Given $f(1)$, $f(5)$ and $f(9)$ estimate $f(0)$ I have a calculus problem.  I am given three values of a function, $f(1)$, $f(5)$ and $f(9)$ and I am asked to estimate $f(0)$.  There is nothing stopping $f(0)$ from obtaining any value.  Let's try restricting to a class of functions, say cubic polynomials.  Let's set $f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 $.  Then we have to solve the simultaneous equation:
$$ \begin{array}{ccccrcrcr}
f(1) &=& a_0 & + & a_1 &  + &  a_2 & + & a_3 \\
f(5) &=& a_0 & + & 5a_1 & + & 5^2 a_2 & + & 5^3 a_3 \\
f(9) &=& a_0 & + & 9a_1 & + & 9^2 a_2 & + & 9^3 a_3
\end{array}
$$
This is three equations in four unknowns.  Therefore $f(0)$ can still be anything.  Let's try writing this in matrix notation:
$$  \left[ \begin{array}{c} f(1) \\ f(5) \\ f(9) \end{array} \right] = \left[ \begin{array}{cccc} 1 & 1 & 1 & 1 \\ 
1 & 5 & 5^2 & 5^3 \\
1 & 9 & 9^2 & 9^3 \end{array} \right]\left[ \begin{array}{c}  a_0 \\ a_1 \\ a_2 \\ a_3\end{array}\right] $$
This is a case of Lagrange interpolation yet it could also be described as an abstract linear algebra problem:
$$ \dim \big\{ f(x) = a_0 + a_1 x + a_2x^2 + a_3x^3 : f(1)=A, f(5) = B, f(9) = C \big\} = 4 - 3 = 1 $$
These solutions do not form a vector space since $f(x) \equiv 0$ is not a solution. We need to find an initial solution and then add multiples of $(x-1)(x-5)(x-9)$.  So it remains to find an initial solution.

We conclude from the process we need one more input for this type of problem is a space of accetable solutions.  What if we use quartic polynoials or fifth degree polynomials?  What are the standard ways of searching for a "best" solution?
Here are some other naÏve approaches.  We could estimate that:
$$ f(1) - f(0) = f'(1)(1-0) $$
Then we could estimate $f'(1)$ using the known values such as:
$$ f'(1) = \frac{f(5) - f(1)}{5-1} $$
These two relations leads to an approximate linear relationship between the three values
$$ f(0) = \frac{5}{4}f(1) - \frac{1}{4}f(5) $$
This is most certainly generally false since $f(0)$ could be anything.  Yet if we presume basic calculus estimates this is what it tell us.
 A: I think you have a misunderstanding or confusion about what Lagrange Interpolation actually means.  
Yes, as you mentioned, value of $f(1), f(5), f(9)$ is insufficient to draw any conclusions when you think the solution is cubic polynomial.
But what Lagrange Interpolation actually suggests is that you can find unique quadratic polynomial if any three values are given. This is not about quality of approximation. If you have 100 data, you can uniquely identify polynomial of degree 99 which passes all the points you have.
If you are confident that the solution is cubic (or higher order) polynomial and you only have three data for interpolation, Lagrange Interpolation tells you nothing since there are infinitely many cubic polynomial passing through $(1, f(1)), (5, f(5)), (9, f(9))$. $f(0)$ can be any value and you can never tell which one is correct.
I suppose that what you actually want is some other form of extrapolation. Lagrange interpolation is one method of doing extrapolation. If you have additional information, there might be a better method.
For example, if you know that your function should be not so steep (mathematically, $|f(t)| < M$ for some constant $M$ and $t$ around where you are interested), those informations can be used to derive better result. 
Without some additional informations like this, you don't have much options on this problem. I would choose using quadratic polynomial and Lagrange interpolation, since that is best option with three input / output data I have (as mentioned above, it is unique)
