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mathmath333

  • one year ago

Functions

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{if}\ f(x)=\dfrac{1}{g(x)}\ \text{then which of the following is correct} \hspace{.33em}\\~\\ &a.)\ f(f(g(f(x)))) =f(g(g(f(f(x)))))\hspace{.33em}\\~\\ &b.)\ f(g(g(f(f(x)))))=f(f(g(g(g(x))))) \hspace{.33em}\\~\\ &c.)\ g(g(f(f(g(f(x))))))=f(f(g(g(f(g(x)))))) \hspace{.33em}\\~\\ &d.)\ f(g(g(g(f(x)))))= g(g(f(f(f(x)))))\hspace{.33em}\\~\\ \end{align}}\)

  2. dan815
    • one year ago
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    lol really

  3. mathmath333
    • one year ago
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    my mind blasted out while typing the options.

  4. dan815
    • one year ago
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    i want to say C

  5. mathmath333
    • one year ago
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    yea option c.) is right

  6. mathmath333
    • one year ago
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    how did u figured so fast.

  7. dan815
    • one year ago
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    okay lets see why the same number of gs and fs makes it equal

  8. dan815
    • one year ago
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    lol just thought itd be nice if the only one with same gs and fs on both sides was the right one

  9. dan815
    • one year ago
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    same number* of

  10. mathmath333
    • one year ago
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    lol , funny judgment instinct

  11. mathmath333
    • one year ago
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    solomon's reaction after seeing this question. |dw:1435247455337:dw|

  12. dan815
    • one year ago
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    can we work out somethign simpler

  13. dan815
    • one year ago
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    f(g(x)) = g(f(x))?

  14. dan815
    • one year ago
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    is that true or what is the relationship between these 2

  15. anonymous
    • one year ago
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    You can pick for simplicity F(x)=x, g(x)=1/x.. I know it's not professional solution but th question who started being silly :P

  16. mathmath333
    • one year ago
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    i was thinking we could assume some values for this function

  17. mathmath333
    • one year ago
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    ya that one 1/x ,x.

  18. dan815
    • one year ago
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    oaky to clean it up a little the x is making it very annoying

  19. dan815
    • one year ago
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    |dw:1435248000948:dw|

  20. dan815
    • one year ago
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    g o (f o g) = (g o f) o g

  21. dan815
    • one year ago
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    do you know if this is true?

  22. mathmath333
    • one year ago
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    that looks liitle relief to eyes

  23. mathmath333
    • one year ago
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    yes it looks true

  24. dan815
    • one year ago
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    are you sure, that looks too convenient

  25. mathmath333
    • one year ago
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    yea i mean this is true [g o (f o g) = (g o f) o g] ,isn't it ?

  26. dan815
    • one year ago
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    are you saying that is true in general or its true for this function

  27. mathmath333
    • one year ago
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    true for this function

  28. dan815
    • one year ago
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    oh interesting!! how did u get that

  29. mathmath333
    • one year ago
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    i put g(x)=x,f(x)=1/x

  30. dan815
    • one year ago
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    thats cheating lol

  31. dan815
    • one year ago
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    @ganeshie8

  32. mathmath333
    • one year ago
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    but that type of cheating is allowed

  33. dan815
    • one year ago
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    hey i got the lat part done i think

  34. mathmath333
    • one year ago
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    u sure this is true ? f o g = f o f

  35. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align}& g(x)=\dfrac{1}{x}\hspace{.33em}\\~\\ & f(x)=x \hspace{.33em}\\~\\ & f(g(x))=f(\dfrac{1}{x})=\dfrac{1}{x} \hspace{.33em}\\~\\ & f(f(x))=f(x)=x \hspace{.33em}\\~\\ & f(g(x))\neq f(f(x)) \hspace{.33em}\\~\\ \end{align}}\)

  36. dan815
    • one year ago
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    i found my mistake plz ignore!

  37. ganeshie8
    • one year ago
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    to me, option \(c\) makes sense only if \(f\) and \(g\) are "inverse functions" and not reciprocals of each other

  38. ganeshie8
    • one year ago
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    *not simply reciprocals of each other

  39. ganeshie8
    • one year ago
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    then we can say the "undo" and "redo" count match on each side but if they're just reciprocals of each other, i don't really see how option c is right

  40. mathmath333
    • one year ago
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    i have check option c.) with substitution \(x\) and \(\dfrac{1}{x}\) it looks correct

  41. dan815
    • one year ago
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    is the answer C?

  42. mathmath333
    • one year ago
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    yes i told earlier

  43. dan815
    • one year ago
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    making sure :)

  44. ganeshie8
    • one year ago
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    does textbook give some explanation ?

  45. mathmath333
    • one year ago
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    yea wait

  46. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{Book: the number of g's and f's should be equal on the LHS and RHS} \hspace{.33em}\\~\\ & \normalsize \text{since both these functions are essentially inverse of each other} \hspace{.33em}\\~\\ \end{align}}\)

  47. mathmath333
    • one year ago
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    i wonder if this is some kind of property

  48. ParthKohli
    • one year ago
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    Multiplicative inverse is definitely not functional inverse. Try to convert all g's into 1/f's and you'll see...

  49. dan815
    • one year ago
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    i wonder what it means essentially

  50. ParthKohli
    • one year ago
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    It means that the author doesn't know how to do it either.

  51. dan815
    • one year ago
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    im wondering if we can show something weaker like f o g o f being 1/g o f o g for multiplicative inverses

  52. dan815
    • one year ago
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    not only is choice C just the same number of f and g but its symmetric too f and g rev so

  53. dan815
    • one year ago
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    complete inverses will obey the rule no matter how the f and g are placed but maybe this relation will hold as long as we have this symmetry

  54. dan815
    • one year ago
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    try picking some functions and checking it out, something that is not inverse

  55. ganeshie8
    • one year ago
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    f(x) = x+1 g(x) = 1/(x+1) f(g(x)) = 1/(x+1)+1 = (x+2)/(x+1) g(f(x)) = 1/(x+1+1) = 1/(x+2) they don't match

  56. ParthKohli
    • one year ago
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    \[f(x) = x+1\]\[g(x) = \frac{1}{x+1}\]\[f\circ g \circ f = f\left(\frac{1}{x+2}\right)= \frac{1}{x+2}+1\]And\[g\circ f \circ g = g\left(\frac{x+2 }{x+1}\right)\]So nah.

  57. ParthKohli
    • one year ago
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    Oh haha.

  58. mathmath333
    • one year ago
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    is the question wrong

  59. ParthKohli
    • one year ago
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    Definitely.

  60. ganeshie8
    • one year ago
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    would the question really make sense if we replace "reciprocal" with "functional inverse" ?

  61. ParthKohli
    • one year ago
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    Yeah. I hope the guy who copied the question didn't make the obvious mistake: using \(\cdots^{-1}\) and \(1/\cdots\) interchangeably.

  62. ganeshie8
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{if}\ f(x) \text{ and g(x)} \ \text{ are inverse functions, then which of the following is correct} \hspace{.33em}\\~\\ &a.)\ f(f(g(f(x)))) =f(g(g(f(f(x)))))\hspace{.33em}\\~\\ &b.)\ f(g(g(f(f(x)))))=f(f(g(g(g(x))))) \hspace{.33em}\\~\\ &c.)\ g(g(f(f(g(f(x))))))=f(f(g(g(f(g(x)))))) \hspace{.33em}\\~\\ &d.)\ f(g(g(g(f(x)))))= g(g(f(f(f(x)))))\hspace{.33em}\\~\\ \end{align}}\)

  63. ParthKohli
    • one year ago
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    Here is what we do: whenever \(f(g(\cdots ))\) or \(g(f(\cdots)) \) occurs, we replace it with \(\cdots\). So (A): LHS \(f(f(x))\) RHS \(f(g(f(x)) = f(x)\) so nah. (B): LHS \(f(g(f(x)) = f(x)\) RHS \(f(g(g(x)) = g(x)\) so nah. (C): LHS \(x\) RHS \(x\) so yes.

  64. ganeshie8
    • one year ago
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    that looks simple !

  65. ParthKohli
    • one year ago
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    Yeah, so the verdict is out: the guy who framed the question didn't think about it.

  66. anonymous
    • one year ago
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    I love functions

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