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Given a function $f$ satisfying the following two conditions for all $x$ and $y$:
(a) $f(x+y)=f(x)\cdot f(y)$,
(b) $f(x)=1+xg(x)$, where $\displaystyle \lim_{x\rightarrow 0}g(x)=1$.
Prove that $f'(x)=f(x)$.

The only thing I know is that $f'(x)=f(x)$ is true for $x=\{0,1\}$ , but how do we know that it's true for all $x$?

  • 15
    Hint: $\frac{f(x+h) - f(x)}h = f(x) \cdot \frac{f(h) - 1}h$ (now use what you now about $f'(0)$).2012-05-22
  • 4
    Where you wrote $x=\{0,1\}$, did you mean $x\in\{0,1\}$? You can write $x=0,1$ (with no braces) and it means the same thing. But "$x=\{0,1\}$" means $x$ is a set with two members: $0$ and $1$.2012-05-22
  • 0
    Note that $ f'=f $ then implies $ f(x)=A\exp(x) $, but from (a) $A=A*A$ and (b) rules out $A=0$ so $f$ is exactly $\exp $.2012-05-23

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