I want to show that in general, mappings with the forms below from $C^\infty (R)$ to $ C^\infty (R)$ are either going to be linear operators or not. Satisfying $A(\alpha x + \beta y) = \alpha Ax + \beta Ay$. Is what I did correct?
(1) $f \rightarrow a f^" +bf^, +cf+d$ (Not a linear operator)
$A(\alpha f_1 + \beta f_2) = a(\alpha f_1 + \beta f_2)^"+b(\alpha f_1 + \beta f_2)^,+c(\alpha f_1 + \beta f_2)+d $
$=a\alpha f_1 ^" +a \beta f_2^" +b\alpha f_1^, + b\beta f_2^, +c\alpha f_1 + c\beta f_2 +d$
$= \alpha Ax + \beta Ay +d$
(2) $f \rightarrow a f^" +bf^, +cf$ (A linear operator)
$A(\alpha f_1 + \beta f_2) = a(\alpha f_1 + \beta f_2)^"+b(\alpha f_1 + \beta f_2)^,+c(\alpha f_1 + \beta f_2) $
$=a\alpha f_1 ^" +a \beta f_2^" +b\alpha f_1^, + b\beta f_2^, +c\alpha f_1 + c\beta f_2 $
$= \alpha Ax + \beta Ay $
(3) $f \rightarrow a f^" +bf^, +cf^n$ (Not a linear operator)
$A(\alpha f_1 + \beta f_2) = a(\alpha f_1 + \beta f_2)^"+b(\alpha f_1 + \beta f_2)^,+c(\alpha f_1 + \beta f_2)^n $
$=a\alpha f_1 ^" +a \beta f_2^" +b\alpha f_1^, + b\beta f_2^, +c\alpha^n f_1^n +...+ c\beta^n f_2^n $
$\neq \alpha Ax + \beta Ay $
It is correct in principle, but if you claim that a map is not linear it is best that you provide an explicit counterexample.
For $(1)$, consider $f \equiv 1$. We have $f \mapsto c + d$ and $2f \mapsto 2c + d \ne 2(c + d)$ as long as $d \ne 0$.
If $d = 0$, then the map is linear.
Similarly, for $(3)$, consider $f \equiv 1$. We have $f \mapsto c$ and $2f \mapsto 2^nc \ne 2c$ as long as $c \ne 0$ and $n \ne 1$.
If $c = 0$ or $n = 1$, then the map is linear.