Question

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

Answer 1

what.... where is the tools drawing ? i cant draw the equations ._.

Answer 2

are you on mobile?

Answer 3

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

Answer 4

what do those brackets mean

Answer 5

actually ⌈x⌉ mean the smallest integer which not less than x, right ?
help meh @dan :)

Answer 6

ceiling function?

Answer 7

yeah.... what's the start to do it, @Loser66 ?

Answer 8

the complete question is find all real a which satisfy f(a) = 15

Answer 9

I really don't know how to solve it :( , wolfram gives us the answer is 2.14, but how ??
@freckles

Answer 10

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

Answer 11

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

Answer 12

Where is the smallest integer bracket at?

Answer 13

I'll just let parentheses be it.

Answer 14

I just used brackets.
I think there is some code you can type in.

Answer 15

\[\lceil x \rceil\]
oh yeah you can use \lceil x \rceil

Answer 16

Oh, I dont know much about symbols.
Thanks for pointing that out.

Answer 17

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

Answer 18

⌈x ⌈x⌉⌉ = 15/x
the right side is integer. so x must be an integer also right ?

Answer 19

no

Answer 20

15/x
I think x is between 0 and 15 though

for example x could be 15/7 see: 15/(15/7)=7

Answer 21

so what I'm saying is x could be an integer or it could be rational

Answer 22

oh, ok.. i see now

Answer 23

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 :(

Answer 24

What is the answer?

Answer 25

7?

Answer 26

not sure. it is an essay question, no choices :v

Answer 27

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

Answer 28

yeah.. maybe a can be more than one answer, but how to get them ?

Answer 29

\[x \cdot [x [x]]=15 \\ \text{ if } x=\frac{15}{7} \text{ then } \\ \frac{15}{7}[ \frac{15}{7}[\frac{15}{7}]]=\frac{15}{7}[ \frac{15}{7}(3)]=\frac{15}{7}[\frac{45}{7}]=\frac{15}{7}(7)=15 \\ \text{ I didn't say } x=7 \\ 7[7[7]]=7[7(7)]=7[49]=7(49) \neq 15\]

Answer 30

so, just using trials any x ?

Answer 31

\[2.46 \cdot [2.46 [2.46]]=2.46[2.46(3)]=2.46(8) \neq 15\]

Answer 32

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

Answer 33

Just kidding.
I dont know enough about ceiling and floor functions.

Answer 34

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 \[x [x[x]]=x[x(x)]=x[x^2]=x(x^2)=x^3 \\ x^3=15 \text{ so I think solving for the real x value here } \\ \text{ will give us a number closed to our solution }\]

Answer 35

close*

Answer 36

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

Answer 37

It doesnt need to be an integer.
The x is outside the ceiling function.

Answer 38

yes i already said x cannot be an integer

Answer 39

x has to be a rational number

Answer 40

I was showing above by contradiction why x cannot be an integer

Answer 41

I take that back.
I'll try a new method (might take forever)

Answer 42

You wana show me a screenshot of this question.

Answer 43

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

Answer 44

I think the answer is somewhere along \[2.99<x<3\]

Answer 45

if you check the three cases above you should see 15/7 is the solution @Shalante

Answer 46

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

Answer 47

oh I thought we were doing ceiling function

Answer 48

Oh yea I reversed it.

Answer 49

That one guy wrote 2.14 at the beginning was correct.

Answer 50

it should be between 2.14 and something like an interval.

Answer 51

you do know 15/7 is approximately 2.14286

Answer 52

I kept thinking ceiling is a restraint meaning it has to be less.

Answer 53

but 2.14 is only an approximation to actual answer which is 15/7

Answer 54

I got 15.00002
0.0
The question asked for all a values.
An interval would help I guess.

Answer 55

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

Answer 56

14.99998 would be the same 15.00002

Answer 57

I thinks theres more than one.
Its close enough.
You know chemistry?
Its like a 0.001 percent error?

Answer 58

Decimals are close enough especially in the thousandth.
But it says find all, so I would use interval.

Answer 59

Are you talking about 15/8 and 15/9?
I dont think it would fit at all.

Answer 60

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

Answer 61

I got 2.999 earlier doing it the floor way by just plugging and chugging.
Its actually not bad either.

Answer 62

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.

Answer 63

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

Answer 64

Actually it does.
Lemme think of more complex ones.

Answer 65

Try 16.36.
I couldnt do it dividing it by 6 to 10.

Answer 66

you are trying to solve x(x(x))=15.4?

Answer 67

or x(x(x))=16.36?

Answer 68

I solved 15.4 already using your method of dividing it like this 15.4/7.
But I cant do it at 16.36

Answer 69

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

Answer 70

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.

Answer 71

16.36/7 16.36/8 16.36/9
Besides plug and chug.

Answer 72

Like a very fast way.

Answer 73

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

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

Answer 74

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.

Answer 75

You already got your answer.
No need to bump this.

Answer 76

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

Answer 77

typo at the red mark, it is
\(f(2.1725) = 2.1725\lceil 2.1725\lceil 2.1725\rceil \rceil =2.1725\lceil 2.1725*3\rceil=2.1725*7 =15.2075\)

Answer 78

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.

Answer 79

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

Answer 80

In general, this is