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

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its called 'squaring or cubing' a function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh ahahah okay thought there was more to it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0got time for another question?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0nothing like a caap lock to set the mood lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay 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
 5 years ago
Best ResponseYou've already chosen the best response.0yes, the sin in a cone

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, 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
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, i remember those :) dampened waves is not my cup o tea, nor is the differentials yet. gonna have to ask someone smarter

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0pretty much you hav to find the peaks and valleys to draw the 'cone' from and determine its linear equation to help out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm okay, well are you good at natural log functions like ln(x) stuff?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0im adequate with log functions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok 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, 0<x<1 and x>0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0this is pretty much a graph with the slope of \({1\over ln(x)}\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so for example when x<0 it looks more like the x part because...type of thing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right so its the inverse of ln(x)?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0not the inverse; that would be the graph of 'e'; its the 'flipped over the xaxis' graph of ln(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh 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<x<1..hm

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we have a vertical asymptote at x=0 becasue the ln(x) function has no value for x=0 and below

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we 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
 5 years ago
Best ResponseYou've already chosen the best response.0it acts like 1/ln(x) for small values of x becasue that part takes control

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0for large values of x; 1/ln(x) plaays less part in it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when you say takes control what specifically do you mean

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0same thing you stated really; the control of the function get governed by the part that becomes dominate for its corresponding value of x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh you mean the y values are bigger is that what you mean by dominate/take over?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when you say it takes control for small values of x you mean negative x values?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0err.. close to 1; since ln(1) = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so wait...When 0<x<1, the graph of the function is closer to 1/ln(x) for smaller values of x because that part takes control, in other words gets closer and closer to 1?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0gets closer and closer to infinity

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when I use my calculator from .1 to .9 are all negative numbers so that can't be right

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0anything less than 1 is gonna be a negative # i think

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right so its not approaching infinity..its approaching negative infinity right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0as 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
 5 years ago
Best ResponseYou've already chosen the best response.0infinity is infinity; if your on the negative side, then yeah, its inf ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When 0<x<1, the graph of the function is closer to 1/ln(x) for smaller values of x because as x gets bigger, y gets infinitely negative. ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and then for larger values of x it looks more like ln(x) than 1/ln(x) when 0<x<1?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0no; stilll 1/ln(x) just on the + side

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0err.. i gotta learn to read better

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0for larger values of x; we get less and less out of it

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.01/ln(1,000,000,000) = .04825.... 1/ln(1,000,000,000,000) = .03619.....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you are taking a fraction of a larger number not a smaller so the difference is not as big?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0something 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
 5 years ago
Best ResponseYou've already chosen the best response.0and finally when x>0 it looks more like the ln(x) part because..hm let me look..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0as x approaches infinity..hm
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