$$s(x)=\begin{cases} 3+ax-9x^2+bx^3,\quad &1\le x\le 2\\ 1+c(x-2)+d(x-2)^2+t(x-2)^3, &2\le x\le 3 \end{cases} $$
This is what I have: $$\begin{cases} s_1= bx^3-9x^2+ax+3 &s_2= t(x-2)^3+d(x-2)^2+c(x-2)+1\\ s_1'=3bx^2-18x+a & s_2'= 3t(x-2)^2+2d(x-2)+c\\ s_1''=6bx-18 & s_2''= 6t(x-2)+2d\\ \end{cases}$$
Then when we have $s_1(2), s_2(2)$:
$$\begin{cases} s_1(2) &= 8b+2a-33\\ s_1^\prime(2) &= 12b-36+a\\ s_1^{\prime \prime}(2) &= 12b-18 \end{cases}$$ and for $s_2(2)$: $$\begin{cases} s_2(2) &= 1\\ s_2^\prime(2) &= c\\ s_2^{\prime \prime}(2) &= 2d \end{cases}$$ and $s_2''(3)=0 \rightarrow 6t+2d=0$ I'm stuck in the next step, do I have it set up right? edit: would I have to solve for $s_1''(1)=0$