anonymous
  • anonymous
given f(x) = x⌈x⌈x⌉⌉ and f(a) = 15. solve for a
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
what.... where is the tools drawing ? i cant draw the equations ._.
Vocaloid
  • Vocaloid
are you on mobile?
anonymous
  • anonymous
no. i just using my laptop. the problem like this given f(x) = x⌈x⌈x⌉⌉ and f(a) = 15. solve for a

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dan815
  • dan815
what do those brackets mean
anonymous
  • anonymous
actually ⌈x⌉ mean the smallest integer which not less than x, right ? help meh @dan :)
Loser66
  • Loser66
ceiling function?
anonymous
  • anonymous
yeah.... what's the start to do it, @Loser66 ?
anonymous
  • anonymous
the complete question is find all real a which satisfy f(a) = 15
Loser66
  • Loser66
I really don't know how to solve it :( , wolfram gives us the answer is 2.14, but how ?? @freckles
freckles
  • freckles
\[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
freckles
  • freckles
\[\frac{15-x}{x}0 \\ \text{ we have } \frac{15-x}{x^2} <[x] \le \frac{15}{x^2}\] I don't know what this means yet if anything can be concluded from it
anonymous
  • anonymous
Where is the smallest integer bracket at?
anonymous
  • anonymous
I'll just let parentheses be it.
freckles
  • freckles
I just used brackets. I think there is some code you can type in.
freckles
  • freckles
\[\lceil x \rceil\] oh yeah you can use \lceil x \rceil
anonymous
  • anonymous
Oh, I dont know much about symbols. Thanks for pointing that out.
anonymous
  • anonymous
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
anonymous
  • anonymous
⌈x ⌈x⌉⌉ = 15/x the right side is integer. so x must be an integer also right ?
freckles
  • freckles
no
freckles
  • freckles
15/x I think x is between 0 and 15 though for example x could be 15/7 see: 15/(15/7)=7
freckles
  • freckles
so what I'm saying is x could be an integer or it could be rational
anonymous
  • anonymous
oh, ok.. i see now
freckles
  • freckles
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 :(
anonymous
  • anonymous
What is the answer?
anonymous
  • anonymous
7?
anonymous
  • anonymous
not sure. it is an essay question, no choices :v
anonymous
  • anonymous
a can be more than one answer. Since you wrote "the complete question is find all real a which satisfy f(a) = 15"
anonymous
  • anonymous
yeah.. maybe a can be more than one answer, but how to get them ?
freckles
  • freckles
\[x \cdot [x [x]]=15 \\ \text{ if } x=\frac{15}{7} \text{ then } \\ \frac{15}{7}[ \frac{15}{7}[\frac{15}{7}]]=\frac{15}{7}[ \frac{15}{7}(3)]=\frac{15}{7}[\frac{45}{7}]=\frac{15}{7}(7)=15 \\ \text{ I didn't say } x=7 \\ 7[7[7]]=7[7(7)]=7[49]=7(49) \neq 15\]
anonymous
  • anonymous
so, just using trials any x ?
freckles
  • freckles
\[2.46 \cdot [2.46 [2.46]]=2.46[2.46(3)]=2.46(8) \neq 15\]
anonymous
  • anonymous
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
anonymous
  • anonymous
Just kidding. I dont know enough about ceiling and floor functions.
freckles
  • freckles
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 \[x [x[x]]=x[x(x)]=x[x^2]=x(x^2)=x^3 \\ x^3=15 \text{ so I think solving for the real x value here } \\ \text{ will give us a number closed to our solution }\]
freckles
  • freckles
close*
freckles
  • freckles
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
anonymous
  • anonymous
It doesnt need to be an integer. The x is outside the ceiling function.
freckles
  • freckles
yes i already said x cannot be an integer
freckles
  • freckles
x has to be a rational number
freckles
  • freckles
I was showing above by contradiction why x cannot be an integer
anonymous
  • anonymous
I take that back. I'll try a new method (might take forever)
anonymous
  • anonymous
You wana show me a screenshot of this question.
freckles
  • freckles
so this is what we have so far \[2
anonymous
  • anonymous
I think the answer is somewhere along \[2.99
freckles
  • freckles
if you check the three cases above you should see 15/7 is the solution @Shalante
anonymous
  • anonymous
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
freckles
  • freckles
oh I thought we were doing ceiling function
anonymous
  • anonymous
Oh yea I reversed it.
anonymous
  • anonymous
That one guy wrote 2.14 at the beginning was correct.
anonymous
  • anonymous
it should be between 2.14 and something like an interval.
freckles
  • freckles
you do know 15/7 is approximately 2.14286
anonymous
  • anonymous
I kept thinking ceiling is a restraint meaning it has to be less.
freckles
  • freckles
but 2.14 is only an approximation to actual answer which is 15/7
anonymous
  • anonymous
I got 15.00002 0.0 The question asked for all a values. An interval would help I guess.
freckles
  • freckles
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
anonymous
  • anonymous
14.99998 would be the same 15.00002
anonymous
  • anonymous
I thinks theres more than one. Its close enough. You know chemistry? Its like a 0.001 percent error?
anonymous
  • anonymous
Decimals are close enough especially in the thousandth. But it says find all, so I would use interval.
anonymous
  • anonymous
Are you talking about 15/8 and 15/9? I dont think it would fit at all.
freckles
  • freckles
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
anonymous
  • anonymous
I got 2.999 earlier doing it the floor way by just plugging and chugging. Its actually not bad either.
anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
Actually it does. Lemme think of more complex ones.
anonymous
  • anonymous
Try 16.36. I couldnt do it dividing it by 6 to 10.
freckles
  • freckles
you are trying to solve x(x(x))=15.4?
freckles
  • freckles
or x(x(x))=16.36?
anonymous
  • anonymous
I solved 15.4 already using your method of dividing it like this 15.4/7. But I cant do it at 16.36
anonymous
  • anonymous
Because I want to know if there is a rule that works on all of them. Or is it just some certain numbers?
anonymous
  • anonymous
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.
anonymous
  • anonymous
16.36/7 16.36/8 16.36/9 Besides plug and chug.
anonymous
  • anonymous
Like a very fast way.
freckles
  • freckles
\[x[x[x]]=16.36 \\ \text{ we know } x \text{ is not an integer } \\ \text{ but so we can get in the neighborhood } \\ \text{ we will solve } x^3=16.36 \\ \text{ which will actually give us a real number and not an integer }\] \[x \approx 2.5 \\ \text{ so this means we know } \\ 2
anonymous
  • anonymous
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.
anonymous
  • anonymous
You already got your answer. No need to bump this.
Loser66
  • Loser66
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
Loser66
  • Loser66
typo at the red mark, it is \(f(2.1725) = 2.1725\lceil 2.1725\lceil 2.1725\rceil \rceil =2.1725\lceil 2.1725*3\rceil=2.1725*7 =15.2075\)
Loser66
  • Loser66
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.
Loser66
  • Loser66
Hahahaha.... That is power of "sleeping". !! I meant after taking a snap, we can solve the problem we couldn't solve before.
Loser66
  • Loser66
In general, this is |dw:1443617046497:dw|
Loser66
  • Loser66
|dw:1443617112164:dw|
Loser66
  • Loser66
|dw:1443617350315:dw|
Loser66
  • Loser66
|dw:1443617476081:dw|
Loser66
  • Loser66
|dw:1443617581563:dw|
Loser66
  • Loser66
If you don't satisfy with it, take half way to the left again, that is x = 2.1293 but it gives you the same result f(2.1293) =14.905. It is not good as f(2.14375)

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