Question

Let $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x$ and $y$. If $f\left(0\right)=1,f\left(3\right)=3$ and $f^{\prime }\left(0\right)=11$, then $f^{\prime }\left(3\right)$ is equal to

• A. $11$
• B. $22$
• C. $33$
• D. $44$

Answer 1

Answer: (C)

$33$

We have,
$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$\Rightarrow f^{\prime }\left(3\right)=\underset{h\to 0}{lim}\frac{f(3+h)-f\left(3\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(3\right)f\left(h\right)-f(3+0)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(3\right)f\left(h\right)-f\left(3\right)f\left(0\right)}{h}$
$=f\left(3\right)\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}$
$=f\left(3\right)\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}$
$=f\left(3\right)f^{\prime }\left(0\right)$
$=3\times 11$
$=33$