Suppose f is a continuous function such that f(1000)=999 and f(x) · f(f(x)) = 1 for all real x. What is f(891)?

Yes, that is indeed a funny function

if f(1000) = 999

then f(999) = 1/999

and f(1/999) = 999

f(x) can never be 0, because that would make 0*f(0) = 1

Also, f(x) can never be 1000, because that would mean that

1000*999 = 1

And since the function is continuous, that means that f(x) is always greater than 0 and less than 1000. Of course, the range might be narrower. It might be as narrow as [1/999,999]

**What is clear is that f(x) = 1/x if x is in the range (which is necessarily smaller than the domain), and the range includes 891. So f(891) = 1/891. Final answer. **

Can such a funny function exist? Sure, many do. All that is required is that it is continuous and that f(x) = 1/x if x is in the function's range. The function can do anything you want outside the range, as long as it is continuous and it maps into the range, which it does by definition. Here is one way to do it with the minimum range of [1/999,999].

f(x) =

when x >= 1000 999

when x < 1/999 999

when 1/999 <= x <= 999 1/x

when 999 < x < 1000 1/999 + (x-999)*(999-1/999)

Graphically, that last little section is just a straight line that connects (999,1/999) to (1000,999)

*Edited on ***August 2, 2013, 9:44 am**