Question

A function $f:R\to R$ satisfies the equation $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x,y\in R,f\left(x\right)\ne 0$. Suppose that the function is differentiable at $x=0$ and $f^{\prime }\left(0\right)=2$. Prove that $f^{\prime }\left(x\right)=2f\left(x\right)$.

Answer 1

Given $f(x+y)=f\left(x\right)f\left(y\right)....\left(1\right)$

we know $f\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}.....\left(2\right)$

Comparing $\left(2\right)$ with $\left(1\right)$

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

$\frac{f\left(x\right)}{f\left(x\right)}=\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}....\left(3\right)$

$\underset{x=0}{lim}\frac{f\left(0\right)}{f\left(0\right)}=\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}$ $[\because givenp(0)=2]$

$\frac{2}{f\left(0\right)}=\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}$ $\left[f\right(0)=0or1]$

$\frac{2}{1}=\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}$ By $0$ it is infinite so take$1$

Sub in $\left(3\right)$

$\frac{f\left(x\right)}{f\left(x\right)}=2\Rightarrow f\left(x\right)=2f\left(x\right)$