what's it called exactly when you square or cube a function not a root because that'd be the opposite but...?

- anonymous

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- amistre64

its called 'squaring or cubing' a function

- anonymous

oh ahahah okay thought there was more to it

- anonymous

got time for another question?

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- amistre64

AN HOUR OR SO

- amistre64

nothing like a caap lock to set the mood lol

- anonymous

okay perfect, ok first of all do you see the graph right before "Forced Vibrations with damping?" on this webite http://www.ltcconline.net/greenl/courses/204/appsHigherOrder/forcedVibrations.htm

- amistre64

yes, the sin in a cone

- anonymous

yes, okay I have a similar problem like the same idea basically but these are the points and i have to find the equation: (-.75, 0) ..(.-5, -.7906)..(-.25, -.8724)..(0,0)..(.25, .90461)... (.5, .94868)...(.75,0)...(1, -1.039)..(1.25, -1.088) ..(1.5,0) ..(1.75, 1.1915) ..(2, 1.2471) and (2.25,0)

- amistre64

yeah, i remember those :) dampened waves is not my cup o tea, nor is the differentials yet. gonna have to ask someone smarter

- amistre64

pretty much you hav to find the peaks and valleys to draw the 'cone' from and determine its linear equation to help out

- anonymous

hmm okay, well are you good at natural log functions like ln(x) stuff?

- amistre64

im adequate with log functions

- anonymous

ok so f(x)=(x)/ln(x) basically how does each algebraic piece, so the only algebra in this is the numerator of x on the top, affect each part of its graph so when x<0, 00

- amistre64

this is pretty much a graph with the slope of \({1\over ln(x)}\)

- anonymous

so for example when x<0 it looks more like the x part because...type of thing

- anonymous

right so its the inverse of ln(x)?

- amistre64

not the inverse; that would be the graph of 'e'; its the 'flipped over the xaxis' graph of ln(x)

- anonymous

oh okay, so how do i explain this, so first when x<0 it acts like the ln(x) function because all y values still come out to be ERROR/undefined, right? and when 0

- amistre64

we have a vertical asymptote at x=0 becasue the ln(x) function has no value for x=0 and below

- anonymous

we have a vertical asymptote at x=0 becasue the ln(x) function has no value for x=0 and below....that is why when x<0 is acts like ln(x)?

- amistre64

it acts like 1/ln(x) for small values of x becasue that part takes control

- amistre64

for large values of x; 1/ln(x) plaays less part in it

- anonymous

when you say takes control what specifically do you mean

- amistre64

same thing you stated really; the control of the function get governed by the part that becomes dominate for its corresponding value of x

- anonymous

oh you mean the y values are bigger is that what you mean by dominate/take over?

- anonymous

when you say it takes control for small values of x you mean negative x values?

- amistre64

i mean lose to 0

- amistre64

err.. close to 1; since ln(1) = 0

- anonymous

so wait...When 0

- amistre64

gets closer and closer to infinity

- anonymous

when I use my calculator from .1 to .9 are all negative numbers so that can't be right

- anonymous

:/

- amistre64

anything less than 1 is gonna be a negative # i think

- anonymous

right so its not approaching infinity..its approaching negative infinity right?

- amistre64

as it approaches 1 from the right it goes thru the roof; from the left it goes thru the floor; to say infinity just tends to imply that its going in either direction really

- amistre64

infinity is infinity; if your on the negative side, then yeah, its -inf ...

- anonymous

When 0

- amistre64

correct

- anonymous

and then for larger values of x it looks more like ln(x) than 1/ln(x) when 0

- amistre64

no; stilll 1/ln(x) just on the + side

- amistre64

err.. i gotta learn to read better

- amistre64

for larger values of x; we get less and less out of it

- anonymous

what do you mean

- amistre64

1/ln(1,000,000,000) = .04825....
1/ln(1,000,000,000,000) = .03619.....

- anonymous

you are taking a fraction of a larger number not a smaller so the difference is not as big?

- amistre64

something like that....
100(.1) = 10
100(.01) = 1
so it still has control over it to an extent but the function still grows to infinity over the long haul

- anonymous

and finally when x>0 it looks more like the ln(x) part because..hm let me look..

- anonymous

as x approaches infinity..hm

- anonymous

:/

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