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this is the graph and this is the function: y = (x^2-1)/x^3
it asks for:
1. x-int:
2. y-int:
3. vertical asymp:
4. horiz. asymp:
5. relative extrema:
6. points of inflection:
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1. there are two x-int, this case is (-1,0) and (1,0) since it crosses the x-axis.
2. it has no y-intercept because it doesnt cross y at all.
3. vertical asymptote is only x=0. however, when x=1, the result is a 0, but that doesnt matter right?
4. horizontal asymptote is y=0 as that if the degree of denominator > degree of numerator = y=0, from a rule.
5. relative extrema, i take the derivative of y = (x^2-1)/(x^3) and used the quotient rule which gave me (-x^4-3x^2)/(x^6) and i set this equal to 0. so i know one point is x=0, i think thats it?
6. points of inflection, i took the first y' then took the derivative of that so i get y'' = (2x^9 + 12x^7)/x^12, however, this is as far as i got. basically ill start putting values into 0 to check? and do i have to do an interval or it could be -infin to +infin?
thanks for looking again. >______< and yes paper, i do read my book, but the examples arent clear so i ask ydg, micro, etc.
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teamliquid: raithed's homework news
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well i did it, dunno if its correct. yes its hw.
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That graph can't be labeled right can it.
And when x=1, y=0, that has nothing to do with asymptote, that's just x intercept.
What do you mean by do an interval for points of inflection, since you are just setting the second derivative to zero, you are going to get discrete values if any shows. If the question asks for points of inflection, or relative extrema or whatnot, unless specified it will be from neg inf to inf.
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the graph is right minus the min/max arrows, i graphed it using a graphing calculator, it asks for x-int and i gave it the point (1,0) as one of the x-intercepts. so for the inflection point i can just set it to 0?
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1.yes
2.yes
3.yes
4.yes, it's easy to check in the figure that those are the only asymptotes
5.the derivative=(3-x^2)/x^4 which has zeroes when x^2=3 <=> x=+-sqrt(3) x=0 is not a zero since the function is not defined there.
6.second derivative=2(x^2-6)/x^5 has zeroes when x=+-sqrt(6), so these are your points of inflexion
since your function is not defined in x=0 you can assume x=/=0 and divide bt x'es in your expressions.
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Yeah, I was just talking about the arrows. Anyway like jtan wrote, you just set the second derivative to zero. Same applies for first derivative and extremas, which jtan also worked out for you.
You are allowed to actually graph it out via calculator? I thought most classes would have you plot it by hand as a part of the question heh.
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so question about number 5, if its not defined does it even have a relative extrema?
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Well, like you said, it is an asymptote at 0, so that by default cannot be a relative extrema. A relative extrema would be a defined value.
EDIT - Actually, I looked at the graph again since the numbers ticked something off, what did you plot there? That graph itself looks wrong at that.
EDIT - Oh, it can be right, what did you set the parameters for the windows to be?
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what numbers are you talking about? you mean how i plotted it and got it? plotted the function of y, then zero to get the intercepts.
okay, so i guess number 5 is done. but how did jitan get (3-x^2)/x^4? and same goes for number 6 with the derivative of 2(x^2-6)/x^5?
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On June 05 2008 12:18 jtan wrote:
1.yes
2.yes
3.yes
4.yes, it's easy to check in the figure that those are the only asymptotes
5.the derivative=(3-x^2)/x^4 which has zeroes when x^2=3 <=> x=+-sqrt(3) x=0 is not a zero since the function is not defined there.
6.second derivative=2(x^2-6)/x^5 has zeroes when x=+-sqrt(6), so these are your points of inflexion
since your function is not defined in x=0 you can assume x=/=0 and divide bt x'es in your expressions.
edit : Actually, nevermind - yes this all correct. The plot should look like this, though : the local max/min are much more apparent when it's plotted properly :
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What are the parameters you set for windows? Since the parts shown for x looks to be very small.
As for those functions, factor the derivatives you came up with, he didn't bother putting the parts factored out where x=0 since it doesn't matter there.
Plot w/ x from -10~10.
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set it to decimal. and he factored?
for number 5, y' = (-x^4-3x^2)/(x^6) --> (3-x^2)/x^4 which is what he got, how?
On June 05 2008 12:51 jgad wrote:Show nested quote +On June 05 2008 12:18 jtan wrote:
1.yes
2.yes
3.yes
4.yes, it's easy to check in the figure that those are the only asymptotes
5.the derivative=(3-x^2)/x^4 which has zeroes when x^2=3 <=> x=+-sqrt(3) x=0 is not a zero since the function is not defined there.
6.second derivative=2(x^2-6)/x^5 has zeroes when x=+-sqrt(6), so these are your points of inflexion
since your function is not defined in x=0 you can assume x=/=0 and divide bt x'es in your expressions. Actually, nevermind - yes this all correct. The plot should look like this, though : the local max/min are much more apparent when it's plotted properly :
but its a rule though, isnt it?
http://cnx.org/content/m13606/latest/
EDIT - it crosses so there's 2 x-int, what are you getting at, i dont understand.
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Look at jgad's graph, better than mine.
And you missed a negative time negative for your first derivative then, I got the same thing as he. Also yours simply can't be right since we'd get imaginary value for local max/min, where the graph clearly indicates some.
jgad was talking about lim x-> (+/-)inf = 0, probably misread the y=0 to be x=0.
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[but its a rule though, isnt it?
Yeah, it's a rule, but throwing rules at a math problem, imo, isn't really the way to learn. Consider how much more interesting it is to look at something like this (forgive my crude scribbles) :
if you factor out the equation you get 1/x - 1/x^3, so you can say that for all values where x>>1 the graph should basically do the same as 1/x all the way out to infinity. It's an easy step and it makes it clear why the rule works. I think the most important thing to take away from a math class isn't so much the ability to remember all the rules to solve the problems they'll give you, but to study the rules and learn why they work - then when you go on to be a mathematical career person, you have a well-worked toolbox that you know inside and out -- then you can use rules to make your job easier and more efficient, but not when you're learning.
The other issue is that, in real life, when you finally decide to go out and flex your mathematical muscle to get something done, you'll probably find that the equations you need for the real world are often brutal, intractable beasts which your rules will be useless against. Then you have to fall back on your core skills to hack out the problem the old-fashioned way. If all you have is a bag of tricks - a few easy rules to beat down docile problems, then you've sort of got a mathematician that's built like a house of cards.
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okay the derivative of #5; ( -x^4 + 3x^2 ) / x^6 i see where he got it from, stupid me.
second derivative of #6;
y = (x^2 - 1) / (x^3)
y' = (-x^4 + 3x^2) / (x^6)
y" = (8x^9 - 12x^7) / x^12 --> 2(x^2-6)/x^5 did jitan factor again? because i cant figure it out.
Ecael, plotting the graph on the calculator, going at -10,10, 2nd calc>zero, im still getting the same results. isnt that how you find the x-int?
EDIT - jgad, so im confused, the x-int is wrong?
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I wasn't talking about the x intercept, nor was jgad, probably. I was talking about how your graph on the OP was difficult to see the local extrema and the inflection point. jgad probably thought the same, since we both posted graphs that emphasized that little change of inflection and the extrema.
EDIT - I check that result, Raithed, you made some mistake somewhere with that second derivative. I also got 2(x^2-6)(x^-5).
Odd, your answer in the OP worked fine after we account for the negative.
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^ yes. In fact, before looking at the equation, I was a bit confused at first and thought he'd messed the horizontal asymptote just from looking at his graph, which looks to have two (at y=±1), though it doesn't.
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OHH, its because i did the graph by hand so it looks bad and sloppy. i would take a picture of my calculator's screen but i wouldve thought "close enough" was good enough. i still dont quite understand how to get the points though(referring to #6). the book has examples with intervals and this one doesnt, so basically i plot points into the original(draw it, etc), then plot points into the second derivative? am i following right?
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For points of inflection, one can define it as points where the second derivative of the function = 0. So you simply set that equation to 0 and solve for x. Intervals doesn't matter in terms of point of inflections, as you are not going to see a point of inflection at infinity. I am not sure what do you mean by ploting points into the original though. If I take it that you are trying to solve it with a calculator, then yeah, you'd simply plot a graph of the second derivative and solve for where that function is 0.
In terms of the original function, the point of inflection changes how the function...um, flows, someone help me with the right term. You'll see a visible change of how the function's slope changes though starting at the point of inflection.
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if its not that big of trouble can you write out how you got 2(x^2-6)(x^-5) ? so basically the second derivative is set to 0.
2(x^2-6)/(x^-5) = 0 and just solve for x, so would it become x^2-6 = 0, x = 3? as one of the points?
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No problem, let me do it out with quotient rule I guess, I used chain since I've always preferred the extra steps over me accidentally forgetting order for quotient rule (bad memory, lol). Oh, for inflection point, look at how you solved it again, x^2 - 6 = 0, x^2 = 6, root(6) =/= 3.
So first derivative = (3-x^2)/(x^4). Quotient rule states that f(x) = u/v, f'(x) = (du)(v) - (dv)(u) / v^2, so (-2x)(x^4) - (3-x^2)(4x^3) / (x^8) = (-2x^5) - (12x^3 - 4x^5) / x^8, factor out a x^3. -2x^2 - 12 + 4x^2 / x^5. or 2x^2 - 12 / x^5, 2((x^2)-6) / x^5.
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ah, okay yeah i did dumb mistakes.
x^2-6 = 0, x^2=6, x = sqrt. 6 <-- is this one of the inflection lines?
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+/- sqrt(6) are the two points of inflection.
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ah.. so am i done or is there more?
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lol, don't have us do all the work. Check yourself as to see what is missing 
(And no, those are the only two, looking at the graphs on the previous page should make that pretty clear)
One really good point that jgad brought up is the usage of graph and simple mathematical rules for you to verify what you know. Knowing the graphical end of it will make checking a lot easier for you, and it is also a quick and reliable double check for you as far as simple mistakes are concerned. A quick look at the graph could've told you that your original derivative had something wrong since you were getting imaginary values for critical points (well, simple logic dictates that it is wrong, but that's beside the point). It'll also give you an idea of about where the critical and inflection points lie in regard to each other.
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thanks a lot Ecael for the help and many others, its 2am, going to bed. <3
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also you can see in the graph that the inflexion points are correct. sqrt(6)=~2.5 and if you look at x=2.5 you see that that's about where the curve changes from convex to concave.
so question about number 5, if its not defined does it even have a relative extrema?
This is a weird question. Your function is not defined IN the point x=0 because you get a division by zero in the function so you get no well defined value there. The function still has local extremas, but it can't have one in x=0 since the function is not defined IN that point.
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5) there are no local max/mins but an absolute max min. and that would be negative infinity and infinity.
6) points of inflection u just take double derivative and see if it's positive or negative.
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There are local extremas, take a look at the function posted instead of the graph, which multiple people have noted as deficient.
That's a cool program, jtan, the powertoy calculator is just not up to par :p
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On June 05 2008 17:58 gwho wrote:
5) there are no local max/mins but an absolute max min. and that would be negative infinity and infinity.
6) points of inflection u just take double derivative and see if it's positive or negative.
wrong
+-sqrt(3) is local max/min
raitheads graph is wrong, he just drew that as an approximate to the graph in his calculator lol
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Sometimes in university, I think to myself, "boy, writing essays sure is fucking boring." Then I remember I haven't done any complicated math in 3 years, and I feel a little better.
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When I was in university, I sometimes would look around at my ~93% male Engineering Physics class and thinking to myself "boy, I wonder where all the girls end up?". Then I took an elective course in neuropsychology and it was like walking into the girls' locker room - wtf?! It was like 93% chicks - and then, going back to my soul-crushing complicated math courses, I felt a little worse 
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