2) (35 pts) In this problem, the last three digits of your student number will be significant. Let $\mathrm{N}_{1}$ be a number equal to the last two digit of your student number, $\mathrm{N}_{2}$ be a number equal to the last three digits of your student number. That is, if the last three digits of your student number is 060 then, $\mathbf{N}_{1}=60$ $\mathrm{N}_{2}=060$ \begin{tabular}{|CC|CC|CC|CC|} \hline $\mathrm{x}$ & $\mathrm{N} 1$ & $\mathrm{-N} 1+2.331$ & $\mathrm{-N} 1+3.56$ & $\mathrm{-N} 1+4.390$ & $\mathrm{-N) 1+7.90$ & $\mathrm{-N} 1+11.987$ & $129.987$ \hline $\mathrm{y}$ & $789.178$ & $669.81$ & $\mathrm{-N) 2+394.234$ & $\mathrm{-N} 2+120.1$ & $\mathrm{-N} 2-100$ & $\mathrm{-N} 2-144.15$ & $-84.7+\mathrm{N} 2$ \hline \end{tabular) Use Lagrange's and Newton Divided Difference's formulas to find the value of the second derivative of $\mathrm{y}$ at $\mathrm{x}=26.587 .$ For the solutions, you must only use a calculator and you have to show all the details. CS.VS. 1587||