Question

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

Answer 1

Have you learned derivatives?

Answer 2

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

Answer 3

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

Answer 4

Very large*

Answer 5

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

Answer 6

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

Answer 7

Are you familiar with these?

Answer 8

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

Answer 9

so false?

Answer 10

@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 \(x\)?

Answer 11

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

Answer 12

@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

Answer 13

i'm confused :/

Answer 14

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

Answer 15

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

Answer 16

watch the language.

Answer 17

Mentioned*

Answer 18

Sorry 😂 i just get nervous quiqly

Answer 19

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 \(x\)

Answer 20

there are no asymptotes for this equation

Answer 21

Alright...
but remember from OS Coc

Be Nice - I will stay positive, be friendly, and not mean
:)

Answer 22

its just a 3rd order polynomial

Answer 23

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

Answer 24

okay... i'm really confused now?

Answer 25

@alekos that's not what asymptotic analysis means

Answer 26

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

Answer 27

https://en.wikipedia.org/wiki/Asymptotic_analysis

Answer 28

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

Answer 29

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

Answer 30

so it's true??

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

well, think of the graph? what happens as you move to the right? as you think of bigger and bigger \(x\)?

Answer 37

it rises

Answer 38

@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

Answer 39

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 \(x^3-3x^2+2x\)

Answer 40

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