= Solution
Assume the scalar <order parameter> can take either sign. The cubic term breaks $m\mapsto-m$ symmetry, and a positive quartic coefficient bounds the <Landau free energy> below. Stationary points satisfy
$$
A'(m)=m(A_2+A_3m+A_4m^2)=0,
$$
so, besides the disordered point $m=0$, the possible ordered points are
$$
m_\pm=\frac{-A_3\pm\sqrt{A_3^2-4A_2A_4}}{2A_4}.
$$
The disordered point has curvature $A''(0)=A_2$ and is locally stable when $A_2>0$. Nonzero <stationary points> first appear when
$$
A_2\le\frac{A_3^2}{4A_4}.
$$
That condition alone does not make them the equilibrium phase: one must compare their <free energies>.
At a nonzero <stationary point>, substitute $A_2=-A_3m-A_4m^2$ to obtain
$$
A(m)=-\frac{A_3m^3}{6}-\frac{A_4m^4}{4}
=-\frac{m^3}{12}(2A_3+3A_4m).
$$
For an ordered state to coexist with $A(0)=0$, this requires
$$
\boxed{m_{\mathrm{coex}}=-\frac{2A_3}{3A_4},\qquad
A_{2,\mathrm{coex}}=\frac{2A_3^2}{9A_4}.}
$$
Both curvatures at <phase coexistence> are positive and equal to $2A_3^2/(9A_4)$. The barrier is at $m=-A_3/(3A_4)$. In fact the potential at <phase coexistence> factorizes:
$$
A(m)=\frac{A_4}{4}m^2\left(m+\frac{2A_3}{3A_4}\right)^2.
$$
The two distinct minima are therefore explicit, with a finite jump in the <order parameter> whenever $A_3\ne0$.
To state the equilibrium inequalities for either sign of $A_3$, put $m=-\operatorname{sgn}(A_3)\rho$ with $\rho\ge0$. This sign always has lower energy than the opposite sign of equal magnitude. Then
$$
A(m)=\rho^2\left[\frac{A_2}{2}-\frac{|A_3|}{3}\rho+\frac{A_4}{4}\rho^2\right].
$$
The bracket's minimum is $A_2/2-A_3^2/(9A_4)$. It follows that
$$
\boxed{\begin{aligned}
A_2>\frac{2A_3^2}{9A_4}&:\quad m=0\text{ is the unique global minimum},\\
A_2=\frac{2A_3^2}{9A_4}&:\quad m=0\text{ and }m_{\mathrm{coex}}\text{ coexist},\\
A_2<\frac{2A_3^2}{9A_4}&:\quad\text{the global minimum is ordered}.
\end{aligned}}
$$
The favored ordered minimum, when it exists, is
$$
m_{\mathrm{ord}}=-\operatorname{sgn}(A_3)
\frac{|A_3|+\sqrt{A_3^2-4A_2A_4}}{2A_4}.
$$
Thus \b[a <first-order phase transition> is possible and occurs when the coefficients cross the coexistence relation]. There is \b[no continuous transition between $m=0$ and the equilibrium ordered state while $A_3$ remains nonzero and $A_4$ remains finite and positive]. The ordered state becomes globally favorable while $A_2$ is still positive, before the disordered curvature can vanish. Although a stationary solution tends to zero at $A_2=0$, it is not the global transition branch. A continuous transition would require eliminating the cubic term, outside the stated $A_3\ne0$ condition.
The <spinodal points> clarify the metastable region: for $2A_3^2/(9A_4)<A_2<A_3^2/(4A_4)$ the ordered minimum is metastable; for $0<A_2<2A_3^2/(9A_4)$ the disordered minimum is metastable. At $A_2=0$ the disordered state loses local stability. These local-stability limits should not be mistaken for the <first-order transition in a cubic-quartic Landau potential>.
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