Find$f$ s.t. $f(1)=2$, $f(2)= 4$, $f(3)= 6$ and $f(4)= \pi$. Find a function where $f(1)=2$, $f(2)= 4$, $f(3)= 6$ and $f(4)= \pi$.
I got $\dfrac16(x-3)(x-2)(x-1)\pi$ as a start to get rid of $\pi$.
 A: Layout: (As hinted by Mark Bennet in the comments). 
Find a polynomial $P_1$ such that $P_1(1)=2$ and $P_1(2)=P_1(3)=P_1(4)=0$.
Find a polynomial $P_2$ such that $P_2(2)=4$ and $P_2(1)=P_2(3)=P_2(4)=0$.
Find a polynomial $P_3$ such that $P_3(3)=6$ and $P_3(1)=P_3(2)=P_3(4)=0$.
Find a polynomial $P_4$ such that $P_4(4)=\pi$ and $P_4(1)=P_4(2)=P_4(3)=0$.
The polynomial $P_1+P_2+P_3+P_4$ as the desired property.
You already found $P_4$, the rest should follow easily.

Alternatively, take the ansatz $P(x)=\alpha x^3+\beta x^2+\gamma x+\delta$.
You get 
$\begin{cases} \alpha +\beta +\gamma +\delta &=2\\
8\alpha +4\beta +2\gamma+\delta &=4\\
27\alpha +9\beta +3\gamma +\delta &=6\\
64\alpha +16\beta +4\gamma +\delta &=\pi\end{cases}$ and you can solve a system, if you're patient enough.
A: [I'm assuming by function you mean cubic polynomial, otherwise this is simple.]
Observe that the first three equations are satisfied by  the function $2x$.
Hence, the polynomial $f(x) - 2x$ has roots $x=1, 2, 3$, or that $f(x) = 2x + A(x-1)(x-2)(x-3)$.
Finally, set $x=4$, we get $\pi = 8 + A\times 3\times 2 \times 1$, or that $ A = \frac{\pi-8}{6}$. Hence, 
$$f(x) = 2x + \frac{\pi - 8} {6} ( x-1)(x-2)(x-3).$$

Lagrange Interpolation would be the general method to approach problems like this. This solution exploits a property from the selected values.
A: Your way of dealing with $\pi$ is very smart, but apparently you only took it because $\pi$ is irrational, and you are expecting to get the "nicer" values of $2,4,6$ in different ways. Here's something you can do, use seperate terms that are zero at $1,2,\pi$ and $6$ at $3$. Try it for yourself. Add all the terms. You already have the requisite insight.

$$\frac{(x-3)(x-2)(x-1)\pi}{6}+\frac{(x-\pi)(x-1)(x-2)6}{2(3-\pi)}+\frac{(x-\pi)(x-1)(x-3)4}{(2-\pi)\times-1}\\+\frac{(x-2)(x-3)(x-\pi)2}{(-1)(-2)(1-\pi)}$$

Simplyfying,

$$\frac{(x-3)(x-2)(x-1)\pi}{6}+\frac{3(x-\pi)(x-1)(x-2)}{(3-\pi)}+\frac{4(x-\pi)(x-1)(x-3)}{(\pi-2)}+\\\frac{(x-2)(x-3)(x-\pi)2}{2(1-\pi)}$$

A: The Lagrange Interpolation method that Git Gud mentions is a way to find a polynomial that goes through a specified (finite) set of points.  Suppose we want a polynomial function $p(x)$ such that $p(x_1)=y_1, p(x_2)=y_2, \ldots, p(x_n)=y_n$, where each $x_i$ is distinct from the others.
Then the following polynomial fulfills the conditions:
$$\sum_{i=1}^n{y_i\frac{(x_1-x)(x_2-x)\cdots (x_{i-1}-x)(x_{i+1}-x)\cdots (x_n-x)}{(x_1-x_i)(x_2-x_i)\cdots (x_{i-1}-x_i)(x_{i+1}-x_i)\cdots (x_n-x_i)}}$$
The idea is that we want to create a polynomial $p_i$ that is equal to 1 when evaluated at $x_i$, but zero when evaluated at the other $x_j$ values.  Thus, our desired polynomial would then just be $\sum_{i=1}^n{y_ip_i(x)}$.  The polynomial $\frac{(x_1-x)(x_2-x)\cdots (x_{i-1}-x)(x_{i+1}-x)\cdots (x_n-x)}{(x_1-x_i)(x_2-x_i)\cdots (x_{i-1}-x_i)(x_{i+1}-x_i)\cdots (x_n-x_i)}$ is precisely such a $p_i$.
Try using this method to create the polynomial: you've already figured out one of the terms in the sum, so just finish it off!
