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Inopeki

  • 2 years ago

TuringTest, functions?

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  1. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    Yes

  3. TuringTest
    • 2 years ago
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    good now, is x a function of y ?

  4. Inopeki
    • 2 years ago
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    No

  5. Inopeki
    • 2 years ago
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    Cause x is the dependent variable

  6. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    y=x+7 y-7=x?

  8. TuringTest
    • 2 years ago
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    so is x a function of y?

  9. TuringTest
    • 2 years ago
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    can we switch their places?

  10. Inopeki
    • 2 years ago
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    They should work both ways.

  11. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    Yes?

  13. TuringTest
    • 2 years ago
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    yes, each x gives exactly one y now the hard part :P is x a function of y?

  14. Inopeki
    • 2 years ago
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    No, cause each y gives us 2 x

  15. Inopeki
    • 2 years ago
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    It could be a function of yx

  16. TuringTest
    • 2 years ago
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    how so? elaborate ;)

  17. zbay
    • 2 years ago
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    |dw:1326051567892:dw| every element of a is maped to an element of b

  18. Inopeki
    • 2 years ago
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    y=x^2 yx=x Therefore each y gives us 2 x It is a function of yx

  19. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    y/x=x actually

  21. Inopeki
    • 2 years ago
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    Lol i rushed it

  22. TuringTest
    • 2 years ago
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    that's not solved for x, you have x on both sides :/

  23. TuringTest
    • 2 years ago
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    take the square root of both sides

  24. Inopeki
    • 2 years ago
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    y1/2=x?

  25. TuringTest
    • 2 years ago
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    is the 1/2 and exponent?

  26. TuringTest
    • 2 years ago
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    an exponent*

  27. Inopeki
    • 2 years ago
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    Yes, didnt you say that yesterday?

  28. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    ALright

  30. TuringTest
    • 2 years ago
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    plus or minus...

  31. Inopeki
    • 2 years ago
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    What?

  32. TuringTest
    • 2 years ago
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    when you solve\[x^2=4\]there are two answers, remember? one positive and one negative. or did you not know that?

  33. Inopeki
    • 2 years ago
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    No, why?

  34. pokemon23
    • 2 years ago
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    TURNING TEST BUDDY :D

  35. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    hi pokemon :)

  37. Inopeki
    • 2 years ago
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    ooooh

  38. Inopeki
    • 2 years ago
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    since negative multiplied by negative is positive

  39. TuringTest
    • 2 years ago
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    exactly :D

  40. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    so that said, back on topic\[r^2=t\]is t a function of r? is r a function of t?

  43. Inopeki
    • 2 years ago
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    r is a function of t but not vice versa

  44. TuringTest
    • 2 years ago
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    actually vice versa, but not vice versa lol

  45. TuringTest
    • 2 years ago
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    which is a function of what? careful.

  46. Inopeki
    • 2 years ago
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    So t is a function of r?

  47. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    Ohh

  49. TuringTest
    • 2 years ago
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    each positive value of t will give two real answers*

  50. Inopeki
    • 2 years ago
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    Yeah, throw me another one

  51. TuringTest
    • 2 years ago
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    what about\[r=\sqrt t\]???

  52. Inopeki
    • 2 years ago
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    sqrt(t) cant be the dependent variable

  53. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    Oh, so if the +/- sign is gone then it can me the other way around?

  55. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    yes because we are only looking at the positive root I'll draw...

  57. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    Actually first the graph of r^2=t:|dw:1326054064202:dw|

  59. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    |dw:1326054308550:dw|^not a function|dw:1326054323340:dw|^is a function

  61. Inopeki
    • 2 years ago
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    because of the plus/minus it can be |dw:1326054372813:dw| or |dw:1326054351716:dw| Depending on if its negative or positive

  62. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    So that was wrong?

  64. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    No its not

  66. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    ut its x, not y. Shouldnt it be horizontal line test?

  68. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    so another way to see it, is|dw:1326055195330:dw|is y a function of x?

  70. Inopeki
    • 2 years ago
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    Yes

  71. TuringTest
    • 2 years ago
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    good is x a function of y?

  72. Inopeki
    • 2 years ago
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    no

  73. TuringTest
    • 2 years ago
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    show please

  74. Inopeki
    • 2 years ago
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    |dw:1326055271855:dw|

  75. TuringTest
    • 2 years ago
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    perfect :)

  76. Inopeki
    • 2 years ago
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    :D

  77. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    y is a function of x and x is a function of y

  79. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    r | t --- 0 | 1 2 | 3 5 | 9 7 | 3 4 | -14 is r(t) a function? is t(r) a function?

  81. Inopeki
    • 2 years ago
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    No Yes

  82. TuringTest
    • 2 years ago
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    correct! very nice, I think you understand the concept of functions :D

  83. Inopeki
    • 2 years ago
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    Seriously? :D

  84. TuringTest
    • 2 years ago
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    yes, you are right :D care to state your rationale?

  85. Inopeki
    • 2 years ago
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    Rationale?

  86. TuringTest
    • 2 years ago
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    why? your reason.

  87. TuringTest
    • 2 years ago
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    how did you know?

  88. Inopeki
    • 2 years ago
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    Oh, for the answer?

  89. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    right, so r(t) does not represent a function, where t(r) does perfect!!!!

  92. Inopeki
    • 2 years ago
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    Yeah!!! :DDD

  93. TuringTest
    • 2 years ago
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    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
    • 2 years ago
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    Ok :D Im going to take a 30min break now :)

  95. TuringTest
    • 2 years ago
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    well deserved, take your time :D

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