Question

Let $f$ be a differentiable function on $R$ and satisfies $f(x+y)=f\left(x\right)+f\left(y\right)\forall x,yϵR$. If $f^{\prime }\left(0\right)=2$, then the value of $f\left(4\right)$ is

Answer 1

$f^{\prime }\left(x\right)={lim}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}={lim}_{h\to 0}\frac{f\left(x\right)+f\left(h\right)-f\left(x\right)}{h}$

$\Rightarrow f^{\prime }\left(0\right)=2$ [$\because f\left(0\right)=0$ by putting $x=y=0$]

$\Rightarrow f\left(x\right)=2x+c$

Putting $x=0,$ we get $c=0$

$\therefore f\left(x\right)=2x\Rightarrow f\left(4\right)=8$