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Question 1178548: Let f(x) = (floor(x(floor x) for x >= 0.
(a) Find all x >= 0 such that f(x) = 1.
(b) Find all x <= 0 such that f(x) = 3.
(c) Find all x>= 0 such that f(x) = 5
(d) Find the number of possible values of f(x) for 0 <= x <= 10.
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
We can find the answers to (a), (b), (c) and (d) by investigating to find ALL the values the function can take on. We will find patterns that make it sufficient to find the possible function values for -10 <= x <= 10.
First consider integer values for x.
If x is an integer (positive or negative), then
floor(x)=x;
x*floor(x) = x*x = x^2;
floor(x(floor(x)) = floor(x^2) = x^2
So at this point we know
f(0)=0
f(1)=1
f(2)=4
f(3)=9
f(4)=16
f(5)=25
f(6)=36
f(7)=49
f(8)=64
f(9)=81
f(10)=100
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Now we need to determine what other values f(x) can have, first for positive values of x.
f(1)=1 and f(2)=4; can f(x) have the value 2 or 3 for positive x?
If f(x) can take on the values 2 or 3 for positive x, then x must be between 1 and 2.
So suppose 1 < x < 2. Then
floor(x)=1;
x*floor(x)=x(1)=x
floor(x(floor(x))=floor(x)=1.
So f(x)=1 for all values of x between 1 and 2; f(x) is never 2 or 3 for positive values of x.
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f(2)=4 and f(3)=9; what values between 4 and 9 can f(x) have for positive x?
If f(x) can take on a value between 4 and 9 for positive x, then x must be between 2 and 3.
So suppose 2 < x < 3. Then
floor(x)=2;
x*floor(x)=x(2)=2x
floor(x(floor(x))=floor(2x).
For x between 2 and 3, 2x is between 4 and 6, so floor(2x) can be 4 or 5.
Specifically, f(x) is 4 for 2 <= x < 2.5, and f(x) is 5 for 2.5 <= x < 3.
So for x between 2 and 3, the possible values of f(x) are 4 and 5; f(x) does not have values 6, 7, or 8 for positive x.
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f(3)=9 and f(4)=16; what values between 9 and 16 can f(x) have for positive x?
If f(x) can take on a value between 9 and 16 for positive x, then x must be between 3 and 4.
So suppose 3 < x < 4. Then
floor(x)=3;
x*floor(x)=x(3)=3x
floor(x(floor(x))=floor(3x).
For x between 3 and 4, 3x is between 9 and 12, so floor(2x) can be 9, 10, or 11.
So for x between 3 and 4, the possible values of f(x) are 9, 10, and 11; f(x) does not have any of the values 12 through 15 for positive x.
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You can continue the analysis if you want. The pattern is this:
For values of x between integers n and n+1, f(x) has values greater than or equal to n^2 and less than n(n+1).
So the values f(x) can have for 0 <= x <= 10 are
0
1
4-5
9-11
16-19
25-29
36-41
49-55
64-71
81-89
100
That's 1+1+2+3+4+5+6+7+8+9+1 = 47 different values.
ANSWER part (d): 47 different values of f(x) for 0 <= x <= 10
ANSWER part (a) (from earlier analysis): f(x)=1 for 1 <= x < 2
ANSWER part (c) (from earlier analysis): f(x)=5 for 2.5 <= x < 3
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We haven't yet done any analysis for negative non-integer values; we need to do that to answer part (b).
f(-1)=1 and f(-2)=4; can f(x) have the value 2 or 3 for negative x?
If f(x) can take on the values 2 or 3 for negative x, then x must be between -1 and -2.
So suppose -2 < x < -1. Then
floor(x)=-2;
x*floor(x)=x(-2)=-2x
floor(x(floor(x))=floor(-2x).
For x between -2 and -1, -2x is between 2 and 4, so floor(2x) can be 2 or 3.
Specifically, f(x) is 2 for -2 <= x < -1.5, and f(x) is 3 for -1.5 <= x < -1.
ANSWER part (b): -1.5 <= x < -1.
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