- #1
Hannisch
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Hi! I've got a problem that's got me a bit.. well, in the end I guess the proper word is sceptic, because it feels a bit like I've made the answer up, which is why I'm taking it up here at all. Anyway:
Determine the domain for f(x) and the asymptotes to the curve y=f(x) for
[tex]f(x) = \frac{|x^3 + x^2|}{x^2 -1}[/tex]
I started with the domain, which was simple enough - it's not defined for -1 or 1. The question I've got here, however, is how I write this in the best way. I mean, I can say that the function is defined in the intervals x>1, -1<x<1 and x<-1, but we've also learned another way of writing:
If a function is defined from -∞ to +∞ the domain is Df = ]-∞,∞[ , but can I write this when the interval is broken like above? I mean, it looks bizarre; Df = ]-∞,-1[ , Df = ]-1,1[ , Df = ]1,∞[ .. ?Anyway, for the asymptotes:
I checked the vertical asymptotes first, which wasn't too hard:
[tex]\lim_{x\to 1} \frac{|x^3 + x^2|}{x^2 -1} = \frac{2}{0} = \infty[/tex]
So, 1 is a vertical asymptote.
[tex]\lim_{x\to -1} \frac{|x^3 + x^2|}{x^2 -1} = \frac{0}{0} [/tex]
-1 isn't an asymptote, is what I've figured from my reading.
Anyway, now we come to the real.. sceptism. We've learned slant asymptotes too. And well.. it's.. odd.
First of all, I came to the conclusion that when x → +∞, x3 + x2 is defintitely positve so:
[tex]\lim_{x\to \infty} \frac{|x^3 + x^2|}{x^2 -1} -(ax+b) = \lim_{x\to \infty} \frac{x^3 + x^2}{x^2 -1} -(ax+b) = \lim_{x\to \infty} \frac{x^2(x + 1)}{x^2(1- \frac{1}{x^2})} -(ax+b) = \lim_{x\to \infty} \frac{x + 1}{1- \frac{1}{x^2}} -(ax+b)[/tex]
And here I start coming up with an answer, and I really feel like I've imagined it.
To find the asymptote I want the whole thing to go to zero and well, my reasoning went something like this:
I first thought that when x → ∞, (1/x2) → 0, so then I thought, well, that sort of leaves me with x + 1 - (ax+b), which goes to zero when it's x+1 - (x+1), which means that it has an asymptote in a=1, b=1. Which is what I feel might be rather imagined. (so, asymptote is y = x + 1.)
So I started searching around and found a webpage where it said that you can determine a slant asymptote by doing long division in this type of situation, so I tried that to
(x3 + x2)/(x2-1)
which in turn gave me:
f(x) = (x+1) + (x+1)/(x2-1), which also means that the asymptote is y=x+1..
And I did the same thing for x → -∞, saying that when this happens x3 + x2 is most definitely negative, so the absolute value of it is -(x3 + x2) and by this got the asymptote (using both aforementioned methods) y = -x - 1.
First: I did not learn the long division method in school, not sure if my teacher will accept it and since I learned it from the internet, is it really a trustworthy method?
Second: If I've done it wrong, help?!
Third: Sorry if this is a confusing post--it was definitely confusing to write!
Homework Statement
Determine the domain for f(x) and the asymptotes to the curve y=f(x) for
[tex]f(x) = \frac{|x^3 + x^2|}{x^2 -1}[/tex]
Homework Equations
The Attempt at a Solution
I started with the domain, which was simple enough - it's not defined for -1 or 1. The question I've got here, however, is how I write this in the best way. I mean, I can say that the function is defined in the intervals x>1, -1<x<1 and x<-1, but we've also learned another way of writing:
If a function is defined from -∞ to +∞ the domain is Df = ]-∞,∞[ , but can I write this when the interval is broken like above? I mean, it looks bizarre; Df = ]-∞,-1[ , Df = ]-1,1[ , Df = ]1,∞[ .. ?Anyway, for the asymptotes:
I checked the vertical asymptotes first, which wasn't too hard:
[tex]\lim_{x\to 1} \frac{|x^3 + x^2|}{x^2 -1} = \frac{2}{0} = \infty[/tex]
So, 1 is a vertical asymptote.
[tex]\lim_{x\to -1} \frac{|x^3 + x^2|}{x^2 -1} = \frac{0}{0} [/tex]
-1 isn't an asymptote, is what I've figured from my reading.
Anyway, now we come to the real.. sceptism. We've learned slant asymptotes too. And well.. it's.. odd.
First of all, I came to the conclusion that when x → +∞, x3 + x2 is defintitely positve so:
[tex]\lim_{x\to \infty} \frac{|x^3 + x^2|}{x^2 -1} -(ax+b) = \lim_{x\to \infty} \frac{x^3 + x^2}{x^2 -1} -(ax+b) = \lim_{x\to \infty} \frac{x^2(x + 1)}{x^2(1- \frac{1}{x^2})} -(ax+b) = \lim_{x\to \infty} \frac{x + 1}{1- \frac{1}{x^2}} -(ax+b)[/tex]
And here I start coming up with an answer, and I really feel like I've imagined it.
To find the asymptote I want the whole thing to go to zero and well, my reasoning went something like this:
I first thought that when x → ∞, (1/x2) → 0, so then I thought, well, that sort of leaves me with x + 1 - (ax+b), which goes to zero when it's x+1 - (x+1), which means that it has an asymptote in a=1, b=1. Which is what I feel might be rather imagined. (so, asymptote is y = x + 1.)
So I started searching around and found a webpage where it said that you can determine a slant asymptote by doing long division in this type of situation, so I tried that to
(x3 + x2)/(x2-1)
which in turn gave me:
f(x) = (x+1) + (x+1)/(x2-1), which also means that the asymptote is y=x+1..
And I did the same thing for x → -∞, saying that when this happens x3 + x2 is most definitely negative, so the absolute value of it is -(x3 + x2) and by this got the asymptote (using both aforementioned methods) y = -x - 1.
First: I did not learn the long division method in school, not sure if my teacher will accept it and since I learned it from the internet, is it really a trustworthy method?
Second: If I've done it wrong, help?!
Third: Sorry if this is a confusing post--it was definitely confusing to write!