anonymous
  • anonymous
what's it called exactly when you square or cube a function not a root because that'd be the opposite but...?
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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amistre64
  • amistre64
its called 'squaring or cubing' a function
anonymous
  • anonymous
oh ahahah okay thought there was more to it
anonymous
  • anonymous
got time for another question?

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amistre64
  • amistre64
AN HOUR OR SO
amistre64
  • amistre64
nothing like a caap lock to set the mood lol
anonymous
  • 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
  • amistre64
yes, the sin in a cone
anonymous
  • 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
  • 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
  • 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
  • anonymous
hmm okay, well are you good at natural log functions like ln(x) stuff?
amistre64
  • amistre64
im adequate with log functions
anonymous
  • 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
  • amistre64
this is pretty much a graph with the slope of \({1\over ln(x)}\)
anonymous
  • anonymous
so for example when x<0 it looks more like the x part because...type of thing
anonymous
  • anonymous
right so its the inverse of ln(x)?
amistre64
  • amistre64
not the inverse; that would be the graph of 'e'; its the 'flipped over the xaxis' graph of ln(x)
anonymous
  • 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
  • amistre64
we have a vertical asymptote at x=0 becasue the ln(x) function has no value for x=0 and below
anonymous
  • 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
  • amistre64
it acts like 1/ln(x) for small values of x becasue that part takes control
amistre64
  • amistre64
for large values of x; 1/ln(x) plaays less part in it
anonymous
  • anonymous
when you say takes control what specifically do you mean
amistre64
  • 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
  • anonymous
oh you mean the y values are bigger is that what you mean by dominate/take over?
anonymous
  • anonymous
when you say it takes control for small values of x you mean negative x values?
amistre64
  • amistre64
i mean lose to 0
amistre64
  • amistre64
err.. close to 1; since ln(1) = 0
anonymous
  • anonymous
so wait...When 0
amistre64
  • amistre64
gets closer and closer to infinity
anonymous
  • anonymous
when I use my calculator from .1 to .9 are all negative numbers so that can't be right
anonymous
  • anonymous
:/
amistre64
  • amistre64
anything less than 1 is gonna be a negative # i think
anonymous
  • anonymous
right so its not approaching infinity..its approaching negative infinity right?
amistre64
  • 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
  • amistre64
infinity is infinity; if your on the negative side, then yeah, its -inf ...
anonymous
  • anonymous
When 0
amistre64
  • amistre64
correct
anonymous
  • anonymous
and then for larger values of x it looks more like ln(x) than 1/ln(x) when 0
amistre64
  • amistre64
no; stilll 1/ln(x) just on the + side
amistre64
  • amistre64
err.. i gotta learn to read better
amistre64
  • amistre64
for larger values of x; we get less and less out of it
anonymous
  • anonymous
what do you mean
amistre64
  • amistre64
1/ln(1,000,000,000) = .04825.... 1/ln(1,000,000,000,000) = .03619.....
anonymous
  • anonymous
you are taking a fraction of a larger number not a smaller so the difference is not as big?
amistre64
  • 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
  • anonymous
and finally when x>0 it looks more like the ln(x) part because..hm let me look..
anonymous
  • anonymous
as x approaches infinity..hm
anonymous
  • anonymous
:/

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