Let $f$ be sufficiently differentiable on $[a,b]$ and write $a=x_0$, $x_1 = \frac{a+b}{2} = x_0+h, x_2 = b = x_1+h$. Prove that
$a)$ there exists a cubic polynomial $q$ such that $f(x_i) = q(x_i)$, $i=0, 1, 2$, $f'(x_1) = q(x_1)$, and that
$b)$ $f(x)-q(x) = \frac{f^{(4)}(\xi)}{4!}(x-x_0)(x-x_1)^2(x-x_2)$.
My solution for part $a)$:
$a)$ Let $p$ be the quadratic that interpolates $f$ at $x_0, x_1, x_2$. Now let $q = p + k(x-x_0)(x-x_1)(x-x_2)$. Note that $q$ is cubic and that $q(x_i) = f(x_i), i = 0, 1, 2$. Also, if $f'(x_1) = q'(x_1)$, then we get:
$q'(x_1) = f'(x_1) = p'(x_1) +k\frac{d}{dx}[(x-x_0)(x-x_1)(x-x_2)]_{x=x_1} = p'_(x_1) - kh^2$. Solving for $k$ yields $k = \frac{p'(x_1)-f'(x_1)}{h^2}$. Note that $h^2 \neq 0$. So the desired polynomial is:
$q(x) = p(x) + (\frac{p'(x_1)-f'(x_1)}{h^2})(x-x_0)(x-x_1)(x-x_2)$.
Is this correct? Also, I'm lost on part $b)$. Any advice greatly appreciated.