Try to give concrete values, let say put x=0 and h=0 in 1 and just observe what's going on.
Question: Let f be a function that is everywhere differentiable and has the following properties.
1.) f(x+h) = (f(x) + f(h)) / (f(-x) + f(-h)) for all real numbers of x and h.
2.) f(x) > 0 for all real numbers of x
3.) f '(0) = -1
PART A
Find the value of f(0).
PART B
Show that f(-x) = 1/f(x) for all real numbers x.
PART C
Using part (b), show that f(x+h) = f(x)f(h) for all real numbers h and x.
PART D
Use the definition of the derivative to find f '(x) in terms of f(x).
Hi guys, this is my first post here.
I have a calculus problem that I'm a bit unsure of how to start, mainly because my teacher is assinging this problem without teaching us the material first (just her style I guess).
I'll give it a shot, anyway. It's not fair that I get help without doing some work.
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Because of information 3), I know that f '(0) = 1. Therefore, f (h) = x - 1. I think info 3 is referring to f(h) since info 2 says f(x) is greater than zero for all real numbers, so that would mean it can't be x - 1 because if x=0, f(x) = -1.
But I'm trying to find f(0) ... not sure how to proceed from here or if it refers to f(x) or f(h).
Okay.
f(0 + 0) = (f(0) + f(0)) / (f(0) f(0)) (Sorry, I haven't memorized the latex codes yet.)
I would assume that the answer is zero, but since f(x) > 0 I can't do that ...
So I'm thinking that f(x+h) must equal negative one (-1) since if f(0) for h is 0, then that leaves f(x)/f(-x).
The part d of the question was incorrect in my original post - I have now edited it.
Part A Attempt
Let x=h=0
If x=1
f(1 + 0) = (f(1) + f(0)) / (f(-1) + f(0))
f(1) = f(1) / f(-1)
I am stuck here .. I'm not even sure if the work 2 lines above this is correct.
Many thanks, skeeter. I definitely need more practice with these kind of problems but I hope that once I keep doing them, they'll become second nature.
Anyway, I guess it's time to use the definition of the derivative:
Now, I'm not sure what to do next. tonio said I would have to use limits. However, do I use h -> 0 or h -> infinity? When I use h -> infinity, I end up getting
which I don't think is right.
I tried thinking about this for 15 minutes but am stumped.
I had a problem exactly like this before, except the value of lim (f(h)-1)/h was given in the problem.
I'm not sure how to apply the earlier parts of the question.
Am I supposed to do something like this:
Let x=1
f'(0) = -1
f'(x) = 1 + -2
Is lim ((f(h) -1) / h)negative 2?
So my effort was wrong again. So now I understand what tonio meant ... I just had to go back to the definition of the derivative and look at the third piece ofinfo (f'(0) = -1).
Thanks skeeter, no wonder you're a MHF expert (and probably tonio as well). Thanks to both of you!
I just have a question, though.
When you go from this step to the next, how come the f(x) part goes outside the "limit"?