Interpolation Error for Non-standard interpolation Consider the quadratic interpolation $Q(y$) on the reference interval based
on values $f(-1), f(1), f''(0)$. Find the interpolation error. Note that $y = 0$ requires special consideration. 
I have calculated that $Q(y)$ should be $$Q(y)=f(-1)*(0.5-0.5*y)+f(1)*(0.5+0.5y)+f''(0)*(-0.5+0.5*y^2)$$ The next step is to construct a new function q(y) which is like the form $$q(y)=f(y)-Q(y)-(\text{something})\times(f(b)-Q(b))$$ for $b \in (-1,1)$. What is the something in function $q(y)$ and how to discuss the point $y=0$?
 A: You are supposed to find a cubic function $g(x)$ such that $g(-1)=g(1)=g^{\prime\prime}(0)=0$. To satisfy the conditions at $x=\pm1$ you need $g(x)=(x+1)(x-1)(x-a)=x^3-ax^2-x+a$ and to satisfy the condition at $x=0$ you need $a=0$ so $g(x)=x^3-x$. Then if you let
$$q(y)=f(y)-Q(y)-\frac{g(y)}{g(b)}\left(f(b)-Q(b)\right)=f(y)-Q(y)-\frac{y^3-y}{b^3-b}\left(f(b)-Q(b)\right)$$
You get $q(-1)=q(b)=q(1)$ and so by Rolle's theorem there are $c$ and $d$ such that $-1<c<b<d<1$ and $q^{\prime}(c)=q^{\prime}(d)=0$. Another application of Rolle's theorem allows us to conclude that there is some $\chi$ such that $c<\chi<d$ and $q^{\prime\prime}(\chi)=0$. Add the fact that also $q^{\prime\prime}(0)=0$ and we are supposed to conclude that there is some $\xi$ such that if $\chi<0$, $\chi<\xi<0$ or if $\chi>0$, $0<\xi<\chi$ and 
$$q^{\prime\prime\prime}(\xi)=f^{\prime\prime\prime}(\xi)-\frac6{b^3-b}\left(f(b)-Q(b)\right)=0$$
So that
$$f(b)-Q(b)=\frac{b^3-b}6f^{\prime\prime\prime}(\xi)$$
for some $-1<\xi<1$
But what about if $\chi=0$? Is it possible that the promised $\chi$ lies on top of the zero we already constructed at $x=0$? And does that mean it's possible that there is only $1$ inflection point for $q(y)$ in $(-1,1)$ and no point in $(-1,1)$ where $q^{\prime\prime\prime}(y)=0$? And does that mean our theorem is false?  
Consider the function
$$f(x)=\begin{cases}-\frac32x^4+x^3-\frac74x+\frac34&\text{if }x<0\\
x^3-\frac74x+\frac34&\text{if }x\ge0\end{cases}$$
Then $f(-1)=f(1/2)=f(1)=0$ and
$$f^{\prime}(x)=\begin{cases}-6x^3+3x^2-\frac74&\text{if }x<0\\
3x^2-\frac74&\text{if }x\ge0\end{cases}$$
and
$$f^{\prime\prime}(x)=\begin{cases}-18x^2+6x&\text{if }x<0\\
6x&\text{if }x\ge0\end{cases}$$
So $f^{\prime\prime}(0)=0$ and finally
$$f^{\prime\prime\prime}(x)=\begin{cases}-36x+6&\text{if }x<0\\
6&\text{if }x\ge0\end{cases}$$
$f^{\prime\prime\prime}(x)\ne0$ because $-36x+6=0$ only when $x=1/6>0$. By construction, $f(x)$ and its first $3$ derivatives are continuous at $x=0$. Also by construction $Q(x)=0$. Then
$$q(x)=f(x)-\frac{f(b)}{b^3-b}(x^3-x)$$
And
$$q^{\prime\prime\prime}(x)=\begin{cases}-36x+6-\frac{6f(b)}{b^3-b}&\text{if }x<0\\
6-\frac{6f(b)}{b^3-b}&\text{if }x\ge0\end{cases}$$
At least for $0<b<1$, $f(b)/(b^3-b)<1$ so in this region the proof of the theorem doesn't work.  
EDIT: $f(3/4)=-9/64$, $(3/4)^3-(3/4)=-7/128$ so the theorem would say that $f^{\prime\prime\prime}(\xi)=18/7$ but that isn't true for any $\xi\in(-1,1)$ because $f^{\prime\prime\prime}(x)\ge6$ for $x\in(-1,1)$ and the error of interpolation is more complicated than this.  
EDIT: After lots of algebra I am getting a result like
$$\left|f(x)-Q(x)\right|<\frac{1-|x|^3}6\max_{x\in(-1,1)}\left|f^{\prime\prime\prime}(x)\right|$$
And as $\epsilon\rightarrow0^+$ the function
$$f(x)=\begin{cases}\frac16x^3+\frac16\epsilon^2x&\text{if }x<-\epsilon\\
-\frac1{24\epsilon}x^4-\frac14\epsilon x^2-\frac1{24}\epsilon^3&\text{if }-\epsilon\le x\le\epsilon\\
-\frac16x^3-\frac16\epsilon^2x&\text{if }\epsilon<x\end{cases}$$
Approaches this bound.
