How to determine $f^{-1}$ for $f(p) = (1 + X)p' + p$ 
Let $ \mathbb{R}_2[X] $ be the vector space of polynomials with real coefficient and a degree less or equal $2$.
$f$ is the map of  $ \mathbb{R}_2[X] $ into $ \mathbb{R}_2[X] $ defined as:
$$ f(p) = (X + 1)p' + p $$
with $ p \in  \mathbb{R}_2[X] $.

*

*Prove $f$ is an endomorphism.

*Prove $f$ is invertible and determine $f^{-1}$.




*

*$f$ is linear because we have: $f(p + \lambda q) = f(p) + \lambda f(q)$ for $p,q \in  \mathbb{R}_2[X] $.

We have: $deg((X + 1)p') \leq 2$ and $deg(p) \leq 2$
$\implies deg(f(p)) \leq 2$
$\implies f(p) \in  \mathbb{R}_2[X]$
$\implies f( \mathbb{R}_2[X] ) \subset  \mathbb{R}_2[X]  $.
Which proves $f$ is an endomorphism.


*We have: $\dim Im(f) = \dim  \mathbb{R}_2[X] $
$\implies f$ is bijective, hence invertible.
This is the part I am stuck in. To determine $f^{-1}$, I need to determine the matrix $A$ of $f$ in the basis $\{1, X, X^2 \}$ and find its inverse $A^{-1}$. Which I don't see how.
Are my answers correct? How can I determine the matrix $A$?

Yor answers are correct. You should elaborate a liitle bit more why $Im(f)=\mathbb R_2[X]$.
We have $f(1)=1$, hence the first column of $A$ is $(1,0,0)^T$.
$f(X)=1+2X$, hence the second column of $A$ is $(1,2,0)^T$.
$f(X^2)=2X+3X^2$, hence the third column of $A$ is $(0,2,3)^T$.
