## mathmath333 one year ago Functions

1. mathmath333

\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

lol really

3. mathmath333

my mind blasted out while typing the options.

4. dan815

i want to say C

5. mathmath333

yea option c.) is right

6. mathmath333

how did u figured so fast.

7. dan815

okay lets see why the same number of gs and fs makes it equal

8. dan815

lol just thought itd be nice if the only one with same gs and fs on both sides was the right one

9. dan815

same number* of

10. mathmath333

lol , funny judgment instinct

11. mathmath333

solomon's reaction after seeing this question. |dw:1435247455337:dw|

12. dan815

can we work out somethign simpler

13. dan815

f(g(x)) = g(f(x))?

14. dan815

is that true or what is the relationship between these 2

15. anonymous

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

i was thinking we could assume some values for this function

17. mathmath333

ya that one 1/x ,x.

18. dan815

oaky to clean it up a little the x is making it very annoying

19. dan815

|dw:1435248000948:dw|

20. dan815

g o (f o g) = (g o f) o g

21. dan815

do you know if this is true?

22. mathmath333

that looks liitle relief to eyes

23. mathmath333

yes it looks true

24. dan815

are you sure, that looks too convenient

25. mathmath333

yea i mean this is true [g o (f o g) = (g o f) o g] ,isn't it ?

26. dan815

are you saying that is true in general or its true for this function

27. mathmath333

true for this function

28. dan815

oh interesting!! how did u get that

29. mathmath333

i put g(x)=x,f(x)=1/x

30. dan815

thats cheating lol

31. dan815

@ganeshie8

32. mathmath333

but that type of cheating is allowed

33. dan815

hey i got the lat part done i think

34. mathmath333

u sure this is true ? f o g = f o f

35. mathmath333

\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

i found my mistake plz ignore!

37. ganeshie8

to me, option $$c$$ makes sense only if $$f$$ and $$g$$ are "inverse functions" and not reciprocals of each other

38. ganeshie8

*not simply reciprocals of each other

39. ganeshie8

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

i have check option c.) with substitution $$x$$ and $$\dfrac{1}{x}$$ it looks correct

41. dan815

42. mathmath333

yes i told earlier

43. dan815

making sure :)

44. ganeshie8

does textbook give some explanation ?

45. mathmath333

yea wait

46. mathmath333

\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

i wonder if this is some kind of property

48. ParthKohli

Multiplicative inverse is definitely not functional inverse. Try to convert all g's into 1/f's and you'll see...

49. dan815

i wonder what it means essentially

50. ParthKohli

It means that the author doesn't know how to do it either.

51. dan815

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

not only is choice C just the same number of f and g but its symmetric too f and g rev so

53. dan815

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

try picking some functions and checking it out, something that is not inverse

55. ganeshie8

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

$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

Oh haha.

58. mathmath333

is the question wrong

59. ParthKohli

Definitely.

60. ganeshie8

would the question really make sense if we replace "reciprocal" with "functional inverse" ?

61. ParthKohli

Yeah. I hope the guy who copied the question didn't make the obvious mistake: using $$\cdots^{-1}$$ and $$1/\cdots$$ interchangeably.

62. ganeshie8

\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

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

that looks simple !

65. ParthKohli

Yeah, so the verdict is out: the guy who framed the question didn't think about it.

66. anonymous

I love functions