Writing a function $f$ when $x$ and $f(x)$ are known I'm trying to write a function. For each possible input, I know what I want for output. The domain of possible inputs is small:
$$\begin{vmatrix}
    x &f(x)\\
    0      & 2\\
    1      & 0\\
    2      & 0\\
    3       &0\\
    4       &0\\
    5       &0\\
    6       &1\\
\end{vmatrix}$$
My thought is to start with a function $g(x)$ that transforms $x$ to a value from which I can subtract a function $h(g(x))$ and receive my desired $f(x)$:
$$\begin{vmatrix}
    x &g(x)& h(g(x)) &f(x)\\

    0 &   7&       5  &  2\\
    6 &   6 &      5 &   1\\
    5  &  5  &     5&    0\\
    4   & 4   &    4   & 0\\
    3 &   3    &   3  &  0\\
    2  &  2     &  2 &   0\\
    1   & 1      & 1&    0
\end{vmatrix}$$
But I'm not sure where to go from there, or even whether I'm heading in the right direction. How should I approach creating this function? Is there a methodical way to go about it, or is it trial and error (and knowledge)? Feel free to suggest modifications to how I'm stating my problem, too. Thanks.

You can generalise the problem: suppose you know the value of $f(x)$ for a particular finite set of values of $x$. (Here, you know the value of $f(x)$ when $x=0,1,2,3,4,5,6$.) Then you can find a possible polynomial function $f$ which takes the given values using the following method.
Suppose you know the value of $f(x)$ when $x=x_1, x_2, \dots, x_n$, and that $f(x_j) = a_j$ for $1 \le j \le n$.
Let
$$P_i(x) = \lambda (x-x_1)(x-x_2) \cdots (x-x_{i-1})(x-x_{i+1}) \cdots (x-x_n)$$
Where $\lambda$ is some constant. That is, it's $\lambda$ times the product of all the $(x-x_k)$ terms with $x-x_i$ left out. Then $P_i(x_k) = 0$ whenever $k \ne i$.
We'd like $P_i(x_i) = 1$: then if we let
$$f(x) = a_1 P_1(x) + a_2 P_2(x) + \cdots + a_n P_n(x)$$
then setting $x=x_j$ sends all the $P_i(x)$ terms to zero except $P_j(x)$, leaving you with $f(x_j) = a_jP_j(x_j) = a_j$, which is exactly what we wanted.
Well we can set $\lambda$ to be equal to $1$ divided by what we get by setting $x=x_j$ in the product: this is never zero, so we can definitely divide by it. So we get
$$P_i(x) = \dfrac{(x-x_1)(x-x_2) \dots (x-x_{i-1})(x-x_{i+1}) \dots (x-x_n)}{(x_i-x_1)(x_i-x_2) \dots (x_i-x_{i-1})(x_i-x_{i+1}) \dots (x_i-x_n)}$$
Then $P_i(x_k) = 0$ if $k \ne i$ and $1$ if $k=i$, which is just dandy.
More concisely, if $f$ is to satisfy $f(x_j)=a_j$ for $1 \le j \le n$ then
$$f(x) = \sum_{j=1}^n \left[ a_j \prod_{i=1}_{i \ne j}^n \frac{x-x_i}{x_j-x_i} \right]$$
This method is called Lagrange interpolation.

So in this case, your $x_1, x_2, \dots, x_7$ are the numbers $0, 1, \dots, 6$ and your $a_1, a_2, \dots, a_7$ are, respectively, $2,0,0,0,0,0,1$. Substituting these into the above formula, we get:
$$\begin{align} f(x) &= 2 \times \dfrac{(x-1)(x-2) \dots (x-6)}{(0-1) (0-2) \dots (0-6)} + 0 \times (\text{stuff}) + 1 \times \dfrac{x(x-1) \dots (x-5)}{6(6-1)(6-2) \dots (6-5)}\\
&= 2 \dfrac{(x-1) (x-2) \dots (x-6)}{720} + \dfrac{x(x-1) \dots (x-5)}{720} \\
&= \dfrac{(x-1)(x-2)(x-3)(x-4)(x-5)}{720} \left[ 2(x-6) + x \right] \\
&= \boxed{\dfrac{(x-1)(x-2)(x-3)(x-4)^2(x-5)}{240}} \end{align}$$
You can check easily that this polynomial satisfies the values in your table.

In fact, in this particular case, all the above machinery wasn't necessary. It's plain that $f(x)=0$ when $x=1,2,3,4,5$, and so $x-j$ must divide $f(x)$ for $j=1,2,3,4,5$, and so
$$f(x) = (x-1)(x-2)(x-3)(x-4)(x-5)g(x)$$
for some polynomial $g(x)$. Since we're only worried about $x=0,6$ beyond this, i.e. $2$ values of $x$, it suggests we have $2$ free parameters in $g(x)$ and hence $g(x)=ax+b$ is linear. That is, we have
$$f(x) = (x-1)(x-2)(x-3)(x-4)(x-5)(ax+b)$$
Substituting $x=0$ and $x=6$, respectively, gives
$$\begin{align}2 &= -5! \cdot b \\ 1 &= 5! \cdot (6a+b) \end{align}$$
and solving simultaneously gives $b=-\dfrac{2}{120}$ and $a = \dfrac{3}{720}$, which (after simplification) yields the desired result.
