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

\(\begin{array}{cc} x_i & f[x_i] & f'[x_i, x_{i + 1}] & & & &\\ \hline 0.30 & 0.29552 \\ & & 0.95534 \\ & & & -0.14200 \\ & & & & -1.0500\\ & & & & & 20.732\\& & & & & & - 431.41\\ 0.30 & 0.29552 \\ & & 0.95250 \\ & & & -0.16300 \\& & & & -0.0134\\& & & & & 0.83866\\ 0.32 & 0.31457 \\ & & 0.94924\\ & & & -0.16367 \\& & & & -0.05533\\ 0.32 & 0.31457 \\ & & 0.94433\\ & & & -0.16533 \\ 0.35 & 0.34290 \\ & & 0.93937\\ 0.35 & 0.34290 \\ \end{array}\)

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. \(H_5 (0.34) = 0.29552 + 0.95534(0.34 - 0.30) - 0.142(0.34 - 0.30)^2\)

\(-1.05(0.34 - 0.30)^2 (0.34 - 0.32) + 20.732(0.34 - 0.30)^2 (0.34 - 0.32)^2\)

\(-431.41(0.34 - 0.30)^2 (0.34 - 0.32)^2 (0.34 - 0.35)\\ H_5 (0.34) = 0.33349\)