anonymous
  • anonymous
The function f(x)=x^3-3x^2+2x rises as x grows very large. True or false?
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Have you learned derivatives?
anonymous
  • anonymous
Kind of? but i'm not really understanding what the question is asking
anonymous
  • anonymous
Yep moreover this question seems a bit vague nobody asks when x gets really larger :p

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anonymous
  • anonymous
Very large*
anonymous
  • anonymous
So maybe they're asking about the lim of x as it temds to infinity
anonymous
  • anonymous
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
anonymous
  • anonymous
Are you familiar with these?
anonymous
  • anonymous
Well my friend in my math we have no very large :) either we've got limx-->+inf of -inf :)
anonymous
  • anonymous
so false?
anonymous
  • anonymous
@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 \(x\)?
anonymous
  • anonymous
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
anonymous
  • anonymous
@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
anonymous
  • anonymous
i'm confused :/
anonymous
  • anonymous
joyraheb's answer to the question and explanation are correct but his other comments are misleading
anonymous
  • anonymous
You idiot we're talking about ordinary functions here who in the hell mention asymptotes?!
UsukiDoll
  • UsukiDoll
watch the language.
anonymous
  • anonymous
Mentioned*
anonymous
  • anonymous
Sorry 😂 i just get nervous quiqly
anonymous
  • anonymous
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 \(x\)
alekos
  • alekos
there are no asymptotes for this equation
UsukiDoll
  • UsukiDoll
Alright... but remember from OS Coc Be Nice - I will stay positive, be friendly, and not mean :)
alekos
  • alekos
its just a 3rd order polynomial
anonymous
  • anonymous
@joyraheb @alekos I think you're both unfortunately confused
anonymous
  • anonymous
okay... i'm really confused now?
anonymous
  • anonymous
@alekos that's not what asymptotic analysis means
UsukiDoll
  • UsukiDoll
well... the only way to shed the light is to really find out what the question is really asking.
anonymous
  • anonymous
https://en.wikipedia.org/wiki/Asymptotic_analysis
anonymous
  • anonymous
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
alekos
  • alekos
there's no asymptotic line such as x=6 or y=-5. It just goes to plus infinity, plain and simple
anonymous
  • anonymous
so it's true??
anonymous
  • anonymous
@mishtea is it true for \(x^3\)? does \(x^3\) 'rise' for larger and larger \(x\)? is it increasing?
anonymous
  • anonymous
@alekos you are still confused; please reread my previous comments and maybe check out the Wikipedia article on asymptotic analysis
UsukiDoll
  • UsukiDoll
maybe let x = any number and see if it increases ¯\_(ツ)_/¯
anonymous
  • anonymous
yes i think because you said earlier that x^3 rises for the larger x and the equation that right
UsukiDoll
  • UsukiDoll
if the value is positive though... anyway I'm just going on the sidelines. This might turn ugly
anonymous
  • anonymous
well, think of the graph?|dw:1437280315128:dw| what happens as you move to the right? as you think of bigger and bigger \(x\)?
anonymous
  • anonymous
it rises
anonymous
  • anonymous
@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
  • anonymous
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 \(x^3-3x^2+2x\)
anonymous
  • anonymous
... yes i have but i'm not a math person ...
anonymous
  • anonymous
thanks @oldrin.bataku! i get it now.
anonymous
  • anonymous
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
anonymous
  • anonymous
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.
anonymous
  • anonymous
Oldrin what do you mean by large x ?
anonymous
  • anonymous
The funny part is that you're explaining limits to her that only apply in such functions
anonymous
  • anonymous
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 \(k>0\)
anonymous
  • anonymous
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
anonymous
  • anonymous
So eventually this large x should be larger than 2
anonymous
  • anonymous
Right? @oldrin.bataku

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