All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math > Calculus
A Funny Function (Posted on 2013-08-01) Difficulty: 3 of 5
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)?

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: A Funny Fallacious Function? | Comment 5 of 7 |
(In reply to A Funny Fallacious Function? by Harry)

I think you are missing something, Harry.

The sample function I gave is certainly continuous.

As for the other requirement, consider x = 999.5, which is on that line you are questioning. In my sample function, f(x) =  (999 + 1/999)/2, which is clearly between 999 and 1/999.  f(f(x)) = 2/(999+1/999).  So f(x)*f(f(x)) = 1.  The same is true for every point between 999 and 1000, because they all have f(x) between 1/999 and 999.

I could have drawn any sort of continuous curve connecting 999 and 1000, as long as f(x) was between f(999) and f(1000)

or consider x = 1,000,000
f(x) = 999
f(f(x)) = 1/999
So f(x)*f(f(x)) = 1

In fact, f(x)*f(f(x)) = 1 for all real x.


  Posted by Steve Herman on 2013-08-04 14:32:26
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information