<s>[INST] 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 $\pi$.
 [/INST] [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.
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