Problem related polynomial interpolation.

Question

Let $f(x)$ be a differentiable function such that $\frac{d^3f}{dx^3}=1$ for all $x\in [0, 3].$ If $p(x)$ be a quadratic polynomial which interpolates $f(x)$ at $x=0, x=2$ and $x=3,$ then $f(1)-p(1)=?$

How to approach ? Please help.

Answer 1

We have $f(x)=ax^2+bx+c$ and $p(x)=dx^2+ex+f$.

From $f(0)=p(0), f(2)=p(2)$ and $f(3)=p(3)$ you get system of linear equations.

Show now that $a=d, b=e$ and $c=f$.

Conclusion ?

Answer 2

When $n+1$ pairs of data points $(x_i, f_i), 0\leq i\leq n$, with all $x_i\in[a, b]$ are interpolated using an $n$-degree interpolation polynomial $P_n(x)$, then for any $x\in [a, b]$, $$f(x)-P_n(x)=\dfrac{f^{n+1}(c)}{(n+1)!}\prod_{i=0}^n(x-x_i)$$ for some $c\in (a, b)$. You should now be able to find $f(1)-P(1)$.