F(x)=[x]-[x^2],F(x)>=0, what would be the limitation of x?

Question

What is d limitation of x?

This is what I have done:
1)Obviously: x>=0

2)consider: 0<=x^2<1 , Then: 0<=x<1
so: [x]-[x^2]=0 ---> D1: [0,1)

3)Consider:1<=x^2<2 , Then; 1<=x<2^(1/2)
so: [x]-[x^2]=0 ---> D2: [1,2^(1/2))

4)Consider: 2<=x^2<3 , Then: 2^(1/2)<=x<3^(1/2)
so: [x]-[x^2]=-1, which is not acceptable

There4 at last, D=[1,2^(1/2))
& f(x)=0

But I want a general solution,if possible,which would be useful 4 solving d same ones.
I mean some thing except considering x in different limitation.Thanx

Thanx Puggy, but I mean:[x]-[x^2]>=0
not:x-x^2>=0
Arent they different? check x=1.2
then:[x]=1 & [x^2]=1
therfore:f(x)=0,which would be acceptable for d given domain.
some thing except checking different domains?

Answer 1

you can see [x] - [x^2] = x - {x} -( x^2 -{x^2}=
x - x^2 + {x^2} - {x} , where {x} is x - [x] and belongs to [0,1).\
So {x^2}-{x} belongs to (-1,1)
Now , when x -x^2 <= -1 (*) the inequality [x] - [x^2] >=0 wouldn't have solutions. Why?
But * takes place for x less then x_1 =(1-sqrt(5))/2 and also for x bigger than x_2 =(1+sqrt(5))/2. So the only ones that colud work are between x_1 and x_2 and, since x canno be negative, between 0 and x_2( which is about 1.6). Now you better proceed as you did at 1) and 2) and you get the solution D.

So I might suggest you to use the formula x-{x} = [x]. I don't know if there is a general approach but you use imagination.
There was other formula like
[x] + [x+1/n] + [x+2/n] +... + [x+(n-1)/n] = [nx].

Answer 2

F(x) = x - x^2.

If we set F(x) >= 0, then

x - x^2 >= 0.

Factor the left hand side, and we get

x(1 - x) >= 0

To obtain our critical values, solve for the *equation*
x(1 - x) = 0. This leads to the solutions x = {0, 1}, so 0 and 1 are our critical values.

What we want to do now is test the values AROUND 0 and 1. We want to test a value
1) less than 0,
2) between 0 and 1,
3) greater than 1.

What we're testing for is positive values (since we want F(x) >= 0 as per the question).

First, we test a value less than 0. Test x = -1:
Then x(1 - x) = -1(1 - (-1)) = -1(2) = -1, which is a negative number. Reject the interval of values less than 0, since we don't want negative numbers.

Test a value between 0 and 1. Test 1/2. Then
x(1 - x) = (1/2)(1 - 1/2) = (1/2)(1/2) = 1/4, which is positive. Therefore, we WANT the interval between 0 and 1.

Test a value greater than 1. Remember that it doesn't matter what value we test, as long as it lies in that interval. Test 100000000. We're going to get a negative number, since
x(1 - x) = [something big] times [something negative] = negative. Reject the interval greater than 1.

Therefore, our answer is x in the interval [0, 1]

Answer 3

first |x^2| is positive no matter what (suppose x is real) so we can rewrite it
|x| - x^2 >=0
x^2 - |x| <=0

when you deal with expressions like |x| you should consider two cases
1) x>=0 then |x| = x {for example |5| = 5}

we have: x^2 - x <=0
x * (x-1) <=0
the interval is [0; 1]
**
the brackets [ ] means we include both numbers as a solutions
the round brackets ( ) - we exclude them

2) x< 0 ==> |x| = -x {eg. |-5| = 5 = - (-5) }

we have x^2 - (x) <=0
x^2 + x <=0
x* (x+1) <=0
the interval is [-1; 0)

so for x>=0 the solution [0; 1]
and for x<0 -- [-1;0)

combine both: [-1; 1] - the final answer

you can check it trying any number:
-0.5 : F(x) = |-0.5| - | (-0.5)^2 | = 0.5 - 0.25 = 0.25 >=0