Exercise 4.8

Proof. We will show that $\mathit{Tp}=q$ where $q$ is the polynomial such that $p(x)-p(3)=(x-3)q(x)$.

Clearly, if $x\ne 3$ then $\mathit{Tp}=q$.

Suppose $x=3$. We have that 3 is a root of $p(x)-p(3)$. Thus, by the same reasoning used in Exercise 6, it follows that

But $p^{\prime }={(p-p(3))}^{\prime }$, hence $\mathit{Tp}(3)=q(3)$. Therefore $\mathit{Tp}=q$ and $\mathit{Tp}$ is indeed a polynomial.

To prove $T$ is linear, let $r,s\in \mathbb{P}(R)$ and $\alpha \in 𝔽$. If $x\ne 3$, then

And

If $x=3$, then $T$ is linear by the properties of derivatives. □