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

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

1. A

$\frac{1}{3}f\left(x\right)$

2. B

$3f\left(x\right)$

3. C

$f\left(x\right)$

4. D

${\left\{f\left(x\right)\right\}}^{3}$

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### Solution:

$f\left(x+y\right)=f\left(x\right)f\left(y\right),\text{\hspace{0.17em}\hspace{0.17em}}{f}^{|}\left(0\right)=3$

$f\text{'}\left(x\right)=\underset{h\to 0}{Lt}\frac{f\left(x+h\right)-f\left(x\right)}{h}$

$=\underset{h\to 0}{Lt}\frac{f\left(x\right).f\left(h\right)-f\left(x\right)}{h}$

$=\underset{h\to 0}{Lt}\frac{f\left(x\right)\left[f\left(h\right)-1\right]}{h}$

$f\left(x+0\right)=f\left(x\right).f\left(0\right)$

$f\left(x\right)=f\left(x\right).f\left(0\right)$

$\therefore f\left(0\right)=1\cdots \left(1\right)$

${f}^{|}\left(x\right)=f\left(x\right).\underset{h\to 0}{Lt}\frac{f\left(h\right)-1}{h}$

$=f\left(x\right).\underset{h\to 0}{Lt}\frac{f\left(h\right)-f\left(0\right)}{h-0}$

$=f\left(x\right).f\text{'}\left(0\right)$

$=f\left(x\right).\left(3\right)$

$=3f\left(x\right)$

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