How Does f'(x) Relate to f(x) in Functional Equations?

chief12
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Homework Statement


function f is differential when x=0,
f'(0) is not equal to zero for all a,b(real Numbers)
f(a+b) = f(a)f(b)

show f'(x) = f'(x)f(x)


Homework Equations





The Attempt at a Solution


f(a+b) = f(a)f(b) for all a,b(real numbers)
f(0), a+b=0
then f(0) = 1

lim x-->0 f(x) = f(0) = 1

then i get stuck
 
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chief12 said:

Homework Statement


function f is differential when x=0,
f'(0) is not equal to zero for all a,b(real Numbers)
f(a+b) = f(a)f(b)

show f'(x) = f'(x)f(x)

that would imply f(x) = 1 for all x... which then gives f'(x) = f'(0) = 0
also f'(0) is not equal to zero for all a,b,... does this mean only for a=-b?

are you sure this is how the question was written? try and write thinsg exactly as they are given...
 
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now guessing at what teh question actually asked... i would start by considering
f(x+0) = f(x) = f(x)f(0)
this shows f(0) = 1
 
then consider f(2x)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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