Consider the provided question, Given, $f\left(x\right)=\mathrm{sin}x$ We construct a divided- difference table, with each of the interpolation point included twice, that means, ${x}_{0}={x}_{1}=0.30,{x}_{2}={x}_{3}=0.32$ and ${x}_{4}={x}_{5}=0.35.$ each of the divided differences $f[{x}_{i},{x}_{i+1}],$ where $f\left(x\right)=\mathrm{sin}x$ and $x}_{i}={x}_{i+1$ is set equal to ${f}^{\prime}\left({x}_{i}\right).$

$\begin{array}{ccccccc}{x}_{i}& f[{x}_{i}]& {f}^{\prime}[{x}_{i},{x}_{i+1}]& & & & \\ 0.30& 0.29552\\ & & 0.95534\\ & & & -0.14200\\ & & & & -1.0500\\ & & & & & 20.732\\ & & & & & & -431.41\\ 0.30& 0.29552\\ & & 0.95250\\ & & & -0.16300\\ & & & & -0.0134\\ & & & & & 0.83866\\ 0.32& 0.31457\\ & & 0.94924\\ & & & -0.16367\\ & & & & -0.05533\\ 0.32& 0.31457\\ & & 0.94433\\ & & & -0.16533\\ 0.35& 0.34290\\ & & 0.93937\\ 0.35& 0.34290\end{array}$

It follows that Hermit interpolating polynomial is,

${H}_{5}(x)=0.29552+0.95532(x-0.30)-0.142(x-0.30{)}^{2}$

$-1.05(x-0.30{)}^{2}(x-0.32)+20.732(x-0.30{)}^{2}(x-0.32{)}^{2}$

$-431.41(x-0.30{)}^{2}(x-0.32{)}^{2}(x-0.35)$

Now, evaluating this polynomial at x = 0.34 using five-digit rounding. ${H}_{5}(0.34)=0.29552+0.95534(0.34-0.30)-0.142(0.34-0.30{)}^{2}$

$-1.05(0.34-0.30{)}^{2}(0.34-0.32)+20.732(0.34-0.30{)}^{2}(0.34-0.32{)}^{2}$

$-431.41(0.34-0.30{)}^{2}(0.34-0.32{)}^{2}(0.34-0.35)\phantom{\rule{0ex}{0ex}}{H}_{5}(0.34)=0.33349$