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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...?

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  1. amistre64
    • 5 years ago
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    its called 'squaring or cubing' a function

  2. anonymous
    • 5 years ago
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    oh ahahah okay thought there was more to it

  3. anonymous
    • 5 years ago
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    got time for another question?

  4. amistre64
    • 5 years ago
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    AN HOUR OR SO

  5. amistre64
    • 5 years ago
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    nothing like a caap lock to set the mood lol

  6. anonymous
    • 5 years ago
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    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

  7. amistre64
    • 5 years ago
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    yes, the sin in a cone

  8. anonymous
    • 5 years ago
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    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)

  9. amistre64
    • 5 years ago
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    yeah, i remember those :) dampened waves is not my cup o tea, nor is the differentials yet. gonna have to ask someone smarter

  10. amistre64
    • 5 years ago
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    pretty much you hav to find the peaks and valleys to draw the 'cone' from and determine its linear equation to help out

  11. anonymous
    • 5 years ago
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    hmm okay, well are you good at natural log functions like ln(x) stuff?

  12. amistre64
    • 5 years ago
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    im adequate with log functions

  13. anonymous
    • 5 years ago
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    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, 0<x<1 and x>0

  14. amistre64
    • 5 years ago
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    this is pretty much a graph with the slope of \({1\over ln(x)}\)

  15. anonymous
    • 5 years ago
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    so for example when x<0 it looks more like the x part because...type of thing

  16. anonymous
    • 5 years ago
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    right so its the inverse of ln(x)?

  17. amistre64
    • 5 years ago
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    not the inverse; that would be the graph of 'e'; its the 'flipped over the xaxis' graph of ln(x)

  18. anonymous
    • 5 years ago
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    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<x<1..hm

  19. amistre64
    • 5 years ago
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    we have a vertical asymptote at x=0 becasue the ln(x) function has no value for x=0 and below

  20. anonymous
    • 5 years ago
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    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)?

  21. amistre64
    • 5 years ago
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    it acts like 1/ln(x) for small values of x becasue that part takes control

  22. amistre64
    • 5 years ago
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    for large values of x; 1/ln(x) plaays less part in it

  23. anonymous
    • 5 years ago
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    when you say takes control what specifically do you mean

  24. amistre64
    • 5 years ago
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    same thing you stated really; the control of the function get governed by the part that becomes dominate for its corresponding value of x

  25. anonymous
    • 5 years ago
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    oh you mean the y values are bigger is that what you mean by dominate/take over?

  26. anonymous
    • 5 years ago
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    when you say it takes control for small values of x you mean negative x values?

  27. amistre64
    • 5 years ago
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    i mean lose to 0

  28. amistre64
    • 5 years ago
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    err.. close to 1; since ln(1) = 0

  29. anonymous
    • 5 years ago
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    so 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?

  30. amistre64
    • 5 years ago
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    gets closer and closer to infinity

  31. anonymous
    • 5 years ago
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    when I use my calculator from .1 to .9 are all negative numbers so that can't be right

  32. anonymous
    • 5 years ago
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    :/

  33. amistre64
    • 5 years ago
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    anything less than 1 is gonna be a negative # i think

  34. anonymous
    • 5 years ago
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    right so its not approaching infinity..its approaching negative infinity right?

  35. amistre64
    • 5 years ago
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    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

  36. amistre64
    • 5 years ago
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    infinity is infinity; if your on the negative side, then yeah, its -inf ...

  37. anonymous
    • 5 years ago
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    When 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. ?

  38. amistre64
    • 5 years ago
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    correct

  39. anonymous
    • 5 years ago
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    and then for larger values of x it looks more like ln(x) than 1/ln(x) when 0<x<1?

  40. amistre64
    • 5 years ago
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    no; stilll 1/ln(x) just on the + side

  41. amistre64
    • 5 years ago
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    err.. i gotta learn to read better

  42. amistre64
    • 5 years ago
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    for larger values of x; we get less and less out of it

  43. anonymous
    • 5 years ago
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    what do you mean

  44. amistre64
    • 5 years ago
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    1/ln(1,000,000,000) = .04825.... 1/ln(1,000,000,000,000) = .03619.....

  45. anonymous
    • 5 years ago
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    you are taking a fraction of a larger number not a smaller so the difference is not as big?

  46. amistre64
    • 5 years ago
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    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

  47. anonymous
    • 5 years ago
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    and finally when x>0 it looks more like the ln(x) part because..hm let me look..

  48. anonymous
    • 5 years ago
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    as x approaches infinity..hm

  49. anonymous
    • 5 years ago
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    :/

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