Using the last equation we define two values for the second derivative, namely $$s''_{i}(x_i)= f_i,$$ and $$s''_{i}(x_{i+1})= f_{i+1},$$ and setting up a straight line between $f_i$ and $f_{i+1}$ we have $$s_i''(x) = \frac{f_i}{x_{i+1}-x_i}(x_{i+1}-x)+ \frac{f_{i+1}}{x_{i+1}-x_i}(x-x_i),$$ and integrating twice one obtains $$s_i(x) = \frac{f_i}{6(x_{i+1}-x_i)}(x_{i+1}-x)^3+ \frac{f_{i+1}}{6(x_{i+1}-x_i)}(x-x_i)^3 +c(x-x_i)+d(x_{i+1}-x).$$