Show that there is no random variable $X$ such that the m.g.f. of $X$ satisfies $M_X(1) = 3$ and $M_X(2) = 4$ Show that there is no random variable $X$ such that the moment generating function of $X$ satisfies $M_X(1)=3$ and $M_X(2) = 4$.
I'm not sure how to prove this, but I'm trying to go through known distributions such as uniform or exponential to see what the graph of the m.g.f. looks like. The definition of the m.g.f. is the expectation $\mathbb{E}[e^{tx}]$.
If we were to consider a uniform distribution, then we would have $\frac{e^t(1-e^{tm}}{m(1-e^t)}$, since is exponential and will never go through the points $(1,3)$ and $(2,4)$.
If we were to consider an exponential distribution, then we would have $\frac{\lambda}{\lambda-t}$ a hyperbola, which won't go through $(1,3)$ and $(2,4)$ either.
Of course, we can consider more distributions, such as binomial, geometric, hypergeometric and so on. However, I can't possibly prove this by going through each and every distribution and checking its m.g.f. Can someone help me with the proof on this one? Thanks

Suppose such an $X$ exists.  Then $Y = e^X$ is well-defined.  But $$\operatorname{Var}[Y] = \operatorname{E}[Y^2] - \operatorname{E}[Y]^2 = \operatorname{E}[e^{2X}] - \operatorname{E}[e^X]^2 = 4 - 3^2 = -5,$$ so no such $Y$ exists, consequently no such $X$ exists.
