# Limits and Continuity

1. Sep 25, 2011

### iRaid

1. The problem statement, all variables and given/known data
Postal charges are $.25 for the first ounce and$.20 for each additional ounce or fraction thereof. Let c be the cost function for mailing a letter weighing w ounces.
a) Is c a continuous function? What is the domain?
b) What is c(1.9)? c(2.01)? c(2.89)?
c) Graph the function c.

2. Relevant equations

3. The attempt at a solution
For a I got:
No. The domain is all real numbers > 0.
B is where I get stuck, I understand the question it's just I cant get what the integer function would be..
C I could do once I have b..

2. Sep 25, 2011

### SammyS

Staff Emeritus
Re: Limits/Continuity

Can you answer B simply by applying what is stated, without resorting to some sort of mathematical "formula" ?

3. Sep 25, 2011

### iRaid

Re: Limits/Continuity

I guess I can answer b, but then how would I graph the function for part c :|

edit: I see what you are getting it, I think I can make a graph too, thanks lol :p

4. Sep 25, 2011

### iRaid

Re: Limits/Continuity

Well I got the answers, but for my own knowledge can you tell me what this function would actually be?

5. Sep 25, 2011

### SammyS

Staff Emeritus
Re: Limits/Continuity

Are you referring to the Greatest Integer function when you say "the integer function" ?

There are many integer functions. Two common ones are the "floor" function (a.k.a. Greatest Integer function), and the "ceiling" function.

6. Sep 25, 2011

### iRaid

Re: Limits/Continuity

Yep ment greatest integer function, I'm pretty sure it's applied here since it says: \$.20 for each additional ounce or fraction thereof

7. Sep 25, 2011

### SammyS

Staff Emeritus
Re: Limits/Continuity

Greatest Integer function, a.k.a. floor function
floor(x) = the greatest integer that's less than or equal to x .

Doesn't work: floor(1.9) = 1

floor(x) + 1 is close, floor(1.9) + 1 = 2 ---- but floor(1) + 1 = 2, not 1 .
Try -floor(-x):
-floor(-(1.9)) = - (-2) = 2 , OK

-floor(-(1)) = - (-1) = 1 , OK

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook