Our total error becomes $$ \left|\epsilon_{\mbox{tot}}\right|\le \frac{2 \epsilon_M}{h^2} + \frac{f_0^{(4)}}{12}h^{2}. $$ It is then natural to ask which value of \( h \) yields the smallest total error. Taking the derivative of \( \left|\epsilon_{\mbox{tot}}\right| \) with respect to \( h \) results in $$ h= \left(\frac{ 24\epsilon_M}{f_0^{(4)}}\right)^{1/4}. $$ With double precision and \( x=10 \) we obtain $$ h\approx 10^{-4}. $$ Beyond this value, it is essentially the loss of numerical precision which takes over.