= Solution
Let $\Delta T=T-T_c$, with $a,c>0$ fixed. The equilibrium <phase diagram> has boundaries
$$
\boxed{\begin{aligned}
\Delta T&=0,&&b>0&&\text{continuous},\\
\Delta T&=\frac{3b^2}{16ac},&&b<0&&\text{first order}.
\end{aligned}}
$$
For either branch, larger $\Delta T$ is disordered and smaller $\Delta T$ is ordered. Both signs of the ordered <magnetization> are symmetry-related <global minima> throughout the ordered region, rather than separate regions selected by temperature and $b$.
The branches meet at the <tricritical point> $(\Delta T,b)=(0,0)$. Directly on $b=0$, minimizing $rm^2+cm^6/3$ gives
$$
\boxed{|m_{\rm eq}|=\left[\frac{a(T_c-T)}c\right]^{1/4}\quad(T<T_c),\qquad \beta_{\rm tri}=\frac14.}
$$
The transition there remains continuous, with a different <mean-field critical exponent>. For $b<0$, the metastability boundaries are $\Delta T=0$ for the disordered state and $\Delta T=b^2/(4ac)$ for the ordered state. The <first-order phase transition> curve lies strictly between them, as shown in the figure. These spinodal curves describe possible hysteretic continuation of local minima, not equilibrium phase boundaries.
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