## Precalculus (6th Edition) Blitzer

$f\left( 2 \right)=6$ , $f\left( 3 \right)=9$ , $f\left( 4 \right)=12$. No, for all functions $f\left( x+y \right)=f\left( x \right)+f\left( y \right)$ doesn’t hold true.
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.