1. Oct 28, 2009

### Mafiaboy

This is an advance function question that our teacher gave us as a bonus question. Right now my percentage is around 79 and if i get this question right i can get at least 83 in my final mark. So it will be very very very appreciated if someone can help me solve this question ASAP, i only have 2 days left to hand it in. Here is the exact format of the question:

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Q: We define a function first:

[X]up = the smallest integer
bigger than X.

[[ so, for example, [-2]up = -1

[-1.4]up = -1 , [-1]up = 0

[0.4]up=1 , [17]up =18

etc. ]]

Here is a relation:

n[X]up = [nX]up

Find the domain of this relation for n and the domain for X. You must Prove your answer

((Just to give another question, as an example, if the relation were:
X2+n2 =4, the domain for X is -2 < X < 2
the domain for n is -2 < n < 2

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2. Oct 29, 2009

### HallsofIvy

Staff Emeritus
I find it difficult to believe that your teacher said, "This is a bonus question. Go find someone to tell you the answer and I will give you an additional 4 points."

Wouldn't it be better (and more honest) to at least try to do the problem yourself?

3. Oct 29, 2009

### flatmaster

Interesting problem. Is n an integer?

4. Oct 29, 2009

### Mafiaboy

c'mmon help me out people, i have to hand it in tomorrow!

if you can't solve don't bother commenting plz.

5. Oct 29, 2009

### Staff: Mentor

It's a general policy in this forum that we don't do your work for you. If you want help, you have to show that you have made an effort on your own.

6. Oct 29, 2009

### Mafiaboy

i said help me solve the question not asking for you to solve. I just need guidance on where to to start and so on and making the question bit clearer.

7. Oct 29, 2009

### Staff: Mentor

OK, here's a start. The relation is n[X]up = [nX]up.

Put some numbers in for n and see what you can conclude from each one as far as the values for x. Describe in words the values of x that can be used on each side separately.

1. 2[x]up = [2x]up
2. -3[x]up = [-3x]up
3. .5[x]up = [.5x]up

8. Oct 30, 2009

### Mafiaboy

ok thanks i guess, though it wasn't that helpful. Anyway lets see what i can come up with.