I was given this function: $$ f(x) = \begin{cases} x^3 & {\text{if}}\ x>0 \\ 0 & {\text{if}}\ x\leq0\ \end{cases} $$
and I was asked to give an upper bound on it's interpolation error expressed only by h when $ x\in[-h,h] $ and using the known points $ f(-x_0)=-h, f(x_1)=0, f(x_2)=h $.
So by newton's interpolation method the interpolation polynomial is: $$ P_2(x) = f(x_0)+f[x_0,x_1,x_2](x-x_0)(x-x_1) $$
and the error is given by: $$ E(x) = f[x_0,x_1,x_2,x](x-x_0)(x-x_1)(x-x_2) $$
But i have no idea of what is the way to get an upper bound on $|E(x)|$ in this case because clearly the third derivative doesn't exist here and therefore i cannot use the formula $f[x_0,x_1,x_2,x]=f'''(\xi)/3!$.
Any ideas?