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

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Homework Help Overview

The discussion revolves around a functional equation involving a differentiable function \( f \) at \( x=0 \) and its derivative \( f'(0) \). Participants are tasked with showing a relationship between \( f'(x) \) and \( f(x) \) based on the equation \( f(a+b) = f(a)f(b) \).

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants explore the implications of the functional equation, with one noting that \( f(0) = 1 \) based on the equation's properties. There is also speculation about the implications of \( f'(0) \) not being zero and whether this condition applies universally or only under specific circumstances.

Discussion Status

The discussion is ongoing, with participants attempting to clarify the problem statement and explore various interpretations of the functional equation. Some guidance has been offered regarding the value of \( f(0) \), but no consensus has been reached on the overall implications or next steps.

Contextual Notes

There is a noted concern about the exact wording of the problem statement, which may affect the interpretation of the conditions given, particularly regarding the behavior of \( f'(0) \).

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...
 
Last edited:
now guessing at what the 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)
 

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