Is this differentiability question solvable?

Question

Let $f(x+y)=f(x)f(y)$ for all $x \in \Bbb R$.

Suppose that $f(3)=3$ then $f'(3)$ is equal to?

Normally, what I'd do is partially differentiate wrt $x$. $$f'(x+y)=f'(x)f(y)$$ On substituting $y=3$, $x=0$, we get $$f'(3)=3f'(0)$$ But we don't know the value of $f'(0)$, nor of any other $f'(x)$.

So is this question solvable? If so, how? Thanks.

The only continuous functions $f$ such that $f(x+y)=f(x)f(y)$ are of the form $f(x)=a^{x}$. — Kavi Rama Murthy, Jun 07 '21 at 09:54
https://math.stackexchange.com/questions/22069/is-there-a-name-for-function-with-the-exponential-property-fxy-fx-cdot-f — Hans Lundmark, Jun 07 '21 at 09:55

score 1 · Accepted Answer · answered Jun 07 '21 at 10:27

Let's put $f'(0) = a$. Then, we obtain $$ f'(x) = a f(x), $$ for all $x \in {\mathbb R}$. Therefore, $$ {\mathrm e^{ax}}\left({\mathrm e^{-ax}}f(x) \right)' = 0, $$ and because ${\mathrm e}^{ax} \neq 0$, $$ {\mathrm e^{-ax}}f(x) = C, $$ where $C$ is a constant. With $f(3) = 3$, we get $$ f(x) = 3{\mathrm e}^{-3a}{\mathrm e}^{ax}. $$ However, as $f(x+y) = f(x)f(y)$ for all $x,y \in {\mathbb R}$, $3{\mathrm e}^{-3a}$ must be $1$ . Therefore, $f(x) = 3^{x/3}$, and $f'(3) = \log 3$.

Is this differentiability question solvable?

1 Answers1