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anonymous

  • one year ago

given f(x) = x⌈x⌈x⌉⌉ and f(a) = 15. solve for a

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  1. anonymous
    • one year ago
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    what.... where is the tools drawing ? i cant draw the equations ._.

  2. Vocaloid
    • one year ago
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    are you on mobile?

  3. anonymous
    • one year ago
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    no. i just using my laptop. the problem like this given f(x) = x⌈x⌈x⌉⌉ and f(a) = 15. solve for a

  4. dan815
    • one year ago
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    what do those brackets mean

  5. anonymous
    • one year ago
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    actually ⌈x⌉ mean the smallest integer which not less than x, right ? help meh @dan :)

  6. Loser66
    • one year ago
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    ceiling function?

  7. anonymous
    • one year ago
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    yeah.... what's the start to do it, @Loser66 ?

  8. anonymous
    • one year ago
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    the complete question is find all real a which satisfy f(a) = 15

  9. Loser66
    • one year ago
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    I really don't know how to solve it :( , wolfram gives us the answer is 2.14, but how ?? @freckles

  10. freckles
    • one year ago
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    \[x [x[x]]=15 \\ [x[x]]=\frac{15}{x} \\ \text{ so } \frac{15}{x} \text{ is an integer } \text{ and we have that } \frac{15}{x}-1<x[x]\le \frac{15}{x} \\ \\ ... \text{ still thinking...}\]

  11. freckles
    • one year ago
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    \[\frac{15-x}{x}<x [x] \le \frac{15}{x} \\ \text{ assume } x >0 \\ \text{ we have } \frac{15-x}{x^2} <[x] \le \frac{15}{x^2}\] I don't know what this means yet if anything can be concluded from it

  12. anonymous
    • one year ago
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    Where is the smallest integer bracket at?

  13. anonymous
    • one year ago
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    I'll just let parentheses be it.

  14. freckles
    • one year ago
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    I just used brackets. I think there is some code you can type in.

  15. freckles
    • one year ago
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    \[\lceil x \rceil\] oh yeah you can use \lceil x \rceil

  16. anonymous
    • one year ago
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    Oh, I dont know much about symbols. Thanks for pointing that out.

  17. anonymous
    • one year ago
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    I'll just use parentheses. I am not sure with this. But I did this \[\sqrt[3]{15}=2.46\] 15/x=(x(x)) Isnt (2.46)=2.46? So (2.46 x 2.46)=6.071 6.071=15/2.46

  18. anonymous
    • one year ago
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    ⌈x ⌈x⌉⌉ = 15/x the right side is integer. so x must be an integer also right ?

  19. freckles
    • one year ago
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    no

  20. freckles
    • one year ago
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    15/x I think x is between 0 and 15 though for example x could be 15/7 see: 15/(15/7)=7

  21. freckles
    • one year ago
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    so what I'm saying is x could be an integer or it could be rational

  22. anonymous
    • one year ago
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    oh, ok.. i see now

  23. freckles
    • one year ago
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    I'm just having problems showing x=15/7 we can certainly show 15/7 is the solution by pluggin it in just having issues coming up with that answer :(

  24. anonymous
    • one year ago
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    What is the answer?

  25. anonymous
    • one year ago
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    7?

  26. anonymous
    • one year ago
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    not sure. it is an essay question, no choices :v

  27. anonymous
    • one year ago
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    a can be more than one answer. Since you wrote "the complete question is find all real a which satisfy f(a) = 15"

  28. anonymous
    • one year ago
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    yeah.. maybe a can be more than one answer, but how to get them ?

  29. freckles
    • one year ago
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    \[x \cdot [x [x]]=15 \\ \text{ if } x=\frac{15}{7} \text{ then } \\ \frac{15}{7}[ \frac{15}{7}[\frac{15}{7}]]=\frac{15}{7}[ \frac{15}{7}(3)]=\frac{15}{7}[\frac{45}{7}]=\frac{15}{7}(7)=15 \\ \text{ I didn't say } x=7 \\ 7[7[7]]=7[7(7)]=7[49]=7(49) \neq 15\]

  30. anonymous
    • one year ago
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    so, just using trials any x ?

  31. freckles
    • one year ago
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    \[2.46 \cdot [2.46 [2.46]]=2.46[2.46(3)]=2.46(8) \neq 15\]

  32. anonymous
    • one year ago
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    For 2.46 \[2\le x <3\] x can be 2.46 then 2.46(2.46) which is \[2.46 \times (2\le x <3)\] x can be 2.46=6.071 6.071=15/2.46 So I got x=2.46

  33. anonymous
    • one year ago
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    Just kidding. I dont know enough about ceiling and floor functions.

  34. freckles
    • one year ago
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    But @Shalante I think your method helped us to find what our solution should be near if we assume x is an an integer then \[x [x[x]]=x[x(x)]=x[x^2]=x(x^2)=x^3 \\ x^3=15 \text{ so I think solving for the real x value here } \\ \text{ will give us a number closed to our solution }\]

  35. freckles
    • one year ago
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    close*

  36. freckles
    • one year ago
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    but solving for x here and plugging x in we will see that 15/x is not an integer so x is not an integer and we can conclude x is a rational near the real cube root value of 15

  37. anonymous
    • one year ago
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    It doesnt need to be an integer. The x is outside the ceiling function.

  38. freckles
    • one year ago
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    yes i already said x cannot be an integer

  39. freckles
    • one year ago
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    x has to be a rational number

  40. freckles
    • one year ago
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    I was showing above by contradiction why x cannot be an integer

  41. anonymous
    • one year ago
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    I take that back. I'll try a new method (might take forever)

  42. anonymous
    • one year ago
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    You wana show me a screenshot of this question.

  43. freckles
    • one year ago
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    so this is what we have so far \[2<x<3 \\ \text{ and } \frac{15}{x} -1<x [x]\le \frac{15}{x} \\ \text{ where } \frac{15}{x} \text{ is an integer } \\ \text{ we also have } x \text{ is rational } \\ x=\frac{m}{n}, \text{ where } m \text{ and } n \text{ are integers not equal to 0}\] \[\frac{m}{n}[\frac{m}{n}[\frac{m}{n}]]=15 \\ \text{ let's see } 2 <\frac{m}{n}<3 \\ \text{ so we have } [\frac{m}{n}]=3 \\ \frac{m}{n}[\frac{m}{n}(3)]=15 \\ \frac{m}{n}[ \frac{3 m}{n}]=15\] \[6<\frac{3m}{n}<9 \text{ since } 2<\frac{m}{n}<3 \text{ so this means } [\frac{3m}{n}] \text{ could be } 7,8, \text{ or } 9 \\ \text{ let's go through the cases } \\ \] \[\text{ assume } [\frac{3m}{n}]=7 \text{ then } \frac{m}{n}(7)=15 \text{ and so } \frac{m}{n}=\frac{15}{7} \\ \text{ assume } [\frac{3m}{n}]=8 \text{ then } \frac{m}{n}(8)=15 \text{ so } \frac{m}{n}=\frac{15}{8} \\ \text{ assume } [\frac{3m}{n}]=9 \text{ then } \frac{m}{n}(9)=15 \text{ so } \frac{m}{n}=\frac{15}{9}\] I just figured out there was no point in writing x as m/n but I don't feel like changing them all back anyways you check your 3 cases to figure out if the solution

  44. anonymous
    • one year ago
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    I think the answer is somewhere along \[2.99<x<3\]

  45. freckles
    • one year ago
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    if you check the three cases above you should see 15/7 is the solution @Shalante

  46. anonymous
    • one year ago
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    Smallest integer of 2.99 is 2 2*2.999=5.998 Smallest integer of 5.998 is 5 5*2.999=14.995 which is close enough. 15/7 is 2.142 2.142 smallest integer is 2 2*2.142=4.284 Smallest integer of 4.284 is 4 2.142*4 is not 15

  47. freckles
    • one year ago
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    oh I thought we were doing ceiling function

  48. anonymous
    • one year ago
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    Oh yea I reversed it.

  49. anonymous
    • one year ago
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    That one guy wrote 2.14 at the beginning was correct.

  50. anonymous
    • one year ago
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    it should be between 2.14 and something like an interval.

  51. freckles
    • one year ago
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    you do know 15/7 is approximately 2.14286

  52. anonymous
    • one year ago
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    I kept thinking ceiling is a restraint meaning it has to be less.

  53. freckles
    • one year ago
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    but 2.14 is only an approximation to actual answer which is 15/7

  54. anonymous
    • one year ago
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    I got 15.00002 0.0 The question asked for all a values. An interval would help I guess.

  55. freckles
    • one year ago
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    well I narrowed it down to 3 cases for a you just check the 3 cases to see which is the answer I think there is only one solution

  56. anonymous
    • one year ago
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    14.99998 would be the same 15.00002

  57. anonymous
    • one year ago
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    I thinks theres more than one. Its close enough. You know chemistry? Its like a 0.001 percent error?

  58. anonymous
    • one year ago
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    Decimals are close enough especially in the thousandth. But it says find all, so I would use interval.

  59. anonymous
    • one year ago
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    Are you talking about 15/8 and 15/9? I dont think it would fit at all.

  60. freckles
    • one year ago
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    I come up with three cases a=15/8 a=15/9 a=15/7 using that a is between 2 and 3 which means the ceiling(a)=3 and so ceiling(3a)=7,8, or 9 since 3a is between 6 and 9 so this gave us three possibilities 7a=15 8a=15 9a=15 but pluggin back in we see only one of these is successful therefore the only solution is a=15/7 since that is the only success

  61. anonymous
    • one year ago
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    I got 2.999 earlier doing it the floor way by just plugging and chugging. Its actually not bad either.

  62. anonymous
    • one year ago
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    I wasnt familiar with greatest integer function until like 25 minutes ago. Thats when I start to understand and keep plugging. Its like guessing a number between 2-3 in the hundredth decimal and they kept telling you if its lower or higher.

  63. anonymous
    • one year ago
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    Not sure if your method worked since the problem is an easier one or thats the way it is. Because I am sure, I would just plug and chug in the more complex one and it does take that long either. like x(x(x))=15.4

  64. anonymous
    • one year ago
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    Actually it does. Lemme think of more complex ones.

  65. anonymous
    • one year ago
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    Try 16.36. I couldnt do it dividing it by 6 to 10.

  66. freckles
    • one year ago
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    you are trying to solve x(x(x))=15.4?

  67. freckles
    • one year ago
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    or x(x(x))=16.36?

  68. anonymous
    • one year ago
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    I solved 15.4 already using your method of dividing it like this 15.4/7. But I cant do it at 16.36

  69. anonymous
    • one year ago
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    Because I want to know if there is a rule that works on all of them. Or is it just some certain numbers?

  70. anonymous
    • one year ago
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    In math besides complex numbers or some other I dont know, almost all rules works on all numbers. Pretty interesting to know for this one.

  71. anonymous
    • one year ago
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    16.36/7 16.36/8 16.36/9 Besides plug and chug.

  72. anonymous
    • one year ago
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    Like a very fast way.

  73. freckles
    • one year ago
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    \[x[x[x]]=16.36 \\ \text{ we know } x \text{ is not an integer } \\ \text{ but so we can get in the neighborhood } \\ \text{ we will solve } x^3=16.36 \\ \text{ which will actually give us a real number and not an integer }\] \[x \approx 2.5 \\ \text{ so this means we know } \\ 2<x<3 \\ \text{ which means } [x]=3 \\ \\ \text{ so back \to the equation } \\ x[x[x]]=16.36 \\ x[x(3)]]=16.36 \\ x[3x]=16.36 \\ \text{ now recall } 2<x<3 \\ \text{ so we have } 6<3x<9 \\ \text{ so this means } \\ [3x]=7, 8, \text{ or } 9 \\ x(7)=16.36 \implies x=\frac{16.36}{7} \\ x(8)=16.36 \implies x=\frac{16.36}{8} \\ x(9)=16.36 \implies x=\frac{16.36}{9}\] we have 3 cases to check: checking: \[\frac{16.36}{7}[\frac{16.36}{7}[\frac{16.36}{7}]]=\frac{16.36}{7}[\frac{16.36}{7}(3)]=\frac{16.36}{7}(8) \neq 16.36 \\ \] \[\frac{16.36}{8}[\frac{16.36}{8}[\frac{16.36}{8}]]=\frac{16.36}{8}[\frac{16.36}{8}(3)]=\frac{16.36}{8}(7) \neq 16.36 \\ \] \[\frac{16.36}{9}[\frac{16.36}{9}[\frac{16.36}{9}]]=\frac{16.36}{9}[\frac{16.36}{9}(2)]=\frac{16.36}{9}(4) \neq 16.36\] So I think there is no real x such that the equation is true since none of the cases worked

  74. anonymous
    • one year ago
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    But 15.4 is not an integer and worked. 15.4/7=2.2 2.2>>3 2.2*3=6.2 6.2>>7 7*2.2=15.4 So this method only works for specific numbers then Maybe there is a pattern to it just like some only work on odd,even, prime, etc. Interesting.

  75. anonymous
    • one year ago
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    You already got your answer. No need to bump this.

  76. Loser66
    • one year ago
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    I have an idea: if x is an integer, then \(\sqrt[3] {15}=2.46\), and we know that \(2\leq x\leq 2.46\) \(f(2) =2\lceil 2\lceil 2\rceil \rceil =2\lceil 2*2\rceil=2*4 =8\) \(f(2.46) = 2.46\lceil 2.46\lceil 2.46\rceil \rceil =2.46\lceil 2.46*3\rceil=2.46*8 =19.86\) therefore, x must go to the left of 2.46. Now, by taking half way, \(\dfrac{2.46-2}{2}= 0.23\), we test 2.23 \(f(2.23)=2.23\lceil 2.23\lceil 2.23\rceil \rceil =2.23\lceil 2.23*3\rceil=2.23*7 =15.61\) The result is still higher than what we want, hence , we need "half way" again. \(\dfrac{2.23-2}{2}= 0.115\), we test 2.115 \(f(2.115)= 2.115\lceil 2.115\lceil 2.115\rceil \rceil =2.115\lceil 2.115*3\rceil=2.115*7 =14.805\) It is a little bit smaller than what we need, but now, we take "half way" to the right from 2.115 \(\dfrac{2.23-2.115}{2}=0.0575, so, we need test 2.115 + 0.0575=2.1725 \(f(2.1725) = 2.1725\lceil 2.1725\lceil 2.1725\rceil \rceil =2.1725\lceil 2.1725*3\rceil=2.1725*7 =15.2075\) it is still a little bit higher than what we need, but now, take half way to the left from 2.1725

  77. Loser66
    • one year ago
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    typo at the red mark, it is \(f(2.1725) = 2.1725\lceil 2.1725\lceil 2.1725\rceil \rceil =2.1725\lceil 2.1725*3\rceil=2.1725*7 =15.2075\)

  78. Loser66
    • one year ago
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    It is still a little bit higher than what we need, now, take half way between 2. 115 and 2.1725, it is 2.14375 Test : \(f(2.14375) =2.14375\lceil 2.14375\lceil 2.14375\rceil \rceil\\ =2.14375\lceil 2.14375*3\rceil=2.14375*7 =15.00625\) I think it is good enough.

  79. Loser66
    • one year ago
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    Hahahaha.... That is power of "sleeping". !! I meant after taking a snap, we can solve the problem we couldn't solve before.

  80. Loser66
    • one year ago
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    In general, this is |dw:1443617046497:dw|

  81. Loser66
    • one year ago
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    |dw:1443617112164:dw|

  82. Loser66
    • one year ago
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    |dw:1443617350315:dw|

  83. Loser66
    • one year ago
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    |dw:1443617476081:dw|

  84. Loser66
    • one year ago
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    |dw:1443617581563:dw|