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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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Colombia677 Posts
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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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