F(11) = 11, f(x + 3) = (11-1) / (11 + 1)

[charbon]

1. I'm having some trouble finding what f(x) should be to complete the desired result.



2. If f(11) = 11 and f(x + 3) = (11 - 1) / (11 + 1) , f(2000) = ?

The Attempt at a Solution

f(11) = 11

f(8 + 3) = (f(8) - 1) / (f(8) + 1) = 11

I'm lost from there...

 

[berkeman]

1. I'm having some trouble finding what f(x) should be to complete the desired result.



2. If f(11) = 11 and f(x + 3) = (11 - 1) / (11 + 1) , f(2000) = ?

The Attempt at a Solution

f(11) = 11

f(8 + 3) = (f(8) - 1) / (f(8) + 1) = 11

I'm lost from there...

Multiply the denominator on the left by the term on the right, distribute terms and solve for f(8).

What is the overall problem statement?

 

[charbon]

Thanks for the reply,

I tried multiplying both ways with the denominator, it's finding something that works with f(8) that's the problem.

This is the whole problem statement. It comes from an issue of The Mathematics Student Journal where you had to solve for f(1979) but it was changed to f(2000).

 

[songoku]

2. If f(11) = 11 and f(x + 3) = (11 - 1) / (11 + 1) , f(2000) = ?

I'm confused. If f(x + 3) = (11 - 1) / (11 + 1), then f(x+3) = 10/12. It's a constant
from what you've done. i guess it should be f(x+3) = [f(x) - 1] / [f(x) + 1]

If so, you can find f (14), then f(17), then f(20), then f(23), so on..
It will be a repeating form. Try it first ^^

 

[charbon]

I'm confused. If f(x + 3) = (11 - 1) / (11 + 1), then f(x+3) = 10/12. It's a constant
from what you've done. i guess it should be f(x+3) = [f(x) - 1] / [f(x) + 1]

If so, you can find f (14), then f(17), then f(20), then f(23), so on..
It will be a repeating form. Try it first ^^

Oh oops, sorry about that mistake.

I guess I should have tried looking for a pattern in the beginning. Thanks a lot for your help. :)

 

[flatmaster]

There's something missing here.

f(11) = 11
f(x + 3) = 5/6
f(2000) = ?

You're never told anything at all about the function in general.

 

[songoku]

There's something missing here.

f(11) = 11
f(x + 3) = 5/6
f(2000) = ?

You're never told anything at all about the function in general.

The information is enough to solve the problem.

If so, you can find f (14), then f(17), then f(20), then f(23), so on..
It will be a repeating form. Try it first ^^