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anonymous
 one year ago
given f(x) = x⌈x⌈x⌉⌉ and f(a) = 15. solve for a
anonymous
 one year ago
given f(x) = x⌈x⌈x⌉⌉ and f(a) = 15. solve for a

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what.... where is the tools drawing ? i cant draw the equations ._.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no. i just using my laptop. the problem like this given f(x) = x⌈x⌈x⌉⌉ and f(a) = 15. solve for a

dan815
 one year ago
Best ResponseYou've already chosen the best response.0what do those brackets mean

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually ⌈x⌉ mean the smallest integer which not less than x, right ? help meh @dan :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah.... what's the start to do it, @Loser66 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the complete question is find all real a which satisfy f(a) = 15

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I really don't know how to solve it :( , wolfram gives us the answer is 2.14, but how ?? @freckles

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[x [x[x]]=15 \\ [x[x]]=\frac{15}{x} \\ \text{ so } \frac{15}{x} \text{ is an integer } \text{ and we have that } \frac{15}{x}1<x[x]\le \frac{15}{x} \\ \\ ... \text{ still thinking...}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[\frac{15x}{x}<x [x] \le \frac{15}{x} \\ \text{ assume } x >0 \\ \text{ we have } \frac{15x}{x^2} <[x] \le \frac{15}{x^2}\] I don't know what this means yet if anything can be concluded from it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Where is the smallest integer bracket at?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'll just let parentheses be it.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4I just used brackets. I think there is some code you can type in.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[\lceil x \rceil\] oh yeah you can use \lceil x \rceil

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, I dont know much about symbols. Thanks for pointing that out.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0⌈x ⌈x⌉⌉ = 15/x the right side is integer. so x must be an integer also right ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.415/x I think x is between 0 and 15 though for example x could be 15/7 see: 15/(15/7)=7

freckles
 one year ago
Best ResponseYou've already chosen the best response.4so what I'm saying is x could be an integer or it could be rational

freckles
 one year ago
Best ResponseYou've already chosen the best response.4I'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 :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not sure. it is an essay question, no choices :v

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a can be more than one answer. Since you wrote "the complete question is find all real a which satisfy f(a) = 15"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah.. maybe a can be more than one answer, but how to get them ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[x \cdot [x [x]]=15 \\ \text{ if } x=\frac{15}{7} \text{ then } \\ \frac{15}{7}[ \frac{15}{7}[\frac{15}{7}]]=\frac{15}{7}[ \frac{15}{7}(3)]=\frac{15}{7}[\frac{45}{7}]=\frac{15}{7}(7)=15 \\ \text{ I didn't say } x=7 \\ 7[7[7]]=7[7(7)]=7[49]=7(49) \neq 15\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so, just using trials any x ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[2.46 \cdot [2.46 [2.46]]=2.46[2.46(3)]=2.46(8) \neq 15\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Just kidding. I dont know enough about ceiling and floor functions.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4But @Shalante I think your method helped us to find what our solution should be near if we assume x is an an integer then \[x [x[x]]=x[x(x)]=x[x^2]=x(x^2)=x^3 \\ x^3=15 \text{ so I think solving for the real x value here } \\ \text{ will give us a number closed to our solution }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4but 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It doesnt need to be an integer. The x is outside the ceiling function.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4yes i already said x cannot be an integer

freckles
 one year ago
Best ResponseYou've already chosen the best response.4x has to be a rational number

freckles
 one year ago
Best ResponseYou've already chosen the best response.4I was showing above by contradiction why x cannot be an integer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I take that back. I'll try a new method (might take forever)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You wana show me a screenshot of this question.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4so 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think the answer is somewhere along \[2.99<x<3\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4if you check the three cases above you should see 15/7 is the solution @Shalante

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Smallest 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.4oh I thought we were doing ceiling function

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh yea I reversed it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That one guy wrote 2.14 at the beginning was correct.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it should be between 2.14 and something like an interval.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4you do know 15/7 is approximately 2.14286

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I kept thinking ceiling is a restraint meaning it has to be less.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4but 2.14 is only an approximation to actual answer which is 15/7

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got 15.00002 0.0 The question asked for all a values. An interval would help I guess.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4well 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.014.99998 would be the same 15.00002

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I thinks theres more than one. Its close enough. You know chemistry? Its like a 0.001 percent error?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Decimals are close enough especially in the thousandth. But it says find all, so I would use interval.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Are you talking about 15/8 and 15/9? I dont think it would fit at all.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4I 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got 2.999 earlier doing it the floor way by just plugging and chugging. Its actually not bad either.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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 23 in the hundredth decimal and they kept telling you if its lower or higher.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Actually it does. Lemme think of more complex ones.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Try 16.36. I couldnt do it dividing it by 6 to 10.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4you are trying to solve x(x(x))=15.4?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I solved 15.4 already using your method of dividing it like this 15.4/7. But I cant do it at 16.36

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Because I want to know if there is a rule that works on all of them. Or is it just some certain numbers?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In math besides complex numbers or some other I dont know, almost all rules works on all numbers. Pretty interesting to know for this one.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.016.36/7 16.36/8 16.36/9 Besides plug and chug.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Like a very fast way.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[x[x[x]]=16.36 \\ \text{ we know } x \text{ is not an integer } \\ \text{ but so we can get in the neighborhood } \\ \text{ we will solve } x^3=16.36 \\ \text{ which will actually give us a real number and not an integer }\] \[x \approx 2.5 \\ \text{ so this means we know } \\ 2<x<3 \\ \text{ which means } [x]=3 \\ \\ \text{ so back \to the equation } \\ x[x[x]]=16.36 \\ x[x(3)]]=16.36 \\ x[3x]=16.36 \\ \text{ now recall } 2<x<3 \\ \text{ so we have } 6<3x<9 \\ \text{ so this means } \\ [3x]=7, 8, \text{ or } 9 \\ x(7)=16.36 \implies x=\frac{16.36}{7} \\ x(8)=16.36 \implies x=\frac{16.36}{8} \\ x(9)=16.36 \implies x=\frac{16.36}{9}\] 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You already got your answer. No need to bump this.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I 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, \(\dfrac{2.462}{2}= 0.23\), 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. \(\dfrac{2.232}{2}= 0.115\), 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{2.232.115}{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

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0typo at the red mark, it is \(f(2.1725) = 2.1725\lceil 2.1725\lceil 2.1725\rceil \rceil =2.1725\lceil 2.1725*3\rceil=2.1725*7 =15.2075\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0It 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.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Hahahaha.... That is power of "sleeping". !! I meant after taking a snap, we can solve the problem we couldn't solve before.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0In general, this is dw:1443617046497:dw