1. Limited time only! Sign up for a free 30min personal 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!

CalcBC-6 Derivative Question (1989)

  1. Oct 14, 2009 #1
    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. jcsd
  3. Oct 14, 2009 #2
    Can you folks see the image?
     
  4. Oct 14, 2009 #3

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: CalcBC-6 Derivative Question (1989)
  1. Derive 6 (Replies: 2)

Loading...