new is 2..cept a negative sign) its as x approaches 2 from the left.. 2(-) the absolute value threw me off 2..so I wasn't sure and put a negative sign infront of the (x-2) so it just became -1 on the bottom... x^2(x-2) on top..they cancel..I knew it was a 4..just wasn't sure if it was a positive or negative...
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  7. #7
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    Since the limit exists, it doesn't matter if we're looking at it from the right or from the left, since it will be 4 regardless. If it was undefined, then you would have 2 different results from the right or left; when equal, the limit exists. This is one of the 3 principles of a limit. If I get some time later, I will try to view some of your other questions.
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  8. #8
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    Well...I'm not so sure of 2..but..
    I'm still thinking a is 5, b is x=3 and c is not a function? because the vertical line test?...
    I dunno..just seems to easy to me..its so easy its hard...

    3 still thinking that as approaches from right is 1 and from left -11..not sure if thats just the answer? discont cause of the jump?

    Edit: Erm..the reason I was getting those numbers, was because I was subsituting k with the value when x = 1...
    I re did it an both come out as 1..so it's continuous at x=2

    Yea, its like..2 now..and I get up at 7..been staring at #4 for like..3 hours now.. Only #4 I really think..
    Im not sure if I understand l'hopitals rule for q 1b...
    Last edited by Xantiox; September 28th 2006 at 11:49 PM.
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  9. #9
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    Well..I handed it in this morn..I had #3 written out already so I think I had finally figured out how to do it (if I got stumped I would try to see how to do the next step), and before I handed it in, I think I saw how the b was -1/2..I think it was because when it would be the - * + for some reason I switched it to +... and I think with the absolute value (d), the - on the outside turns the - on the inside to a positive...? Or..maybe I'm not quite seeing d :P but whatever I saw this morning it made sense lol..
    I find it funny cause a guy was asking me if I got c..which was like the first 1 I did and dead easy to me..

    If I get 100 on that assignment..I'm gonna die..considering how long it took me to do it..I did my chem in like 5 min..I got 2 wrong tho...1 cause I misread 1mg as 1 g so made my halflife calc off..and for some reason..I had this right..it was which is INCORRECT, and I had it right..but before I handed it in..I noticed that (and a correct statement was above it) I changed the answer w/o looking at q...
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