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anonymous
 one year ago
The function f(x)=x^33x^2+2x rises as x grows very large.
True or false?
anonymous
 one year ago
The function f(x)=x^33x^2+2x rises as x grows very large. True or false?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Have you learned derivatives?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Kind of? but i'm not really understanding what the question is asking

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep moreover this question seems a bit vague nobody asks when x gets really larger :p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So maybe they're asking about the lim of x as it temds to infinity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually, a lot of people ask about what happens as x gets 'very large'  that's the entire point of asymptotics. these are limits \(x\to\pm\infty\)  you have encountered them as horizontal asymptotes or 'end behavior'. it is a common topic in precalculus treatments of polynomial and rational functions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Are you familiar with these?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well my friend in my math we have no very large :) either we've got limx>+inf of inf :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@mishtea the point is that, for sufficiently large \(x\), \(x^3\) will grow faster and faster than any terms of lower powers of \(x\), like \(x^2,x,1\). so notice that if we factor out \(x^3\) we get $$x^33x^2+2x=x^3(13/x+2/x^2)$$ you can immediately see that if we let \(x\) grow and grow larger and larger, \(3/x\) and \(2/x^2\) would quickly become negligibly small, so the whole function would behave a lot like $$x^33x^2+2x=x^3(13/x+2/x^2)\sim x^3(1)=x^3$$ where \(\sim\) means "behaves like for large \(x\)" does \(x^3\) 'rise' for large \(x\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Anyway well the answer is true because as we as x of this function f(x) tends to infinity the answer is +inf so its surely true

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@joyraheb that's actually very, very incorrect; sure, large is contextdependent, but once again the existence of the entire field of asymptotics disagrees with you

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0joyraheb's answer to the question and explanation are correct but his other comments are misleading

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You idiot we're talking about ordinary functions here who in the hell mention asymptotes?!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry 😂 i just get nervous quiqly

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0note this is the same reason why the limit of rational functions works out as it does: $$\frac{ax^n+O(x^{n1})}{bx^n+O(x^{n1})}=\frac{a+O(1/x)}{b+O(1/x)}\sim\frac{a}b$$for large \(x\)

alekos
 one year ago
Best ResponseYou've already chosen the best response.0there are no asymptotes for this equation

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0Alright... but remember from OS Coc Be Nice  I will stay positive, be friendly, and not mean :)

alekos
 one year ago
Best ResponseYou've already chosen the best response.0its just a 3rd order polynomial

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@joyraheb @alekos I think you're both unfortunately confused

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay... i'm really confused now?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@alekos that's not what asymptotic analysis means

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0well... the only way to shed the light is to really find out what the question is really asking.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I've already explained what the question is really asking  it wants to know about endbehavior, which is essentially asking about asymptotic equivalence, i.e. knowing that \(x^3\) dominates as \(x\to\infty\), and since \(x^3\) 'rises' for large \(x\) then it follows that \(x^33x^2+2x\) will, too

alekos
 one year ago
Best ResponseYou've already chosen the best response.0there's no asymptotic line such as x=6 or y=5. It just goes to plus infinity, plain and simple

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@mishtea is it true for \(x^3\)? does \(x^3\) 'rise' for larger and larger \(x\)? is it increasing?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@alekos you are still confused; please reread my previous comments and maybe check out the Wikipedia article on asymptotic analysis

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0maybe let x = any number and see if it increases ¯\_(ツ)_/¯

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes i think because you said earlier that x^3 rises for the larger x and the equation that right

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0if the value is positive though... anyway I'm just going on the sidelines. This might turn ugly

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, think of the graph?dw:1437280315128:dw what happens as you move to the right? as you think of bigger and bigger \(x\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@mishtea have you studied limits, studying variations.... If not then you wouldn't understand because its a very simple question if you have taken these

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0right, and since we've shown that for large \(x\), that since \(x^3\) grows so much faster than the \(3x^2+2x\) part, we know that in the 'long run' the behavior of \(x^33x^2+2x\) will be like that of \(x^3\). so if you know that \(x^3\) will rise for large \(x\), then so will \(x^33x^2+2x\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0... yes i have but i'm not a math person ...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks @oldrin.bataku! i get it now.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for sake of example, if it were, say, \(5x^33x^2+2x\), then it would behave like \(5x^3\) in the long run  this *falls* for large \(x\) because of the negative coefficient. the same dominance rules apply, though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is why for a polynomial, if you want to know about the endbehavior (rise to the left, fall to the right, etc.  things like that), you only need to look at the term of highest degree, because it will dominate the others in the long run.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oldrin what do you mean by large x ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The funny part is that you're explaining limits to her that only apply in such functions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sufficiently large \(x\), i.e. there exists some \(X_0\) such that we have a property hold for all \(x>X_0\). in terms of dominance, it follows that \(g\) dominates \(f\) if there exists some point \(X_0\) beyond which (i.e. for all \(x>X_0\)) we can guarantee \(kg(x)>f(x)\) no matter how small we choose \(k>0\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is written $$f\in o(g)\Leftrightarrow \forall k>0\ \exists X_0\ \forall x>X_0 (kg(x)>f(x))$$ in computer science we call this littleo notation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So eventually this large x should be larger than 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right? @oldrin.bataku
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