The function f:R→Rsatisfies the equation f(x+y)=f(x)f(y)∀x,y∈R . if f'(0)=3, then f(x)=

$f (x + y) = f (x) f (y), f^{|} (0) = 3$

$f' (x) = \underset{h \to 0}{L t} \frac{f (x + h) - f (x)}{h}$

$= \underset{h \to 0}{L t} \frac{f (x) . f (h) - f (x)}{h}$

$= \underset{h \to 0}{L t} \frac{f (x) [f (h) - 1]}{h}$

$f (x + 0) = f (x) . f (0)$

$f (x) = f (x) . f (0)$

$∴ f (0) = 1 \dots (1)$

$f^{|} (x) = f (x) . \underset{h \to 0}{L t} \frac{f (h) - 1}{h}$

$= f (x) . \underset{h \to 0}{L t} \frac{f (h) - f (0)}{h - 0}$

$= f (x) . f' (0)$

$= f (x) . (3)$

$= 3 f (x)$

The function $f : R \to R$ satisfies the equation $f (x + y) = f (x) f (y) \forall x, y \in R$ . if $f' (0) = 3$ , then $f (x) =$