= Solution
For $a\in A$, $m\in M$,
$$
(f\cdot x)(am)=f(am)x=af(m)x=a(f\cdot x)(m).
$$
Additivity and complex linearity follow from those of $f$ and multiplication in $A$, so $f\cdot x$ is an <module homomorphism>. In particular the action stays inside $V$. Associativity gives
$$
((f\cdot x)\cdot y)(m)=f(m)xy=(f\cdot(xy))(m),\qquad (f\cdot1)(m)=f(m).
$$
Both distributive laws hold pointwise. These are exactly the axioms for a <right module>, proving \b[$\operatorname{Hom}_A(M,A)$ is a right $A$-<module>].
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