# The Mysery Function

1. Sep 26, 2006

### nealh149

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. Sep 26, 2006

### dmoravec

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.

3. Sep 26, 2006

### nealh149

Thanks a bunch, i'm surprised i didn't think of that. Hopefully next time.

4. Sep 27, 2006

### nealh149

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.