## mathmath333 one year ago functions

1. mathmath333

\large \color{black}{\begin{align} & \normalsize \text{Let }\ f(x) \text{be function such that } \hspace{.33em}\\~\\ & f(x)f(x+1)=-f(x-1)f(x-2)f(x-3)f(x-4) ,\ \ x\geq 0 \hspace{.33em}\\~\\ & f(83)=81 \hspace{.33em}\\~\\ & f(77)=9 \hspace{.33em}\\~\\ & f(102)=? \hspace{.33em}\\~\\ & a.)\ 27 \hspace{.33em}\\~\\ & b.)\ 54 \hspace{.33em}\\~\\ & c.)\ 729 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{Data insufficient} \hspace{.33em}\\~\\ \end{align}}

2. mathmath333

it should have a fast way , the time given to solve such problem is average 2 min

3. dan815

that negative sign should really be there right, on the right side of the eqn

4. mathmath333

yes its there

5. dan815

is it C?

6. dan815

orr Data insufficient

7. freckles

@mathmath333 I always like your questions. Can I ask where do you get them from?

8. dan815

here is something we do know |dw:1435263889685:dw|

9. mathmath333
10. dan815

and every multiplication of 2 of them can be rewritten recursively

11. mathmath333

i have just checked into solution set , it has given solution for $$f(81)$$ and not $$f(102)$$ as described in the question

12. dan815

f(77)=9 f(83)=9^2 f(102)=? and 9^3 is 729 if this turns up

13. dan815

it could be lol, i feel like the separation is too much cant write anythign recursively

14. mathmath333

lol i think the question is this book has a typo \large \color{black}{\begin{align} & \normalsize \text{Let }\ f(x) \text{be function such that } \hspace{.33em}\\~\\ & f(x)f(x+1)=-f(x-1)f(x-2)f(x-3)f(x-4) ,\ \ x\geq 0 \hspace{.33em}\\~\\ & f(83)=81 \hspace{.33em}\\~\\ & f(77)=9 \hspace{.33em}\\~\\ &\color{red}{ f(81)}=? \hspace{.33em}\\~\\ & a.)\ 27 \hspace{.33em}\\~\\ & b.)\ 54 \hspace{.33em}\\~\\ & c.)\ 729 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{Data insufficient} \hspace{.33em}\\~\\ \end{align}}

15. mathmath333

27 is given as the answer.

16. dan815

okay lets see so 3^2, 3^3 and 3^4 for f(77), f(77+4), f(77+6)

17. freckles

actually I think we get $(f(81))^2=729$

18. freckles

let me post my work

19. dan815

can we do another question which we know doesnt have a typo xD

20. mathmath333

\large \color{black}{\begin{align} & f(82)f(83)=-f(81)f(80)f(79)f(78) \ -\color{red}{(1)}\hspace{.33em}\\~\\ & f(81)f(82)=-f(80)f(79)f(78)f(77) \ -\color{red}{(2)}\hspace{.33em}\\~\\ \end{align}} dividing 1 and 2 works

21. dan815

oh beauty

22. freckles

$f(82)f(83)=-f(81)f(80)f(79)f(78) \\ f(82) \cdot 81 =-f(81)f(80)f(79)f(78) \\ 81=\frac{-f(81)f(80)f(79)f(78)}{f(82)} \\ \text{ now we plug \in } 81 \\ f(81)f(82)=-f(80)f(79)f(78)f(77) \\ f(81)=\frac{-f(80)f(79)f(78)}{f(82)} f(77) \\ \text{ multiply both sides by} f(81) \\ f(81) \cdot f(81)=\frac{-f(81)f(80)f(79)f(78)}{f(82)} f(77)$

23. dan815

thats really cool, give me another one!

24. freckles

we already said that one thingy was 81 so you have $(f(81))^2=81 f(77)$

25. freckles

and I agree with @dan815 more please!

26. anonymous

I better go to sleep! very clever @freckles :-)

27. mathmath333

well i only post the one's i stuck at, let's see if i have some tough ones

28. mathmath333

\large \color{black}{\begin{align} & \normalsize \text{let}\ f(x)=121-x^2,\ g(x)=|x-8|+|x+8| \hspace{.33em}\\~\\ & \normalsize \text{and}\ h(x)=\text{min}(f(x),g(x)). \hspace{.33em}\\~\\ & \normalsize \text{What is the number of integer values of x for which } \hspace{.33em}\\~\\ & h(x)\ \normalsize \text{is equal to a positive integral value? } \hspace{.33em}\\~\\ & a.)\ 17 \hspace{.33em}\\~\\ & b.)\ 19 \hspace{.33em}\\~\\ & c.)\ 21 \hspace{.33em}\\~\\ & d.)\ 23 \hspace{.33em}\\~\\ \end{align}}

29. mathmath333

@dan815 @freckles

30. freckles

ok I'm here and looking now

31. mathmath333

yep good luck

32. Michele_Laino

by induction, I got this: $\Large {\left\{ {f\left( n \right)} \right\}^2} = f\left( {n + 2} \right) \times f\left( {n - 4} \right)$

33. Michele_Laino

setting n=81, we get: $\Large {\left\{ {f\left( {81} \right)} \right\}^2} = f\left( {83} \right)f\left( {77} \right)$ namely the result of @freckles

34. freckles

I have to come back and look this one crab time sorry

35. dan815

i dont quite understand what does it mean min(f(x),g(x))

36. mathmath333

u need to graph that f(x) and g(x) and find the intersection , and as it asks minimum u have to choose the lowest part with respect to that

37. mathmath333

it can also be done without graphing

38. dan815

oh ok i see now

39. mathmath333

example min(f(x),g(x)) for |dw:1435266966304:dw| f(x)=-x,g(x) =-x^2+5

40. mathmath333

|dw:1435267032841:dw|

41. dan815

i think to start off look at g1(x) = |x-8| and g2(x) = |x+8|

42. dan815

this will give us a constant when 0<|x| < 8

43. dan815

it will be 16 in that domain

44. dan815

from -8 to 8

45. mathmath333

it looks line this|dw:1435267447526:dw|

46. dan815

yes exactly

47. dan815

|dw:1435267527733:dw|

48. mathmath333

|dw:1435267567368:dw|

49. dan815

u want the number of integers of x in that intersection?

50. mathmath333

|dw:1435267618964:dw|

51. dan815

between the intersection?

52. dan815

|dw:1435267649589:dw|

53. dan815

or just the min on the abs value?

54. mathmath333

well i m still confused on interpreeting the quetion let me think

55. dan815

ya me too the min(f(x),g(x)) still dunno what that means exacttllyy

56. dan815

there are 17 points from that flat line, for hte integers then we have y=2x and y=-2x lines from the abs value

57. mathmath333

anser given is 21

58. dan815

yeah there will be 2 more from the intersection on each side

59. dan815

we see 11^2-x^2=2x x^2+2x-11^2=0 solve for the roots and see how many integers we can fit

60. dan815

you get sqrt 122 -1 so its greater than 10 barely that means another 2 from the right side and another 2 from the left side

61. dan815

17+4=21

62. mathmath333

it is not giving rational roots hmm

63. dan815

we just want to see the bound

64. mathmath333

yea we need to count the number line of x of the intersection part.

65. dan815

tbh i think the question is worded wrong it should just say.. how many integer solutions are there that are less than the quadratic

66. mathmath333

these are framed to confuse students majority are confuzing

67. dan815

lol thats annoying -.- they should confuse us with hard questions, not easy ones worded badly

68. mathmath333

my head spins now going to sleep

69. dan815

okay cya! thanks for the questions

70. mathmath333

thnks

71. dan815

looks like you already found this but ill leave this here, just in case on how we got the equations 2x and -2x for the lines

72. dan815

|dw:1435268655727:dw|

73. dan815

|dw:1435268725528:dw|

74. freckles

there are 21 integers between -11 and 11 (not including the endpoints) because at the endpoints h=0 which isn't a positive number

75. freckles

like i didn't want to include the endpoints because I want h positive

76. freckles

and anything outside the interval I mentioned h would be negative

77. freckles

|dw:1435269002394:dw|

78. freckles

for example h(-10)=g(-10) (since f(-10)>g(-10)) ... h(-5)=g(-5) (since f((-5)>g(-5)) ... h(0)=g(0) (since f(0)>g(0)) ..so on... h(x)=g(x) for integer solutions between -10 and 10 (inclusive) h(x)=f(x) for integers solutions on (-inf,-11] union [11,inf) but h(x)<=0 there so what we want to look at is: h(x)=g(x) for integer solutions between -10 and 10 (inclusive) since h(x)>0 here $|x-8|+|x+8|=121-x^2 \\ \text{ from our graph one solution occurs on } (-\inf,-8) \\ -(x-8)-(x+8)=121-x^2 \\ (x-1)^2=122 \\ x= 1 \pm \sqrt{122} \\ x=1-\sqrt{122} \text{ is only valid on that interval } \\ x \approx -10.045 \\ |x-8|+|x+8|=121-x^2 \\ \text{ the other solution \to this occurs on } (8,\infty) \\ x-8+x+8=121-x^2 \\ (x+1)^2=122 \ x=-1 \pm \sqrt{121} \\ x=-1 +\sqrt{122} \text{ is only valid on that interval } \\ x=10.045$ |dw:1435269711175:dw|

79. freckles

anyways just saying all of this just in case you guys were still confused on the h=min(f,g) thing

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