1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The Mysery Function

  1. Sep 26, 2006 #1
    The Mysery Function!!!

    This is a problem I have for practice.

    Let f be a function that is differentiable everywhere and has the following properties.

    (i) f(x+h) = [f(x)+f(h)] / [f(-x)+f(-h)

    (ii) f(x) > 0 for all real numbers x

    (iii) f '(0) =-1

    (a) find the value of f(0)

    (a) show that f(-x) = 1/f(x) for all real numbers x

    These first two parts were fine

    a. f(0) = f(1 + -1) = 1

    b. f(x) = f(a+h)
    f(-x) = f(-a-h) = [f(-a)+f(-h)]/[f(a)+f(h)] = 1/f(a+h) = 1/f(x)

    I don't even know where to start on the next part though

    (c) Using part b show that f(x+h) = f(x)f(h) for all real numbers x and h.

    Any help would be great
  2. jcsd
  3. Sep 26, 2006 #2
    try plugging the answer for b [ f(-x)=1/f(x)] into property i for both f(-x) and f(-h). It should simplify out after some algebra.
  4. Sep 26, 2006 #3
    Thanks a bunch, i'm surprised i didn't think of that. Hopefully next time. :smile:
  5. Sep 27, 2006 #4
    Ok there's another part of this problem that's giving me trouble.

    (d) use the definition of the derivative to find f ' (x) in terms of f(x)

    I know that f ' (x) = [f(x+h) - f(x)] / h as h approached zero

    I tried changing the first part of the numerator to f(x)f(h) and it would change to just f(x) because f(0) = 1. The numerator is zero on top, but i still can't get it.

    Any help here would be great.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: The Mysery Function
  1. A function (Replies: 19)

  2. Functions ? (Replies: 18)