CalcBC-6 Derivative Question (1989)

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The discussion centers around the mathematical problem of proving that if h=0, then f(-x)=1/f(x) for a differentiable function f with specific properties. The properties include f(x + h) = (f(x) + f(h)) / (f(-x) + f(-h)), f(x) > 0 for all real numbers x, and f'(0) = -1. Participants clarify that f(-x) cannot equal zero due to property (iii), emphasizing the need to avoid assuming constant values for f(x). The approach to solving the problem involves utilizing the definition of the derivative at zero.

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Homework Statement



http://img96.imageshack.us/img96/1278/70542563.th.jpg
http://img96.imageshack.us/img96/1278/70542563.jpg

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
 
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Can you folks see the image?
 
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) [tex]f(x + h) = \frac{f(x) + f(h)}{f(-x) + f(-h)}[/tex] 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:
[tex]f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}[/tex]
 

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