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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

its called 'squaring or cubing' a function

2. anonymous

oh ahahah okay thought there was more to it

3. anonymous

got time for another question?

4. amistre64

AN HOUR OR SO

5. amistre64

nothing like a caap lock to set the mood lol

6. 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

7. amistre64

yes, the sin in a cone

8. 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)

9. amistre64

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

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

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

12. amistre64

im adequate with log functions

13. 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, 0<x<1 and x>0

14. amistre64

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

15. anonymous

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

16. anonymous

right so its the inverse of ln(x)?

17. amistre64

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

18. 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<x<1..hm

19. amistre64

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

20. 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)?

21. amistre64

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

22. amistre64

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

23. anonymous

when you say takes control what specifically do you mean

24. 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

25. anonymous

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

26. anonymous

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

27. amistre64

i mean lose to 0

28. amistre64

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

29. anonymous

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

gets closer and closer to infinity

31. anonymous

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

32. anonymous

:/

33. amistre64

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

34. anonymous

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

35. 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

36. amistre64

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

37. anonymous

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

correct

39. anonymous

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

40. amistre64

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

41. amistre64

err.. i gotta learn to read better

42. amistre64

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

43. anonymous

what do you mean

44. amistre64

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

45. anonymous

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

46. 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

47. anonymous

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

48. anonymous

as x approaches infinity..hm

49. anonymous

:/

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