What degree should my hermite polynomial be I constructed with a degree 6 polynomial but apparently even a degree 5 would suffice. 

$$\array{f(x)&=&ax^6&+&bx^5&+&cx^4&+&dx^3+ex^2+fx+g \\
f'(x)&=& &&6ax^5&+&&&\cdots \\
f''(x)&=&&&&&30ax^4&+&\cdots}$$
So it is now just a matter of substitution in the values from the table and solving a set of linear equations?   
By making the degree higher than it needs to be, am I going to just make the algebra unnecessarily longer or will it lead to a system of equations with no unique solutions? 
 A: $\newcommand{\R}{\mathbb{R}}$.
Let $W=Sym({\R^*})$ be the algebra of polynomials in one variable with real coefficients. 
Let $e_1$ be a basis for $\R$ and $x$ be the function $\R \to \R$ given by $x(e_1)=1$.
Thus if you didn't like the description above, $W$ is just the vector space generated by the monoid generated by $x$.
In any event if you add any element in the ideal in $W$ of polynomials that vanish, and whose first derivative vanishes on the set $\{0,0.5,1\}$, you will get another polynomial that interpolates the values in the table.  These are in fact all of the polynomials that interpolate the values in the table.
The surprising part is that this set is an ideal in $W$.  Here is why 
The set of such polynomials will be a subspace of polynomials of the form $g(x)(x)(x-0.5)(x-1)$ by remainder theorem.  To find the subspace we use the following table
$\begin{array}
x & 0 & 0.5 & 1\\
g'(x)& g(x)(x-1)(x-0.5)|_{0} & g(x)x(x-0.5)|_1 & g(x)x(x-1)|_0.5 \\
g'(x) & g(0)0.5 & g(1)0.5 & -g(0.5)0.25
\end{array} $
 so the condition that a polynomial must have its derivatives 0 at the given x values is that $g(0)=0, g(1)=0, g(2)=0$.
Thus the set of all polynomials that interpolates that values will need to have factors of $x(x-0.5)(x-1)$ in $g(x)$.  Thus we know we got all of the polynomials that interpolate the values given above and they are of the form $f(x)(x(x-0.5)(x-1))^2$ so in particular it is an ideal.
This is a very interesting question because it gives one a reason to consider ideals that are not radical like $\langle (x(x-0.5)(x-1))^2 \rangle$ in $W$.
Keep up the good work!
(If you want a more direct answer to the question, then in the special case where you assume your polynomial is of degree 6, the kernel of the LHS of the linear system that you create will be the coefficients of $(x(x-0.5)(x-1))^2$, $d/dx((x(x-0.5)(x-1))^2)$ when these are expanded out.)
