Numerical differentiation by interpolation

Question

In numerical differentiation by using Lagrange interpolating polynomial (Numerical Methods For Scientific And Engineering Computation By M.K. Jain , ch. # 5) the error may be obtained by using the relation : $\frac{1}{(n+1)!}\frac{d^j} {dx^j}f^(n+1)(\zeta)$= $ \frac{j!} {(n+j+1)!}f ^{(n+j+1)}(\eta_j)$ where j = 1, 2... and $ min(x_0, x_1,...,x_n,x) <\zeta_j <max(x_0, x_1,...,x_n,x)$ with the help of above mentioned formulas, the quadratic interpolation error is found as; $E''_2(x_0) = \frac{1}{3!}(2x_0 -x_1 - x_2)f^3(\zeta) +1/24 (x_0 - x_1)(x_0 -x_2[f^4(\eta_1) + f^4(\eta_2])$ where $ \zeta, \eta_1, \eta_2 \in (x_0, x_2)$ Kindly explain how the quadratic interpolation error is found and what is the reason for last open interval i.e. $ \zeta, \eta_1, \eta_2 \in (x_0, x_2)$ ?