Question

Let $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x$ and $y$. If $f\left(7\right)=2$ and $f^{\prime }\left(0\right)=3$, then $f^{\prime }\left(7\right)$ is equal to

• A. $5$
• B. $6$
• C. $0$
• D. none of these

Answer 1

The correct option is B $6$

Given, $f\left(7\right)=2,f^{\prime }\left(0\right)=3$ Also, $f(x+y)=f\left(x\right)f\left(y\right)$

$\therefore f^{\prime }\left(7\right)=\underset{h\to 0}{lim}\frac{f(7+h)-f\left(7\right)}{h}$

$=\underset{h\to 0}{lim}\frac{f\left(7\right)f\left(h\right)-f\left(7\right)f\left(0\right)}{h}$, Since $f\left(0\right)=1$

$=f\left(7\right)\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}=f\left(7\right)f^{\prime }\left(0\right)=2\cdot 3=6$