Alternative formula for the Bernoulli pmf? If I understand correctly, a Bernoulli pmf just needs to assign a probability $p$ if there is a success $(x=1)$, and $1-p$ otherwise $(x = 0)$. Rather than the usual formula, can't the following function also satisfy this condition and represent a the pmf of a Bernoulli RV?
$
f(x) = xp + (1-x)(1-p).
$
So if $f(x=1) = p, f(x=0) = 1-p.$
What would be wrong in formulating it this way?

Your alternative form is often written braced form as
$$
f(x)=\begin{cases} p & \text{if $x=1$} \\
                   1-p & \text{if $x=0$}
      \end{cases} $$
and there is nothing wrong with that. It might be useful, for instance, for programming and for elementary exposition.
But if you want to do any form of algebra or calculus, it is inconvenient, so the other form is preferred. But both forms are equally valid, it is only a pragmatic question of what works best for whatever you are doing with it.
