The function f(x)=x^3-3x^2+2x rises as x grows very large. True or false?

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The function f(x)=x^3-3x^2+2x rises as x grows very large. True or false?

Mathematics
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Have you learned derivatives?
Kind of? but i'm not really understanding what the question is asking
Yep moreover this question seems a bit vague nobody asks when x gets really larger :p

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Other answers:

Very large*
So maybe they're asking about the lim of x as it temds to infinity
actually, 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
Are you familiar with these?
Well my friend in my math we have no very large :) either we've got limx-->+inf of -inf :)
so false?
@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 $$x^3-3x^2+2x=x^3(1-3/x+2/x^2)$$ 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 $$x^3-3x^2+2x=x^3(1-3/x+2/x^2)\sim x^3(1)=x^3$$ where \(\sim\) means "behaves like for large \(x\)" does \(x^3\) 'rise' for large \(x\)?
Anyway 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
@joyraheb that's actually very, very incorrect; sure, large is context-dependent, but once again the existence of the entire field of asymptotics disagrees with you
i'm confused :/
joyraheb's answer to the question and explanation are correct but his other comments are misleading
You idiot we're talking about ordinary functions here who in the hell mention asymptotes?!
watch the language.
Mentioned*
Sorry 😂 i just get nervous quiqly
note this is the same reason why the limit of rational functions works out as it does: $$\frac{ax^n+O(x^{n-1})}{bx^n+O(x^{n-1})}=\frac{a+O(1/x)}{b+O(1/x)}\sim\frac{a}b$$for large \(x\)
there are no asymptotes for this equation
Alright... but remember from OS Coc Be Nice - I will stay positive, be friendly, and not mean :)
its just a 3rd order polynomial
@joyraheb @alekos I think you're both unfortunately confused
okay... i'm really confused now?
@alekos that's not what asymptotic analysis means
well... the only way to shed the light is to really find out what the question is really asking.
https://en.wikipedia.org/wiki/Asymptotic_analysis
I've already explained what the question is really asking -- it wants to know about end-behavior, 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 \(x^3-3x^2+2x\) will, too
there's no asymptotic line such as x=6 or y=-5. It just goes to plus infinity, plain and simple
so it's true??
@mishtea is it true for \(x^3\)? does \(x^3\) 'rise' for larger and larger \(x\)? is it increasing?
@alekos you are still confused; please reread my previous comments and maybe check out the Wikipedia article on asymptotic analysis
maybe let x = any number and see if it increases ¯\_(ツ)_/¯
yes i think because you said earlier that x^3 rises for the larger x and the equation that right
if the value is positive though... anyway I'm just going on the sidelines. This might turn ugly
well, think of the graph?|dw:1437280315128:dw| what happens as you move to the right? as you think of bigger and bigger \(x\)?
it rises
@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
right, 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 \(x^3-3x^2+2x\) will be like that of \(x^3\). so if you know that \(x^3\) will rise for large \(x\), then so will \(x^3-3x^2+2x\)
... yes i have but i'm not a math person ...
thanks @oldrin.bataku! i get it now.
for sake of example, if it were, say, \(-5x^3-3x^2+2x\), 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
this is why for a polynomial, if you want to know about the end-behavior (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.
Oldrin what do you mean by large x ?
The funny part is that you're explaining limits to her that only apply in such functions
sufficiently 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 \(k>0\)
this 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 little-o notation
So eventually this large x should be larger than 2
Right? @oldrin.bataku

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