Answer
See below
Work Step by Step
From Theorem 6.4.5:
$$[T(p(x))]_C=[T]^C_B[p(x))]_B$$
a) Computing $[T]^C_B$
$T(1)=1'=0\\
T(x)=1\\
T(x^2)=(x^2)'=2x\\
T(x^3)=(x^3)'=3x^2\\
T(x^4)=(x^4)'=4x^3$
then we have $T(1)=0=0.1+0.x+0.x^2+0.x^3\\
T(x)=1=1.1+0.x+0.x^2+0.x^3\\
T(x^2)=2x=0.1+2.x+0.x^2+0.x^3\\
T(x^3)=3x^2=0.1+0.x+3.x^2+0.x^3\\
T(x^4)=4x^3=0.1+0.x+0.x^2+4.x^3$
Therefore, $[T(1)]_C=\begin{bmatrix}
0 \\ 0 \\ 0 \\ 0
\end{bmatrix}\\
[T(x)]_C=\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0
\end{bmatrix}\\
[T(x^2)]_C=\begin{bmatrix}
0 \\ 2 \\ 0 \\ 0
\end{bmatrix}\\
[T(x^3)]_C=\begin{bmatrix}
0 \\ 0 \\ 3 \\ 0
\end{bmatrix} \\ [T(x^4)]_C=\begin{bmatrix}
0 \\ 0 \\ 0 \\ 4
\end{bmatrix}$
Hence, $[T]_B^C=\begin{bmatrix}
0 & 1 & 0 & 0 &0 \\
0 & 0 & 2 & 0 & 0\\ 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 4 \end{bmatrix}$
We can notice that $[p(x)]_B=\begin{bmatrix}
3 \\
-4\\ 6 \\ 6 & 2 \end{bmatrix}$
Consequently, $[T(v)]_C=[T]^C_B[p(x)]_B=\begin{bmatrix}
0 & 1 & 0 & 0 &0 \\
0 & 0 & 2 & 0 & 0\\ 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 4 \end{bmatrix}\begin{bmatrix}
3 \\
-4\\ 6 \\ 6 & 2 \end{bmatrix}=\begin{bmatrix}
-4 \\
12 \\ 18 \\ -8
\end{bmatrix}\\
\rightarrow T(p(x))=-4+12x+18x^2-8x^3$
b) From part (a), we obtain: $T(p(x))=T(3-4x+6x^2+6x^3-2x^4)=(3-4x+6x^2+6x^3-2x^4)'=-4+12x+18x^2-8x^3$
