## Inopeki Group Title TuringTest, functions? 2 years ago 2 years ago

1. TuringTest Group Title

Right, so now that we've covered a lot of that let's look for something tricky: starting simple: y=x+7 is y a function of x?

2. Inopeki Group Title

Yes

3. TuringTest Group Title

good now, is x a function of y ?

4. Inopeki Group Title

No

5. Inopeki Group Title

Cause x is the dependent variable

6. TuringTest Group Title

x is the independent variable, but the independent variable can be a function of the dependent variable sometimes. It can work both ways. Not always, though. You must check. to find out try solving for x above

7. Inopeki Group Title

y=x+7 y-7=x?

8. TuringTest Group Title

so is x a function of y?

9. TuringTest Group Title

can we switch their places?

10. Inopeki Group Title

They should work both ways.

11. TuringTest Group Title

right, so in y=x+7 we can write one variable as a function of the other at will x=y-7 so we can change the way we look at this and call it f(x)=x+7=y or g(y)=y-7=x they are both functions and both equivalent. now trickier: what about y=x^2 is y a function of x ?

12. Inopeki Group Title

Yes?

13. TuringTest Group Title

yes, each x gives exactly one y now the hard part :P is x a function of y?

14. Inopeki Group Title

No, cause each y gives us 2 x

15. Inopeki Group Title

It could be a function of yx

16. TuringTest Group Title

how so? elaborate ;)

17. zbay Group Title

|dw:1326051567892:dw| every element of a is maped to an element of b

18. Inopeki Group Title

y=x^2 yx=x Therefore each y gives us 2 x It is a function of yx

19. TuringTest Group Title

nice try but not, the algebra is wrong btw zbay's drawing show essentially the same concept as the tables I made y=x^2 solve for x do you know how?

20. Inopeki Group Title

y/x=x actually

21. Inopeki Group Title

Lol i rushed it

22. TuringTest Group Title

that's not solved for x, you have x on both sides :/

23. TuringTest Group Title

take the square root of both sides

24. Inopeki Group Title

y1/2=x?

25. TuringTest Group Title

is the 1/2 and exponent?

26. TuringTest Group Title

an exponent*

27. Inopeki Group Title

Yes, didnt you say that yesterday?

28. TuringTest Group Title

yes, just making sure, you should write it as y^(1/2)=x to avoid confusion that is right but you forgot one little detail about taking a square root of both sides of an equation...

29. Inopeki Group Title

ALright

30. TuringTest Group Title

plus or minus...

31. Inopeki Group Title

What?

32. TuringTest Group Title

when you solve$x^2=4$there are two answers, remember? one positive and one negative. or did you not know that?

33. Inopeki Group Title

No, why?

34. pokemon23 Group Title

TURNING TEST BUDDY :D

35. TuringTest Group Title

look at two different ways of getting the answer:$x^2=y$look at y=4:$2^2=4$$(-2)^2=4$so x=2 and x=-2 BOTH correspond to y=4 This is fine for y as a function of x, but means that x cannot be a function of y, because one y corresponds to multiple x.

36. TuringTest Group Title

hi pokemon :)

37. Inopeki Group Title

ooooh

38. Inopeki Group Title

since negative multiplied by negative is positive

39. TuringTest Group Title

exactly :D

40. TuringTest Group Title

so whenever you have to take the square root like that we must write$x^2=y\to x=\pm\sqrt y$which shows that x is not a function of y sometimes we can ignore this and only look at the positive root to make it a function, but we may have to adjust things accordingly to do that. Here is something many tutors on Open Study still don't know so listen up...

41. TuringTest Group Title

If you are asked What is $\sqrt4$there is only one answer, 2 but if you are are asked $x^2=4$what is x? you must look at both plus and minus:$x^2=4\to x=\pm2$ Many tutors here think that you always need +/- when you look at a square root. The truth is you only do that when you have to TAKE the square root of both sides of an equation. There, be a step ahead ;)

42. TuringTest Group Title

so that said, back on topic$r^2=t$is t a function of r? is r a function of t?

43. Inopeki Group Title

r is a function of t but not vice versa

44. TuringTest Group Title

actually vice versa, but not vice versa lol

45. TuringTest Group Title

which is a function of what? careful.

46. Inopeki Group Title

So t is a function of r?

47. TuringTest Group Title

yes, because it's okay for t to be the dependent variable. one r gives one t If we turn it around and solve for r however, we have to use +/-sqrt so that means r won't be a function of t.$r^2=t\to r=\pm\sqrt t$so each value of t except zero will give us two real answers. not a function

48. Inopeki Group Title

Ohh

49. TuringTest Group Title

each positive value of t will give two real answers*

50. Inopeki Group Title

Yeah, throw me another one

51. TuringTest Group Title

what about$r=\sqrt t$???

52. Inopeki Group Title

sqrt(t) cant be the dependent variable

53. TuringTest Group Title

actually it can because I took away the +/- sign remember what I said about how many tutors are wrong about square roots? in the case of$\sqrt4=2$we have one answer, but for$x^2=4\to x=\pm2$we have two answers. so this is equivalent to the first case, we are only looking at the /positive/ square root.

54. Inopeki Group Title

Oh, so if the +/- sign is gone then it can me the other way around?

55. TuringTest Group Title

it is the act of TAKING the square root that introduces the +/-, which is what makes it not a function. If we don't have to actually take the square root, then it is just positive. so$r=\sqrt t$is a function

56. TuringTest Group Title

yes because we are only looking at the positive root I'll draw...

57. TuringTest Group Title

I made our graph with r as the vertical direction so we could use the vertical line test. Think of how that works.|dw:1326053937990:dw|I'll draw in r=sqrt(t)...

58. TuringTest Group Title

Actually first the graph of r^2=t:|dw:1326054064202:dw|

59. TuringTest Group Title

Now the graph of just$r=\sqrt t$|dw:1326054150478:dw|see how we just use the positive part? notice how r^2=t does not pass the vertical line test with t as the independent variable, so r is not a function of t however in our last graph we clearly do have r=sqrt(t) with r as a function ot t.

60. TuringTest Group Title

|dw:1326054308550:dw|^not a function|dw:1326054323340:dw|^is a function

61. Inopeki Group Title

because of the plus/minus it can be |dw:1326054372813:dw| or |dw:1326054351716:dw| Depending on if its negative or positive

62. TuringTest Group Title

but that would be splitting the graph up from it's original form if we start Let me try to convey the idea graphically and with the table...

63. Inopeki Group Title

So that was wrong?

64. TuringTest Group Title

Nothong really I just want to clear up the idea of what it means to break up the function like that. say we have$y=x^2$|dw:1326054620838:dw|you would agree that y is a function of x, yes?

65. Inopeki Group Title

No its not

66. TuringTest Group Title

because it passes the vertical line test|dw:1326054781944:dw|but what if we changed the coordinates on the graph put x vertical and y horizontal|dw:1326054851735:dw|even though the function hasn't changed, by changing the axis and using the vertical line test we show that y is a function of x, but not vice versa

67. Inopeki Group Title

ut its x, not y. Shouldnt it be horizontal line test?

68. TuringTest Group Title

no, vertical line test for a graph of independent variable on the horizontal dependent on the vertical however doing a horizontal line test would amount to the same thing as changing the axes if you noticed ;)|dw:1326055073462:dw|same result, we see that y is a function of x, but not the other way around.

69. TuringTest Group Title

so another way to see it, is|dw:1326055195330:dw|is y a function of x?

70. Inopeki Group Title

Yes

71. TuringTest Group Title

good is x a function of y?

72. Inopeki Group Title

no

73. TuringTest Group Title

74. Inopeki Group Title

|dw:1326055271855:dw|

75. TuringTest Group Title

perfect :)

76. Inopeki Group Title

:D

77. TuringTest Group Title

last thing is easier, tables... x | y ---- 2 | 2 5 | 1 0 | 3 4 | 6 is y a function of x ? is x a function of y ?

78. Inopeki Group Title

y is a function of x and x is a function of y

79. TuringTest Group Title

excellent actually I should be more careful with my language. With the table we can only say that this is potentially a function because we don't necessarily have all the values. That's a technicality though. ok last one...

80. TuringTest Group Title

r | t --- 0 | 1 2 | 3 5 | 9 7 | 3 4 | -14 is r(t) a function? is t(r) a function?

81. Inopeki Group Title

No Yes

82. TuringTest Group Title

correct! very nice, I think you understand the concept of functions :D

83. Inopeki Group Title

Seriously? :D

84. TuringTest Group Title

yes, you are right :D care to state your rationale?

85. Inopeki Group Title

Rationale?

86. TuringTest Group Title

87. TuringTest Group Title

how did you know?

88. Inopeki Group Title

89. TuringTest Group Title

yeah r | t --- 0 | 1 2 | 3 5 | 9 7 | 3 4 | -14 is r(t) a function? is t(r) a function WHY?

90. Inopeki Group Title

r | t --- 0 | 1 2 | 3<-\ 5 | 9 These make t have several values of the same variable (2,7) I dont see 7 | 3<-/ duplicates on the other side. 4 | -14

91. TuringTest Group Title

right, so r(t) does not represent a function, where t(r) does perfect!!!!

92. Inopeki Group Title

Yeah!!! :DDD

93. TuringTest Group Title

so now you have three interpretations of a function: graphical, numerical, and mathematical. try to harmonize them in your mind, they will be your best friends :)

94. Inopeki Group Title

Ok :D Im going to take a 30min break now :)

95. TuringTest Group Title

well deserved, take your time :D