Greatest Integer Problem

[gr3gg0r]

I just completed Trigonometry and College Algebra, and i'm heading into Calculus, so i thought i would get a head start on the material. So right now i'm working out of an old calculus book i got at the library. Then i came across this problem:

A dial-direct long distance call between two cities costs $1.04 for the first 2 minutes and $0.36 for each additional minute or fraction thereof.
a.) Use the greatest integer function to write the cost C of a call in terms of the time t (in minutes). Graph the cost function and discuss its continuity.
b.) Find the cost of a 9-minute call.

I didn't think this would be so hard, so i worked at it a little bit, the only part i had trouble with was coming up with the correct function, graphing and solving for 9 minutes is easy after that. I got:
C(t) = 1.04, 0 < t <= 2
1.04 + 0.36 * greatest integer of (t - 2 ), t > 2

So i doulble checked in the back of the book and it says the answer is:
C(t) = 1.04, 0 < t <= 2
1.04 + 0.36 * greatest integer of (t - 1), t > 2, t is not an integer
1.04 + 0.36(t - 2), t > 2, t is an integer

i don't understand why there needs to be to seperate functions for it being an integer or not. I thought with greatest integer, if the number is already an integer, it would leave it alone, and also, why only t - 1 in the second one, shouldn't it be t - 2?

Anyways, i hope someone will help me out. Thanks!

[Tjl]

I just completed Trigonometry and College Algebra, and i'm heading into Calculus, so i thought i would get a head start on the material. So right now i'm working out of an old calculus book i got at the library. Then i came across this problem:

A dial-direct long distance call between two cities costs $1.04 for the first 2 minutes and $0.36 for each additional minute or fraction thereof.
a.) Use the greatest integer function to write the cost C of a call in terms of the time t (in minutes). Graph the cost function and discuss its continuity.
b.) Find the cost of a 9-minute call.

I didn't think this would be so hard, so i worked at it a little bit, the only part i had trouble with was coming up with the correct function, graphing and solving for 9 minutes is easy after that. I got:
C(t) = 1.04, 0 < t <= 2
1.04 + 0.36 * greatest integer of (t - 2 ), t > 2

So i doulble checked in the back of the book and it says the answer is:
C(t) = 1.04, 0 < t <= 2
1.04 + 0.36 * greatest integer of (t - 1), t > 2, t is not an integer
1.04 + 0.36(t - 2), t > 2, t is an integer

i don't understand why there needs to be to seperate functions for it being an integer or not. I thought with greatest integer, if the number is already an integer, it would leave it alone, and also, why only t - 1 in the second one, shouldn't it be t - 2?

Anyways, i hope someone will help me out. Thanks!

Your book is right. You need two functions because you have two sets of parameters.

The first function is (t-1) because of the greatest integer function. Take a call of 5.1 minutes for example. It should be equivalent to a 6 minute call which is 2.48 dollars. Now (5.1-1) = 4.1. The greatest integer of 4.1 is 4. So you have the final function of 1.04 + (.36*4) = 2.48.

[NateTG]

i don't understand why there needs to be to seperate functions for it being an integer or not. I thought with greatest integer, if the number is already an integer, it would leave it alone, and also, why only t - 1 in the second one, shouldn't it be t - 2?

Because the greatest integer function is the greatest integer less than or equal to the number. What you want is the least integer function (which apparently isn't readily available to you).

The other issue (the need for a special function for the t is an integer case) has to do with how the 'greatest integer' function is defined. Does it produce the greatest integer less than a number, or does it produce the greatest integer less than or equal to a number? Depending on how it's set up, you will need a special case for some problems.

[Tjl]

So basically my answer reformatted :biggrin:

[gr3gg0r]

thanks guys, i think i understand it now, i was making the mistake that 5.1 mins needed to be charged as 6 mins instead of 5 mins. Thanks again!