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

1. anonymous

Have you learned derivatives?

2. anonymous

Kind of? but i'm not really understanding what the question is asking

3. anonymous

Yep moreover this question seems a bit vague nobody asks when x gets really larger :p

4. anonymous

Very large*

5. anonymous

So maybe they're asking about the lim of x as it temds to infinity

6. 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

7. anonymous

Are you familiar with these?

8. anonymous

Well my friend in my math we have no very large :) either we've got limx-->+inf of -inf :)

9. anonymous

so false?

10. 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 $$x$$?

11. 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

12. 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

13. anonymous

i'm confused :/

14. anonymous

joyraheb's answer to the question and explanation are correct but his other comments are misleading

15. anonymous

You idiot we're talking about ordinary functions here who in the hell mention asymptotes?!

16. UsukiDoll

watch the language.

17. anonymous

Mentioned*

18. anonymous

Sorry 😂 i just get nervous quiqly

19. 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 $$x$$

20. alekos

there are no asymptotes for this equation

21. UsukiDoll

Alright... but remember from OS Coc Be Nice - I will stay positive, be friendly, and not mean :)

22. alekos

its just a 3rd order polynomial

23. anonymous

@joyraheb @alekos I think you're both unfortunately confused

24. anonymous

okay... i'm really confused now?

25. anonymous

@alekos that's not what asymptotic analysis means

26. UsukiDoll

well... the only way to shed the light is to really find out what the question is really asking.

27. anonymous
28. 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

29. alekos

there's no asymptotic line such as x=6 or y=-5. It just goes to plus infinity, plain and simple

30. anonymous

so it's true??

31. anonymous

@mishtea is it true for $$x^3$$? does $$x^3$$ 'rise' for larger and larger $$x$$? is it increasing?

32. anonymous

@alekos you are still confused; please reread my previous comments and maybe check out the Wikipedia article on asymptotic analysis

33. UsukiDoll

maybe let x = any number and see if it increases ¯\_(ツ)_/¯

34. anonymous

yes i think because you said earlier that x^3 rises for the larger x and the equation that right

35. UsukiDoll

if the value is positive though... anyway I'm just going on the sidelines. This might turn ugly

36. anonymous

well, think of the graph?|dw:1437280315128:dw| what happens as you move to the right? as you think of bigger and bigger $$x$$?

37. anonymous

it rises

38. 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

39. 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 $$x^3-3x^2+2x$$

40. anonymous

... yes i have but i'm not a math person ...

41. anonymous

thanks @oldrin.bataku! i get it now.

42. 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

43. 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.

44. anonymous

Oldrin what do you mean by large x ?

45. anonymous

The funny part is that you're explaining limits to her that only apply in such functions

46. 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 $$k>0$$

47. 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

48. anonymous

So eventually this large x should be larger than 2

49. anonymous

Right? @oldrin.bataku