Answer
$f\left( 2 \right)=6$ , $f\left( 3 \right)=9$ , $f\left( 4 \right)=12$. No, for all functions $f\left( x+y \right)=f\left( x \right)+f\left( y \right)$ doesn’t hold true.
Work Step by Step
By using the given relation $f\left( x+y \right)=f\left( x \right)+f\left( y \right)$ and $f\left( 1 \right)=3$
For $f\left( 2 \right)$ ,
$\begin{align}
& f(x+y)=f(x)+f(y) \\
& f\left( 2 \right)=f\left( 1+1 \right) \\
& =f\left( 1 \right)+f\left( 1 \right) \\
& =3+3 \\
& =6
\end{align}$
For $f\left( 3 \right)$ ,
$\begin{align}
& f(x+y)=f(x)+f(y) \\
& f\left( 3 \right)=f\left( 1+2 \right) \\
& =f\left( 1 \right)+f\left( 2 \right) \\
& =3+6 \\
& =9
\end{align}$
For $f\left( 4 \right)$ ,
$\begin{align}
& f(x+y)=f(x)+f(y) \\
& f\left( 4 \right)=f\left( 1+3 \right) \\
& =f\left( 1 \right)+f\left( 3 \right) \\
& =3+9 \\
& =12
\end{align}$
$f\left( x+y \right)=f\left( x \right)+f\left( y \right)$ does not hold true for all the functions.
Hence, the given values of the functions are: $f\left( 2 \right)=6$ , $f\left( 3 \right)=9$ , $f\left( 4 \right)=12$. And, for all functions, $f\left( x+y \right)=f\left( x \right)+f\left( y \right)$ is not true.
