Additive function differentiable at 0 Let $f:\mathbb{R}\to \mathbb{R}$ be a function such that $f(x+y)=f(x)+f(y)$ where $x,y\in \mathbb{R}$. If $f(x)$ is differentiable at $x=0$, then

$f(x)$ is differentiable only in a finite interval containing zero


$f’(x)$ is constant for all $x\in \mathbb{R}$


$f(x)$ is constant for all $x\in \mathbb{R}$


$f(x)$ is differentiable except at finitely many points

To pick the right option, the function must be determined first. However, I have not been able to do from the given data, not even a bit. I need a hint to begin the solution process
Multiple correct
 A: Well $f(x + y) = f(x) + f(y) $ for any $x, y \in \mathbb{R}$. Then $f(x) = cx ~$ for any $x \in \mathbb{Q}.$ Now $f$ is differentiable at $x = 0$, i.e., $f(x)$ is cont at $x = 0$.
$\boxed{\blacksquare~ \mathbf{EDIT:}~ \text{The map } f \text{ is in } \mathscr{C}^{0}(\mathbb{R}), \text{ this can be proved form the continuity of } f 
\text{ at any single }x_0 \in \mathbb{R}, \text{ here }0 = x_0}$
(It's Cauchy's Functional Equation indeed and it's solution mechanism is standard i.e., $$\mathbb{N} \to \mathbb{Z} \to \mathbb{Q} \xrightarrow{\text{if } f ~\in ~\mathscr{C}^0 } \mathbb{R}$$ which is quite trivial)
Then for any given $\epsilon > 0$, $\exists~ \delta > 0$ such that   $$  \lvert f(x) - f(0) \rvert < \frac{\epsilon}{2}  \quad \text{when }~  \lvert x \rvert < \frac{\delta}{2} $$
Then for any arbitrary $y \in \mathbb{R}$, we have that
$$ \lvert f(x) - f(y) \rvert  \leqslant  \lvert f(x) - f(0) \rvert + \lvert f(y) - f(0)\rvert < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon  $$
When $$ \lvert x - y\rvert \leqslant \lvert x \rvert + \lvert y \rvert  < \frac{\delta}{2} + \frac{\delta}{2} = \delta $$
Hence $f$ is continuous for any $y \in \mathbb{R}$.
Now it's equivalent to Cauchy's Functional Equation and hence solution is $\color{red}{f(x) = cx,} \color{blue}{\text{ for any } x \in \mathbb{R}} $
Then $f(x) \in \mathscr{C}^1[\mathbb{R}] \implies$ only the $2$nd option is true. $\blacksquare$.

Regrets: This maybe is an overkill of the simple problem. Just note the steps

*

*$f$ satisfies the given functional equation and $f$ is cont at some $x_0 \in 
\mathbb{R} \implies f \in \mathscr{C}^0(\mathbb{R})$.


*$f(x + y) = f(x) + f(y)$ for any $x, y \in \mathbb{R}$ and $f \in \mathscr{C}^0$ $\implies f(x) = cx ~\forall~ x \in \mathbb{R}$.

This is enough for the problem.
