# CalcBC-6 Derivative Question (1989)

1. Oct 14, 2009

### Loppyfoot

1. The problem statement, all variables and given/known data

http://img96.imageshack.us/img96/1278/70542563.th.jpg [Broken]
http://img96.imageshack.us/img96/1278/70542563.jpg [Broken]

The problem statement is as follows:
If h=0, show that f(-x)=1/f(x)

So here is what I have gotten. f(x+0) = (f(x) + f(0)) / f(-x) + f(-0)
What does f(-0) equal? Is it 0, or 1?

Because if f(-x)=0, then after some simplifying, I get f(-x)=(f(x)+1) / f(x)

Thanks

Last edited by a moderator: May 4, 2017
2. Oct 14, 2009

### Loppyfoot

Can you folks see the image?

3. Oct 14, 2009

### CompuChip

I didn't see it at first, but it says:

Let f be a function that is everywhere differentiable and that has the following properties:
(i) $$f(x + h) = \frac{f(x) + f(h)}{f(-x) + f(-h)}$$ for all real numbers h and x
(ii) f(x) > 0 for all real numbers x
(iii) f'(0) = -1

I must admit I don't see the answer right away. Anyway I can tell you that f(-x) is not zero, because of property (iii). In fact you shouldn't assume a value for f(x) at all, because that would mean that f is a constant function.

Probably you should use that it is differentiable:
$$f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}$$