4
Greatest integer function
By: Ahmad Ghandehari
F(x)  x
1-1. x  means, the greatest integer which is less than or equal to x.
For each x belongs to the set of real numbers, and n belongs to the set of integer
numbers, we can write:
x   n  n  x  n  1
(1)
It means:
a): x   n  n  x  n  1 and
b): n  x  n  1  x   n .
Examples:
a): x   4  4  x  5
b): x   4   4  x  3
c): 1  x  2 
x   1
d):  2  x  1  x   2
e): 2  x  2.3 
f ):
x   2
2  x  3  x   1
g): 1/ 7  1 , 1.001  1 , 1.999  1
Example 1:
 x  1  x  1
3
 2 then find the interval of x variation.
If, 2 
 2   2 
Solution:
 x  1  x  1
 x  1
 x  1
2
 3
2  
2  
 2 



 2   2 
 2 
 2 
2
x 1
 1   4  x  1  2   3  x  1
2
1-2. For each n belongs to the set of integer numbers we can write.
(2)
x  n   x   n
Examples:
a) x  2  x   2
1
b) x  5  x   5
c) x  x   x   x   2x 
Problem 1:
If, 3( x   2)  x   22 , then find the interval of x variation.
Solution:
3(x   2)  x   22  3x   6  x   22  4x   6  22 
4x   6  22  4x   16  x   4  4  x  5 .
Problem 2:
 x  5
x  7 x  9
 2

 22 then find the interval of x variation.
If , 3

 2 
 2   2 
Solution:
 x  5
 ( x  5)  2   ( x  5)  4 
3
 2

  
  22 
2
2
 2 

 x  5
x  5  x  5

3
 2
 1  
 2  22 

 2 
 2
  2

 x  5
 x  5
 x  5
3
 2
2
 2  22 


 2 
 2 
 2 
 x  5
 x  5
 x  5
6
 4  22  6 
 18  
3 


 2 
 2 
 2 
3
x5
 4  6  x  5  8  1 x  3
2
Problem 3:
Supposing
 2 y  1
 x   x 2   x 3  , then find the interval of y
2  1.4 and 

 2 
variation.
Solution:
x  2  1  x  1.4  1  x  2.4  x   2.4  2
x 2  (2.4) 2  x 2  5.8 
x 2   5.8  5
x 3  x ( x 2 )  x 3  2.4(5.8)  x 3  13.92
 2 y  1
2
3
 2   x   x   x 
2
x 3   13.92  13
 2 y  1
 2 y  1
 2   2  5  13   2   20 
20 
2y  1
41
43
 21  40  2 y  1  42  41  2 y  43 
y
.
2
2
2
Problem 4:
If , x   log 1  log 2  log 3  ...  log 100 , then find the interval of x variation.
Solution:
log 1  log 2  log 3    log 9  0
log 10  log 11  log 12  log 99  1
log 100  2  2
x   0  1  1  1  ...  31  2
44444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444414444444444444444444444444442
x   90  2  x   92  92  x  93 .
Problem 5:
How many integer roots has the equation  x   3! 1
Solution:
We know that 0!=1!=1 , so
a) x   3  0  x   3  x  3
xI
b) x   3  1  x   4  x  4
xI
Problem 6:
Find the amount of area which is restricts to the graph of x . y  1 .
Solution:
We know that (1)(1)  1 and (1)( 1)  1 so
x   1  1  x  2
a) x . y  1  
y  1  1  y  2
x   1   1  x  0
b) x . y  1  
y  1   1  y  0
A1  1 and A2  1 so A1  A2  2 square unit.
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Problem 7:
1

Prove that 2 x   x    x  
2

Solution:
Let x  n  p , n  I and 0  p  1
x   n  p  n  p  n  0  n
2x   2n  2p  2n  2p
1 
1
1


 x  2   n  p  2   n  p  2 
1

We want to prove that 2 x   x    x  
2

1
1


so 2n  2p   n  n  p    2p   p  
2
2


Now we have to prove that


1
2p  p  2 

 0  2p  1  2p  0
1 
1
1
a):let, 0  p  ;  1

2   p   1  p    0
2
2

2
1

Therefore 2p   p  
2

 2p  1
 1  2p  2
1

1 3
b):let,  p  1 ; 
 1
1 p 
 p    1
2

2 2
2


1

There fore 2p   p  
2

Problem 8:
 2 x  1
If 2 x   x   
 , then find the interval of x variation.
 x 1 
Solution:
1

We prove in problem 7; 2 x   x    x  
2

 2 x  1
2x   x    x  1  


4


1
 ( x  1)  x 
 
x

1


x   x  2   x   

x 
1 

x


1




2  
x  1
 x 
1

x


1




2 
 x  1
 x 
x
1  
  0 , so
x 1
x

1


1
1
1
3

 x  2   1  1  x  2  2  2  x  2 .
0
Problem 9:
1 
1

If, 2 x     x    2 , then find the interval of x variation.
2 
2

Solution:
1 
1  
1

 x  2    x  2  1   x  2   1 , therefore,
1 
1

2x    x    1  2 
2 
2

1
1
1
1
3


3 x    3   x    1  1  x   2 
x .
2
2
2
2
2


Problem 10:
x x
If    , then find all xs.
3 2
Solution:
x
 k  x  2k and k  I
2
x
x
 3   k  k  3  k  1  3k  x  3k  3
x  2k therefore 3k  2k  3k  3 
Let
3k  2k
k  0
k  0
 
 

2k  3k  3
k  3
k  3  0
k  I ;  3  k  0  k  2 ,  1 , 0
x  2k  x  4 ,  2 , 0 .
5
Problem 11:
If x 2  3  4 x  , then find all xs.
Solution:
x 2  3  0  4x   0  x  0
let x   k and k  I
2
2
2
x   k  k  x  k  1  k  x  (k  1) 
k 2  3  x 2  3  (k  1)2  3
k 2  3  4k  k 2  2k  1  3
 k 2  3  4k
k  3  4k  k  2k  4   2

k  2k  4  4k
 k 2  4k  3  0
 (k  1)( k  3)  0
 
 2
2
(k  1)  3  0 it is always true
k  2k  4  0
2
2
(k  1)( k  3)  0  1  k  3  k  1, 2 , 3
x 2  3  4k , x  0
k 1 ;
x2  3  4  x2  1  x  1
k2 ;
x2  3  8  x2  5  x  5
k 3 ;
x 2  3  12  x 2  9  x  3 .
Problem 12:
 0 when x  I
We know that x    x   
 1 when x  I
Now if 2x    x   5 then find the interval of x variation.
Solution:
a) Let x  I , therefore x    x   0   x   x 
2x    x   5
2x   x   5  x   5  x  5 .
b) Let x  I therefore x    x   1   x   1  x 
2x    x   5
2x   1  x   5  x   6  6  x  7 so
x belongs to the set of (6 , 7)  5
6
1-3. Greatest integer function properties
F( x)  x 
This is the graph of f ( x)  x  when  2  x  4
The domain of f ( x)  x  is the set of real numbers.
The range of this function is the set of integer numbers.
In any integer numbers this function does not limit.
If x  I then this function does not continuous and f ( x ) does not exists.
If x  I then this function has limit and is continuous and differentiable and
f ( x )  0
6- f ( x)  x  is increasing function.
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